Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Inequality
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality
$$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$
I stumbled upon this question some days ago and been trying AM-GM to find the solution but so far have been unsuccessful.
|
Ok, here we go.
Let $f(a,b,c,\lambda)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a^3+b^3+c^3}{3}+\lambda(a+b+c-3)$.
$\nabla f=0$ means:
*
*$\frac{\partial f}{\partial a}=-a^2-\frac{1}{a^2}+\lambda=0$
*$\frac{\partial f}{\partial b}=-b^2-\frac{1}{b^2}+\lambda=0$
*$\frac{\partial f}{\partial c}=-c^2-\frac{1}{c^2}+\lambda=0$
Now,
$ \frac{\partial f}{\partial b}- \frac{\partial f}{\partial a}=0 <=> a^2-b^2+\frac{1}{a^2}-\frac{1}{b^2}=0 <=>$
$(a-b)(a+b)+(\frac{1}{a}-\frac{1}{b})(\frac{1}{a}+\frac{1}{b})=0 <=>$
$(a-b)[(a+b)-\frac{1}{ab}(\frac{1}{a}+\frac{1}{b})]=0 <=>$
$(a-b)(a+b)(1-\frac{1}{(ab)^2})=0$
And of course the same for all other combinations.
So $a=b$ or $ab=1$. Also, $b=c$ or $bc=1$.
Suppose $a=b$. $b=c$ doesn't work (we can test to see it's not a minimum) so we must have $bc=1$.
Hence we search a value of $a$ with $2a+\frac{1}{a}=3$. This is $a=\frac{1}{2}$. We can test to see this is the minimum (with a value of $\frac{7}{4}$), and we're done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/891931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
}
|
How to prove: $\left(\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}}}-1\right)^{4}=5$? Question:
show that: the beautiful ${\tt sqrt}$-identity:
$$
\left({2 \over \sqrt{\vphantom{\Large A}\, 4\ -\ 3\,\sqrt[4]{\,5\,}\
+\ 2\,\sqrt[4]{\,25\,}\ - \,\sqrt[4]{\,125\,}\,}\,}\ -\ 1\right)^{4}
=5
$$
Can you someone have methods to prove this by hand? (Maybe this problem have many methods?because this result is integer. It's a surprise to me.)
Thank you
Because I found this
$$4\ -\ 3\sqrt[4]{\,5\,}\ +\ 2\sqrt[4]{\,25\,}\ -\ \sqrt[4]{\,125\,}$$ is not square numbers.
|
[This is a paraphrase on JimmyK4542's elegant answer earlier]
Let $$b=-5^{\frac 14}$$
Then the long expression under the long square root sign becomes an arithmetico-geometric series:
$$k=4+3b+2b^2+b^3$$
Multiplying by $b$:
$$\begin{align}
b\cdot k&=\quad \qquad 4b+3b^2+2b^3+b^4\\
&=\quad \qquad 4b+3b^2+2b^3+5 \end{align}$$
Subtracting:
$$\begin{align}
(b-1)k&=1+b+b^2+b^3\\
&=\dfrac{b^4-1}{b-1}\\
&=\dfrac 4{b-1}\\
k&=\dfrac 4{(b-1)^2}\\
\sqrt{k}&=\dfrac 2{1-b}\\
\left( \dfrac 2{\sqrt{k}} -1\right)^4&={(-b)}^4=5
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/892739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
}
|
By plugging $p=1-q$, into the $3$ equations show that $x=y=z$ By plugging $p=1-q$, into the 3 equations:
$$\begin{cases} z=py+qx \\ x=pz+qy \\ y=px+qz \end{cases}$$ show that $\boxed{x=y=z}$
This is from the final part of question 7 in this STEP paper,
and is following the advice of another students solution , only i cannot get to the required result despite the advice.
Any one able to get to $\boxed{x=y=z}$ by substituting $p=1-q$?
Kind regards,
|
The equations are equivalent to:
$$\begin{cases} z=q(x-y)+y \\ x=q(y-z)+z \\ y=q(z-x)+x \end{cases}$$
Substituting $z$ on the third equation we get:
$$y = -q^2(y-x) + q(y-x) + x \Rightarrow (y-x)(q^2-q+1) = 0$$
Similarly, $(x-z)(q^2-q+1) = 0$ and $(z-y)(q^2-q+1) = 0$.
So either $x = y = z$ or $q^2 -q +1 = 0$, but there's no real number $q$ that satisfies that equation. Therefore:
$$\boxed{x = y = z}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/893808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Verifying an antiderivative found in any integral table If $a > 0$, and $0 < b < c$.
\begin{equation*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
= \frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert .
\end{equation*}
(This is the antiderivative given in any calculus book.) With calculations that I made, I showed that
\begin{align*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
&= \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert
,
\end{align*}
and since the sum of an antiderivative of $1/[b + c\sin(ax)]$ and a constant is another antiderivate of $1/[b + c\sin(ax)]$, and since
\begin{equation*}
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b}
\qquad \text{and} \qquad
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b}
\end{equation*}
are reciprocals of each other,
\begin{align*}
&\int \frac{1}{b + c\sin(ax)} \, {\mathit dx} \\
&\qquad \qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert \\
&\qquad \qquad \qquad\qquad - \frac{1}{a\sqrt{c^{2} - b^{2}}} \ln\left( \frac{c + \sqrt{c^{2} - b^{2}}}{b} \right) \\
&\qquad \qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\sin(ax) + \dfrac{c + \sqrt{c^{2} - b^{2}}}{b} + \dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert \\
&\qquad\qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
\right\vert .
\end{align*}
Furthermore, we have the following trigonometric identity:
\begin{align*}
&\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)} \\
&\qquad\qquad =\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
} .
\end{align*}
So, the antiderivative given in any calculus book is the same function as the second antiderivative that I obtained:
\begin{align*}
&\frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert \\
&\qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
\right\vert .
\end{align*}
Here are my questions. Is it evident to anyone that the function on the right side of the trigonometric identity simplifies to the function on the left side? (It is surprising that the two sides are equal. The numerators and denominators on each side are "almost" the same: the numerator and denominator on the right side have two more terms than those on the left side.) If it is not evident, can someone give me calculations, starting from integration using the technique of trigonometric substitution, that show that the antiderivative of $1/[b + c\sin(ax)]$ is the function that is found in the integral tables of any calculus book - calculations that avoid all the algebraic manipulations?
|
@Tunk-Fey makes a good point and the two expressions are not necessarily equal as they can differ by a constant. However, surprisingly, the two sides are equal in this case. Just ''rationalize'' the numerator of the expression inside the logarithm of the right side. To be more precise, perform the following computation:
\begin{equation}
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}\times 1,
\end{equation}
where we write 1 as
\begin{equation}
1=
\frac{
b + c\sin(ax) - \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
{
b + c\sin(ax) - \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
.
\end{equation}
You will then find that the denominator becomes:
$$
2\,b \left( 1+\cos \left( ax \right) \right) \left( c\sin \left( ax
\right) +b \right)
$$
and the numerator becomes:
$$
2\,b \left( 1+\cos \left(a x \right) \right) \left( \sqrt {c^2-b^2 }\cos \left(a x \right) +b\sin \left( ax
\right) +c \right)
$$
If you divide these two expressions, you do get the argument of the logarithmic function on the left side.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/897548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
How to solve $(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$
Solve $$(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$$
I have a very similar problem like this on my homework, and I have no clue how to set it up or even start. How could I set this up?
|
$$(x-3) \frac{dy}{dx}+y=0 \Rightarrow \frac{dy}{dx}=\frac{-y}{x-3} \Rightarrow -\frac{dy}{y}=\frac{dx}{x-3} \Rightarrow \int \left ( \frac{-1}{y} \right)dy=\int \frac{1}{x-3} dx \\ \Rightarrow -\ln |y| =\ln |x-3|+c \Rightarrow e^{-\ln |y|}=e^{\ln |x-3|+c} \Rightarrow \frac{1}{|y|}=C |x-3| \Rightarrow |y|=\frac{c'}{|x-3|}\\ \Rightarrow y= \pm \frac{c'}{x-3} \Rightarrow y=\frac{A}{x-3} $$
So, for our non-homogeneous problem we have:
$$y(x)=\frac{A(x)}{x-3}$$
$$(x-3) \frac{dy}{dx}+y=6e^x \Rightarrow (x-3)\frac{A'(x)(x-3)-A(x)}{(x-3)^2}+\frac{A(x)}{x-3}=6e^x \\ \Rightarrow A'(x)-\frac{A(x)}{x-3}+\frac{A(x)}{x-3}=6e^x \\ \Rightarrow A'(x)=6e^x \Rightarrow A(x)=6e^x+D$$
Therefore,the solution is:
$$y(x)=\frac{6e^x+D}{x-3}+\frac{A}{x-3}=\frac{6e^x+E}{x-3}, \text{where E=A+D is a constant}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/898034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Trigonometric formula simplifies to $\sin x\cos x[\tan x+\cot x]$ Again, I have a little trouble figuring out how we got from the first step to the next one. It would be really appreciated if someone could help me out.
$$
\begin{split}LHS &= \cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi +x)\left[\cot\left(\frac{3\pi}{2}-x\right) +\cot(2\pi+x) \right] \\ &= \sin x\cos x[\tan x+\cot x]\end{split}
$$
I've tried the following: $\cot(3\pi/2)$ is $0$, so it would leave out $-\cot x$, that's where I lose the thread. $\cos(3\pi/2)$ is 0,
so that leaves out $\cos x$ and $\cos2\pi$ is one and that's where I lose the thread.
Note: that this is not an assignment question, these are all solved examples, only for my practice. Source image
|
Using the formula $$\cos{(a+b)}=\cos{(a)} \cos {(b)}-\sin{(a)} \sin{(b)}$$ we get the following:
$$\cos{ \left ( \frac{3 \pi}{2}+x \right )}=\cos{ \left ( \frac{3 \pi}{2}\right )} \cos{(x)}-\sin{ \left ( \frac{3 \pi}{2} \right )} \sin{(x)}=0 \cdot \cos{(x)}-(-1) \cdot \sin{(x)}=\sin{(x)}$$
$$\cos{(2 \pi+x)}=\cos{(2 \pi)} \cos{(x)}-\sin{(2 \pi )} \sin{(x)}=\cos{(x)}$$
Use the formula $$\cot{(a+b)}=\frac{\cos{(a)} \cos{(b)}}{\sin{(a)} \cos{(b)}+\cos{(a)} \sin{(b)}}-\frac{\sin{(a)} \sin{(b)}}{\sin{(a)} \cos{(b)}+\cos{(a)} \sin{(b)}} \\ \text{ and } \\
\cot{(a-b)}=-\frac{\cos{(a)} \cos{(b)}}{\cos{(a)} \sin{(b)}-\sin{(a)} \cos{(b)}}-\frac{\sin{(a)} \sin{(b)}}{\cos{(a)} \sin{(b)}-\sin{(a)} \cos{(b)}}$$
to calculate $\cot{\left (\frac{3\pi}{2}-x \right )}$ and $\cot{(2 \pi+x)}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/898382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Sum of the series $\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$ How do I find the sum of the following infinite series:
$$\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$$
I think the sum can be converted to definite integral and calculated but I don't know how to proceed from there.
|
Using generalized Binomial Theorem, $$(1+z)^n=1+nz+\frac{n(n-1)}{2!}z^2+\frac{n(n-1)(n-2)}{3!}z^3+\cdots$$ for $|z|<1$
$\displaystyle S=\sum_{r=1}^\infty\dfrac{2\cdot6\cdots(4r-2)}{5\cdot10\cdots5(r+1)}$
Observe that there are $r$ terms in the numerator unlike the denominator which has $r+1$ terms
So, we write multiplying the numerator with the previous term of $2$ of the Arithmetic Series $2,6,10,\cdots,$
$\displaystyle -2S=\sum_{r=1}^\infty\dfrac{(-2)2\cdot6 \cdots(4r-2)}{5\cdot10\cdots5(r+1)}$
Dividing each multiplicand of the numerator by the common difference of $-2,2,6,\cdots$ and taking out $5$ from each multiplicand of the denominator,
$\displaystyle -2S=\sum_{r=1}^\infty\frac{4^{r+1}}{5^{r+1}}\cdot\dfrac{-\dfrac24\cdot\dfrac24\cdot\dfrac64\cdots\dfrac{4r-2}4}{(r+1)!}$
Multiplying each multiplicand of the numerator by $-1$ as the formula needs $r(r-1)(r-2)\cdots,$
$\displaystyle -2S=\sum_{r=1}^\infty\left(\frac45\right)^{r+1}\cdot\dfrac{\frac12\left(\frac12-1\right)\cdot\left(\frac12-2\right)\cdots\left(\frac12-r\right)}{(r+1)!}\cdot(-1)^{r+1}$
Observe that here the denominator and the exponent of $-\dfrac45$ of the first term $=2!,$ and compare with the generalized Binomial expansion
$\displaystyle\implies -2S+1+\frac12\left(-\frac45\right)$
$\displaystyle=1+\frac12\left(-\frac45\right)+\sum_{r=1}^\infty\dfrac{\frac12\left(\frac12-1\right)\cdot\left(\frac12-2\right)\cdots\left(\frac12-r\right)}{(r+1)!}\left(-\frac45\right)^{r+1}$
$\displaystyle\implies -2S+1+\frac12\left(-\frac45\right)=\left(1-\frac45\right)^{\dfrac12}$
Hope things are manageable from here
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/900687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Direct formula for area of a triangle formed by three lines, given their equations in the cartesian plane. I read this formula in some book but it didn't provide a proof so I thought someone on this website could figure it out. What it says is:
If we consider 3 non-concurrent, non parallel lines represented by the equations :
$$a_1x+b_1y+c_1=0$$
$$a_2x+b_2y+c_2=0$$
$$a_3x+b_3y+c_3=0$$
Then the area of the triangle that these lines will enclose is given by the magnitude of :
$$\frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3}$$
Where $C_1,C_2,C_3$ are the co-factors of $c_1,c_2,c_3$ respectively in the above matrix.
What I'm wondering is, where did this come from? And why isn't it famous? Earlier we had to calculate areas by finding the vertices and all but this does it in a minute or so and thus deserves more familiarity.
|
Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":
$$\begin{align}
x \cos\theta + y \sin\theta - p &= 0 \\
x \cos\phi + y \sin\phi - q &= 0 \\
x \cos\psi + y \sin\psi - r &= 0
\end{align}$$
with $\theta$, $\phi$, $\psi$ and $p$, $q$, $r$ (and $A$, $B$, $C$ and $a$, $b$, $c$) as in the figure:
Then
$$C_1 = \left|\begin{array}{cc}
\cos\phi & \sin\phi \\
\cos\psi & \sin\psi
\end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$
Likewise,
$$C_2 = \sin B \qquad C_3 = \sin C$$
Moreover,
$$D := \left|\begin{array}{ccc}
\cos\theta & \sin\theta & - p \\
\cos\phi & \sin\phi & - q \\
\cos\psi & \sin\psi & - r
\end{array}\right| = - \left( p C_1 + q C_2 + r C_3 \right) = - \left(\;p \sin A + q \sin B + r \sin C\;\right)$$
Writing $d$ for the circumdiameter of the triangle, the Law of Sines tells us that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
Therefore,
$$\begin{align}
D &= - \left( \frac{ap}{d} + \frac{bq}{d} + \frac{cr}{d} \right) \\[4pt]
&= -\frac{1}{d}\left(\;ap + b q + c r\;\right) \\[4pt]
&= -\frac{1}{d}\left(\;2|\triangle COB| + 2|\triangle AOC| + 2|\triangle BOA| \;\right) \\[4pt]
&= -\frac{2\;|\triangle ABC|}{d}
\end{align}$$
Also,
$$C_1 C_2 C_3 = \sin A \sin B \sin C = \frac{a}{d}\frac{b}{d}\sin C= \frac{2\;|\triangle ABC|}{d^2}$$
Finally:
$$\frac{D^2}{2C_1C_2C_3} = \frac{4\;|\triangle ABC|^2/d^2}{4\;|\triangle ABC|/d^2} = |\triangle ABC|$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/901819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 0
}
|
How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with degree 3 such that $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots
Note that $\dfrac{\pi}{12}$, $\dfrac{9\pi}{12}$, $\dfrac{17\pi}{12}$ are solution of equation $\cos3\theta=\dfrac{1}{\sqrt{2}}$ and $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are distinct number.
We have $\cos3\theta=4\cos^3\theta-3\cos\theta$. Let $x=\cos\theta$, therefore
$\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots of $4x^3-3x=\dfrac{1}{\sqrt{2}}.$
I want method similar to this to find $P(x)$.
Thank you.
|
By a tedious expansion of $P(x)=(x-r_1)(x-r_2)\ldots$ that other answers have covered or by using Vieta's formulas, you can find that
$$P(x)=x^4+\left[-\sin\left(\frac{\pi}{24}\right)-\sin\left(\frac{7\pi}{24}\right)-\sin\left(\frac{11\pi}{24}\right)-\sin\left(\frac{13\pi}{24}\right)\right]x^3$$
$$\ldots+\left[\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{7\pi}{24}\right)+\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)+\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)\right.$$
$$\left.\ldots+\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)+\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)+\sin\left(\frac{13\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)\right]x^2$$
$$\ldots+\left[-\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)-\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)\right.$$
$$\left.\ldots-\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)-\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)\right]x$$
$$\ldots+\left[\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)\right]$$
Using Wolfram Alpha, we can find that $\sin\left(\frac{\pi}{24}\right)=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{3}}}$, $\sin\left(\frac{7\pi}{24}\right)=\frac{1}{2}\sqrt{2+\sqrt{2-\sqrt{3}}}$, $\sin\left(\frac{13\pi}{24}\right)=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{3}}}$ & $\sin\left(\frac{19\pi}{24}\right)=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{3}}}$. You could then substitute these in (I'd be interested to see what it simplifies to). Please note that with an expression so long, I'm bound to have made a mistake somewhere. Thanks for the interesting question.
Edit: Using $\sin(\ldots)$ in terms of $e^{(i\ldots)}$, I've managed to express the coefficient of $x^3$ as $\frac{1}{2}(i-1)\left(e^{i\pi/24}+e^{5i\pi/24}\right)-\frac{1}{2}(i+1)\left(e^{-i\pi/24}+e^{-5i\pi/24}\right)$ though can't think where to go from there.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/905303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
}
|
How to find ${\large\int}_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx$? Please help me to find a closed form for this integral:
$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx
$$
Routine textbook methods for this complicated integral fail.
|
Here is another Feynman's way to evaluate the integral. Set $x=\frac1t$, then
$$
\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x}\ dx=\int_0^1\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt.
$$
Now consider
$$
\mathcal{I}(\alpha)=\int_0^1t^\alpha\cdot\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt.
$$
Hence
\begin{align}
\frac{d^2\mathcal{I}}{d\alpha^2}&=\int_0^1\frac{t^\alpha}{1+t^2}(t-1-t\ln t)\ dt\\
&=\int_0^1\sum_{n=0}^\infty(-1)^nt^{2n+\alpha}(t-1-t\ln t)\ dt\\
&=\sum_{n=0}^\infty(-1)^n\int_0^1\left(t^{2n+\alpha+1}-t^{2n+\alpha}-t^{2n+\alpha+1}\ln t\right)\ dt\tag1\\
&=\sum_{n=0}^\infty(-1)^n\left[\frac{1}{2n+\alpha+2}-\frac{1}{2n+\alpha+1}+\frac{1}{(2n+\alpha+2)^2}\right]\tag2\\
&=-\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]+\frac14\left[\psi\left(\frac{\alpha+1}{4}\right)-\psi\left(\frac{\alpha+3}{4}\right)\right]\\
&\quad+\frac1{16}\left[\psi_1\left(\frac{\alpha+2}{4}\right)-\psi_1\left(\frac{\alpha+4}{4}\right)\right]\tag3\\
\frac{d\mathcal{I}}{d\alpha}&=-\ln\Gamma\left(\frac{\alpha+2}{4}\right)+\ln\Gamma\left(\frac{\alpha+4}{4}\right)+\ln\Gamma\left(\frac{\alpha+1}{4}\right)-\ln\Gamma\left(\frac{\alpha+3}{4}\right)\\
&\quad+\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]\tag4\\
&=\ln\Gamma\left[\frac{(\alpha+1)(\alpha+4)}{(\alpha+2)(\alpha+3)}\right]+\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]\tag5\\
\mathcal{I}(\alpha)&=\int\ln\Gamma\left[\frac{(\alpha+1)(\alpha+4)}{(\alpha+2)(\alpha+3)}\right]\ d\alpha+\ln\Gamma\left(\frac{\alpha+2}{4}\right)-\ln\Gamma\left(\frac{\alpha+4}{4}\right).
\end{align}
Since $0<t<1$, then as $\alpha\to\infty$, implying $\mathcal{I}(\alpha)\to0$ and $\mathcal{I}'(\alpha)\to0$. Thus
$$
\mathcal{I}(0)=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x}\ dx=\color{blue}{\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G},
$$
where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.
Notes :
$\displaystyle(1)\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots$
$\displaystyle(2)\ \ \sum_{k=0}^\infty\frac{(-1)^{k}}{(z+k)^{m+1}}=\frac1{(-2)^{m+1}m!}\left[\psi_{m}\left(\frac{z}{2}\right)-\psi_{m}\left(\frac{z+1}{2}\right)\right]$
$\displaystyle(3)\ \ \psi_{m}(z) := \frac{d^m}{dz^m} \psi(z) = \frac{d^{m+1}}{dz^{m+1}} \ln\Gamma(z)$
$\displaystyle(4)\ \ \int\ln\Gamma(z)\ dz=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log \text{G}(1+z)+C$,
$\qquad$where $\text{G}(\cdot)$ is the Barnes G-function and $C$ is a constant of integration. Also for $a,b>0$
$$
\int\ln\Gamma\left(\frac{x+a}{b}\right)\ dx=b\ \psi^{(-2)}\left(\frac{x+a}{b}\right)+C.
$$
$\displaystyle(5)\ \ $As $\alpha\to\infty$, both of sides tend to $0$, hence the constant of integration is $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/905653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 3,
"answer_id": 0
}
|
If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$ If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$.
|
Hint: Just use $\sin^2\theta = 1 - \cos^2\theta$, get the eqn in $(a-b)^2 = 0$ form.
Then, simply use the fact that $\sec\theta = \frac1{\cos\theta}$
Spoiler :
!> $$\sin^2 \theta + 2\cos\theta - 2 = 0 \\
\implies 1 - \cos^2\theta + 2\cos\theta -2 = 0\\
\implies -(\cos^2\theta - 2 \cos\theta + 1^2) = 0\\
\implies (\cos\theta - 1)^2 = 0\\
\implies \cos\theta = 1\\
\implies \cos^3 \theta = 1^3 \quad\dots(1)\\
\implies \sec^3\theta = 1 \quad\dots(2)$$
From $(1)$ and $(2)$, the answer is $1+1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/905980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Algebra equation with functions, constraints and a graph. Consider the function $f:[1,3]\to\mathbf R$, $f(x)=-x^4+8x^3+ax^2+bx+d$, where $a$, $b$, $d$ are real constants. Find the values of $d$ for which $f$ has 3 stationary points between $x=1$ and $x=3$ and $f(1)=f(3)=0$.
|
$f(1)=f(3)=0$, then $7+a+b+d=0$ and $135+9a+3b+d=0$. We know that $a=(d-57)/3$ and $b=12-4d/3$. Since there're three roots for the first derivative, by Rolle's theorem there are two roots for second derivative between $[1,3]$. Namely, $-12x^2+48x+2a=0$ has two roots in $[1,3]$, that is $2\pm \sqrt{4+\frac{a}{6}}\in[1,3]$. Solving the equation we know that $-24\le a< -18$, which implies $-15\le d< 3$. We may now solve for the first derivative. Then $f'(1),f'(2-\sqrt{4+\frac{a}{6}})$,$f'(2+ \sqrt{4+\frac{a}{6}}),f'(3)$ have different signs. These can be written as $-2(9+d)/3,-2\sqrt{2}/27(15+d)^{3/2},2\sqrt{2}/27(15+d)^{3/2},2(9+d)/3$. The second one is definitely negative, so the first one should be positive, which means $-15\le d< -9$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/906848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How prove this inequality $\sum\limits_{cyc}\frac{1}{a+3}-\sum\limits_{cyc}\frac{1}{a+b+c+1}\ge 0$ show that:
$$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$
where $abcd=1,a,b,c,d>0$
I have show three variable inequality
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$.
Prove that
$$\frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$$
also see:can see
：http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=243
from this equality,I have see a nice methods:
I think this Four varible inequality is also true
First,Thank you Aditya answer,But I read it your solution,it's not true
|
We have
$$\begin{align}
\frac3{a+b+c+1} &- \frac1{a+3} - \frac1{b+3} - \frac1{c+3}\\
&= \sum_{cyc}^{a, b, c} \left(\frac1{a+b+c+1}-\frac1{a+3} \right) \\
&= \frac1{a+b+c+1} \sum_{cyc}^{a, b, c}\frac{2-b-c}{a+3} \\
&= \frac1{(a+b+c+1)\prod_{cyc}^{a, b, c}(a+3)} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2) \\
&\le \frac1{1\times3^3} \sum_{cyc}^{a, b, c} (18-6a-4ab-a^2b-ab^2-6a^2)
\end{align}$$
using $a, b, c > 0$. Summing over four such inequalities, we get
$$3\sum_{cyc}\frac1{a+b+c+1} - 3\sum_{cyc} \frac1{a+3} \\ \le \frac2{27}\left( 108-\sum_{cyc}\left(9(a+a^2)+4(2ab+bd)+(a^2b+ab^2+a^2c+ac^2+b^2d+bd^2)\right) \right)$$
Now by AM-GM and the constraint, we have that $\sum_{cyc} a^mb^n \ge 4\sqrt[4]{(abcd)^{m+n}}=4$ for all $m, n \ge 0$, so RHS $\le 0$ and we are done.
P.S. the method looks general, though I wouldn't want to write down the cyclic sums for more variables!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/907336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
}
|
Solve a limit with radicals I don't know how to solve this limit. What should I do?
$$
\lim_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}}
$$
Thank you!!
|
As David H commented, rationalizing the denominator would be a good starting point.
Now, if you know Taylor series, the problem starts to be simple since, around $x=0$, you have $$\sqrt{x^3+2x+1}=1+x+O\left(x^2\right)$$ $$\sqrt{x^2-3x+1}=1-\frac{3 x}{2}+O\left(x^2\right)$$ $$\sqrt{4x^2-3x+1}=1-\frac{3 x}{2}+O\left(x^2\right)$$ $$\sqrt{2x^3+6x^2+5x+1}=1+\frac{5 x}{2}+O\left(x^2\right)$$ So, numerator is $$ \frac{5 x}{2}+O\left(x^2\right)$$and denominator is $$-4 x+O\left(x^2\right)$$
I am sure that you can take from here.
All of the above has been done using the fact that, if $x$ is small $$\sqrt{a+bx+cx^2+dx^3+ex^4+\cdots}\simeq \sqrt{a+bx}=\sqrt{a}+\frac{b x}{2 \sqrt{a}}+O\left(x^2\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/907412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
}
|
To find maximum value If $A>0,B>0$ and $C>0$ and further it is known that $A+B+C=\frac{5\pi}{4}$,then find the maximum value of $\sin A+\sin B+\sin C$
|
$\sin{A}+\sin{B}+\sin{C}=2\sin{\frac{A+B}{2}}\cos{\frac{A-B}{2}}+\sin{C}\le 2\sin{\frac{A+B}{2}}+\sin{(\frac{5\pi}{4}-(A+B))}.$ Equality holds when $A=B$. Thus we consider the function $$f(x)=2\sin{x}+\sin{\left(\frac{5\pi}{4}-2x\right)}=2\sin{x}+\sin{\left(2x-\frac{\pi}{4}\right)}.$$
defined on $(0,\frac{5\pi}{8})$. Then $$f'(x)=2\cos{x}+2\cos{\left(2x-\frac{\pi}{4}\right)}$$
It achieves zero when $x=2x+\frac{3\pi}{4}+2k\pi$ or $x=2k\pi-2x+\frac{5\pi}{4}$.
That is, $0<x<\frac{5\pi}{12},f'>0$ and $5\pi/12<x<5\pi/8,f'<0$. Thus $f$ achieves maximum at $x=5\pi/12$. Namely, $A=B=C=5\pi/12$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/907695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
}
|
How to find the sum of sequence $ 1+4+4^2+\cdots+4^{X+Y} $? I see the following sequence and it's:
$$h=1+4+4^2+\cdots+4^{X+Y}=\frac{4^{X+Y+1}-1}{4-1}$$
how we get this sequence?
I know this is a primary question but I confused :)
|
This is an example of a geometric series. Let's say we try to sum:
$$S=1+r+r^2+r^3+\dotsb+r^n$$
In your example, $r=4$, and $n=X+Y$.
There is a well known "trick" for solving this. Multiply by $r$:
$$Sr=r+r^2+r^3+\dotsb+r^n+r^{n+1}$$
Notice how this is the same thing as $S$, except without the $1$ at front and with an extra $r^{n+1}$ at the end. In fact:
$$Sr=S-1+r^{n+1}$$
Solving:
\begin{align}
Sr&=S-1+r^{n+1} \\
Sr-S&=-1+r^{n+1} \\
S(r-1)&=r^{n+1}-1 \\
S&=\frac{r^{n+1}-1}{r-1}
\end{align}
$$\fbox{$1+r+r^2+r^3+\dotsb+r^n=\dfrac{r^{n+1}-1}{r-1}$}$$
Plugging in $r=4$ and $n=X+Y$, we get your answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/907874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Hard Definite integral involving the Zeta function Prove that: $$\displaystyle \int_{0}^{1}\frac{1-x}{1-x^{6}}{\ln^4{x}} \ {dx} = \frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54} $$
I was able to simplify it a bit by substituting ${y = -\ln{x}}$ and some further mathematical manipulation but was not able to get the correct form.
|
Note that $\frac{1-x}{1-x^6}=\sum_{k=0}^\infty (x^{6k}-x^{6k+1})$. And the integration $\int_0^1 x^n \ln{x}^4=\partial_n^4 \int_0^1 x^n dx=\frac{24}{(n+1)^5}$. We have
$$LHS = 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$
Use the discrete Fourier, and denote $\xi=\exp(i\frac{\pi}{3})$, $\xi_i =\xi^i$. Then
$$LHS=24\sum_{i=1}^6a_i\sum_{k=1}^\infty \frac{\xi_i^k}{k^5}=24\sum_{i=1}^6 a_i Li_5(\xi_i).$$
where $a_1=a_5=\frac{1}{6}$,$a_2=-\frac{i}{2\sqrt{3}}=-a_4$,$a_6=0$,$a_3=-\frac{1}{3}$.Thus we use the summation formula for polylogarithm
$$Li_5(\xi_2)+(-1)^5 Li_5(\xi_4)=-\frac{(2\pi i)^5}{5!}B_5(\frac{1}{3})=\frac{4i\pi^5}{729}.$$
$B$ is the Bernoulli polynomial. Also, $$Li_5(\xi_1)+Li_5(\xi_5)=(1-\frac{2}{2^5}-\frac{3}{3^5}+\frac{6}{6^5})\sum_{k=1}^\infty\frac{1}{k^5}=\frac{25}{27}\sum_{k=1}^\infty\frac{1}{k^5}.$$
Also note that $L_5(-1)=-(1-\frac{2}{2^5})\zeta(5).$ We conclude that
$$24\sum_{i=1}^6 a_i Li_5(\xi_i)=24\frac{-i}{2\sqrt{3}}\cdot\frac{4i\pi^5}{729}+24(\frac{1}{6}\frac{25}{27}+\frac{1}{3}\cdot\frac{15}{16})\zeta(5)=\frac{16\pi^5}{243\sqrt{3}}+\frac{605}{54}\zeta(5)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/909977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
}
|
How many ways to generate unique multiplication result from given set? From set {2, 2, 3, 5}, I can have 8 ways to generate unique multiplication result, which:
- two number multiplication: 2*2, 2*3, 2*5, 3*5
- three number multiplication: 2*2*3, 2*2*5, 2*3*5
- four number multiplication: 2*2*3*5
Then, how many unique multiplication result I can generate for following set
{ 11, 11, 11, 11, 11, 7, 7, 7, 7, 7, 5, 5, 5, 5, 3, 3, 3, 2, 2, 2}
What is the formula?
*Edit: only prime numbers are allowed in the set
|
Every positive integer has a unique prime factorization. So your products will all be of the form
$$ 11^a\cdot 7^b\cdot 5^c\cdot 3^d\cdot 2^e $$
where $0\le a,b,c,d,e$ and $a,b\le 5;\; c\le 4;\; d,e\le 3$. Which gives you
$$ 6\cdot 6\cdot 5\cdot 4\cdot 4 = 2880 $$
possible numbers of this form. But that includes the case of zero factors (in which case you have $a=b=c=d=e=0$ and the product is $1$) as well as the cases of one factor ($a+b+c+d+e=1$, $5$ alternatives). Therefore your actual number is
$$ 6\cdot 6\cdot 5\cdot 4\cdot 4 - 1 - 5 = 2874 $$
This generalizes to other multisets in an obvious way. If prime number $i$ (its actual value is irrelevant) occurs $k_i$ times, and you have $n$ such factors in total, then the number of distinct non-trivial products is
$$ (k_1+1)\cdot(k_2+1)\cdots(k_n+1) - 1 - n =
\left(\prod_{i=1}^n(k_i+1)\right) - 1 - n $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/910716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the equation of normal line to the graph of given function
Give the equation of the normal line to the graph of
$$y = 2x \sqrt{x^2+8} + 2$$
at the point $(0,2)$
What I've done so far is:
Taken the derivative and got
$$(2x^2)/\sqrt{ x^2+8} + 2\sqrt{x^2+8}$$
I have no idea if this is right, it was pretty hard to get the derivative of that. Before I go on any further, is this derivative right?
Source: http://online.math.uh.edu/apcalculus/exams/AP_AB_version1_1.htm - #9
|
$$f(x)=2x \sqrt{x^2+8}+2 \Rightarrow f'(x)=2\sqrt{x^2+8}+\frac{x}{\sqrt{x^2+8}}2x=2\sqrt{x^2+8}+\frac{2x^2}{\sqrt{x^2+8}}=\frac{2(x^2+8)+2x^2}{\sqrt{x^2}8}=\frac{4x^2+16}{\sqrt{x^2+8}}$$
At the point $(0,2)$, $f'(0)=\frac{16}{\sqrt{8}}=\frac{16}{2 \sqrt{2}}=\frac{8}{\sqrt{2}}$.
The slope of the normal line is $\frac{-1}{f'(0)}=-\frac{\sqrt{2}}{8}$.
Therefore, the equation of the normal line at the point $(0,2)$ is the following:
$$y-y_1=m(x-x_1) \Rightarrow y-2=-\frac{\sqrt{2}}{8}(x-0) \Rightarrow 8y-16=-\sqrt{2}x \Rightarrow 8y+\sqrt{2}x=16$$
$$8y+\sqrt{2}x=16 \Rightarrow 8 \sqrt{2}y+2x=16 \sqrt{2} \Rightarrow 4 \sqrt{2}y+x=8 \sqrt{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/911494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Prove $\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ if $a^2+b^2+c^2=1$
Ff $a,b,c$ are positive real numbers that $a^2+b^2+c^2=1$ ,Prove: $$\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$$
Additional info: I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done: I tried to change LHS to something more easy to work but I was not successful. For example $$\frac{ab}{1+c^2}=\frac{1}{2}\left(\frac{a^2+b^2+2ab}{1+c^2}-\frac{a^2+b^2}{1+c^2}\right)$$
that was not useful. Any hint for starting step is appreciated.
|
Write $x=a^2$, $y=b^2$, and $z=c^2$. Then, $x+y+z=1$. A first consequence of this is
$$
(1+z)\geq 2\sqrt{xy+z}.\tag{*}
$$
This is true because
$$
(1+z)^2-(2\sqrt{xy+z})^2=(1+z)^2-4z-4xy\\
=(1-z)^2-4xy=(x+y)^2-4xy=(x-y)^2\geq 0.
$$
Analogous to (*), we also have
$$
(1+y)\geq2\sqrt{xz+y},\quad (1+x)\geq 2\sqrt{yz+x}.
$$
This implies
$$
\sum_{\text{cyc}}\frac{ab}{1+c^2}=\sum_{\text{cyc}}\frac{\sqrt{xy}}{1+z}\leq\frac{1}{2}\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}.
$$
So the claim follows if we can show $\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}\leq\frac{3}{2}$. But that has already been done here, so we are done!
p.s. For completeness, I'll produce the argument from the link here: first, observe
$$
xy+z=xy+(1-x-y)=(1-x)(1-y)=(y+z)(x+z).
$$
Then, it follows from the AM-GM inequality that
$$
\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}=\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{(x+z)(y+z)}}\leq\sum_{\text{cyc}}\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{3}{2}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/912905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4]{xy^3}$$
$$3x+z=x+x+x+z\ge 4\sqrt[4]{x^3z}$$
but this is not true,because not all four equalities can hold at once.
This problem is from china middle school test,so I think have without Lagrange methods,so I think this inequality have AM-GM inequality
|
$$f(x,y,z)=\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
$$\nabla f= \frac{(3 (2+5 y) (5+2 z) (x^2-y z))}{(x^2 y z)} \hat e_x+\frac{(-((2 x-15 y^2) (3 x+z) (5+2 z))}{(x y^2 z))}\hat e_y+\frac{(-((x+3 y) (2+5 y) (15 x-2 z^2))}{(x y z^2))}\hat e_z$$
We get 4 real solutions:
$$(5/6,-5/18,-5/2),(1,-\sqrt{2/15},\sqrt{15/2}),(1,\sqrt{2/15},\sqrt{15/2}),(6/5,-2/5,-18/5)$$
Out of which only $(1,\sqrt{2/15},\sqrt{15/2})$ is acceptable due to $x,y,z\ge0$
Now the value of function at this point is:
$$f(1,\sqrt{2/15},\sqrt{15/2})=241+44\sqrt{30}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/913664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
}
|
How did Ulam and Neumann find this solution? In the book "Chaos, Fractals and Noise - Stochastic Aspects of
Dynamics" from Lasota and Mackey the operator $P: L^1[0,1] \to L^1[0,1]$
$$ (Pf)(x) = \frac{1}{4\sqrt{1-x}} \left[ f\left(\frac{1}{2}\left(1-\sqrt{1-x}\right)\right) + f\left(\frac{1}{2}\left(1+\sqrt{1-x}\right)\right)\right]$$
was considered, where $L^1$ denotes the space of Lebesgue integrable functions.
It is stated in the book that Ulam and Neumann showed that
$$ f^*(x) =\frac{1}{\pi\sqrt{x(1-x)}} $$
solves
$$ Pf^* = f^*,$$ in the article "On combination of stochastic and deterministic processes, Bull. Amer. Math. Soc. vol. 53 (1947) p. 1120.". I went to the library of my university and I actually found the book "vol 53 of Bull. Amer. Math. Soc.", but this book only contains the abstract of the article, not the article itself.
So my question is: Where do I find the article - or how did they find $f^*$ ?
Edit: I am interested how Ulam and Neumann actually constructed $f^*$.
|
I don't have the reference for the derivation, but I can proof, that $f^*$ is indeed an eigenfunction of $P$ with eigenvalue $1$, i.e. $Pf^*=f^*$.
Proof
\begin{align}
(Pf^*)(x)&= \frac{1}{4\sqrt{1-x}} \Bigl( f\Bigl(\frac12 (1-\sqrt{1-x})\Bigr)+
f\Bigl(\frac12 (1+\sqrt{1-x})\Bigr)\Bigr)
\\
&=\frac{1}{4\sqrt{1-x}} \Bigl(
\frac1\pi \frac{1}{\sqrt{\frac12 (1-\sqrt{1-x})(1-\frac12 (1-\sqrt{1-x})}}\Bigr)
+
\frac1\pi \frac{1}{\sqrt{\frac12 (1+\sqrt{1-x})(1-\frac12 (1+\sqrt{1-x})}}\Bigr)
\Bigr)
\\
&=\frac{1}{4\pi\sqrt{1-x}}\Bigl(
\frac{1}{\sqrt{\frac12 -\frac12 \sqrt{1-x})(\frac12 +\frac12 \sqrt{1-x})}}\Bigr)
+\frac{1}{\sqrt{\frac12 +\frac12 \sqrt{1-x})(\frac12 -\frac12 \sqrt{1-x})}}\Bigr)
\\ &=
\frac{1}{4\pi\sqrt{1-x}}\Bigl(
\frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr)
+ \frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr)
\\ &=
\frac{2}{4\pi\sqrt{1-x}}\Bigl(
\frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr)
\\ &=
\frac{2}{4\pi\sqrt{1-x}}\Bigl(
\frac{1}{\sqrt{\frac14 -\frac14({1-x})}}\Bigr)
\\ &=
\frac{2}{\pi\sqrt{1-x}}\Bigl(
\frac{1}{\sqrt{1-({1-x})}}\Bigr)
\\ &=
\frac{2}{2\pi\sqrt{1-x}}\Bigl(
\frac{1}{\sqrt{x}}\Bigr)=f^*
\end{align}
Okay there are a lot of brackets missing/wrong and maybe this is of no use. The message was only, that it is trivial to proof, that $Pf^*=f^*$. Simply plug in the definition of $f^*$ and use binomial expansion for $(a-b)(a+b)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/913761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
}
|
Simpler closed form for $\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}$ I'm trying to find a closed form of this sum:
$$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}.\tag{1}$$
WolframAlpha gives a large expressions containing multiple generalized hypergeometric functions, that is quite difficult to handle. After some simplification it looks as follows:
$$S=\frac{\pi^{3/2}}{3}-\sqrt{\pi}-\frac{\sqrt{\pi}}{324}\left[9\,_3F_2\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)\\+3\,_4F_3\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)+\,_5F_4\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)\right].\tag{2}$$ I wonder if there is a simpler form. Elementary functions and simpler special funtions (like Bessel, gamma, zeta, polylogarithm, polygamma, error function etc) are okay, but not hypergeometric functions.
Could you help me with it? Thanks!
|
Another possible closed form of $S$ is the following. It containts also a generalized hypergeometric function, but just one.
$$S = \frac{\sqrt{\pi}}{648} {_6F_5}\left(\begin{array}c\ 1,\frac32,\frac32,\frac32,\frac32,\frac32\\2,\frac52,\frac52,\frac52,\frac52\end{array}\middle|\,\frac14\right).$$
WolframAlpha's simplification gives back your form.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/915054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 2,
"answer_id": 0
}
|
Simplify rational expression How do I simplfy this expression?
$$\dfrac{\frac{x}{2}+\frac{y}{3}}{6x+4y}$$
I tried to use the following rule $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$
But I did not get the right result.
Thanks!!
|
$$\frac{\frac{x}{2}+\frac{y}{3}}{6x+4y}$$
Start by simplifying the numerator. Specifically, add the two fractions.
$$\frac{\frac{x}{2}+\frac{y}{3}}{6x+4y}=\frac{\frac{3x}{6}+\frac{2y}{6}}{6x+4y}=\frac{\frac{3x+2y}{6}}{6x+4y}$$
Then, since the fraction bar means division, you have:
$$\frac{\frac{3x+2y}{6}}{6x+4y}=\frac{3x+2y}{6}\div(6x+4y)$$
And the rest is just the division of two fractions.
$$\frac{3x+2y}{6}\div(6x+4y)=\frac{3x+2y}{6}\times\frac{1}{6x+4y}=\frac{3x+2y}{36x+24y}$$
However, we're not done. We need to factor the numerator and denominator and simplify.
$$\frac{3x+2y}{36x+24y}=\frac{3x+2y}{12(3x+2y)}=\frac{1}{12}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/915154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equation*} \dfrac{5}{3} = 1 + \dfrac{2}{3}\text{.}
\end{equation*} This is because when 5 is divided by 3, 3 goes into 5
once (hence the $1$ term) and there is a remainder of $2$ (hence the
$\dfrac{2}{3}$ term). Note the following: every division problem can
be decomposed into an integer (the $1$ in this case) plus a fraction,
with the denominator being what you divide by (the $3$ in this
case).
So, when $n$ is divided by 14, the remainder is 10. This can be
written as \begin{equation*} \dfrac{n}{14} = a + \dfrac{10}{14}
\end{equation*} where $a$ is an integer.
We want to find the remainder when $n$ is divided by 7, which I'll
call $r$. So \begin{equation*} \dfrac{n}{7} = b + \dfrac{r}{7}\text{,}
\end{equation*} where $b$ is an integer.
Here's the key point to notice: notice that \begin{equation*}
\dfrac{n}{7} = \dfrac{2n}{14} = 2\left(\dfrac{n}{14}\right)\text{.}
\end{equation*} This is because $\dfrac{1}{7} = \dfrac{2}{14}$.
Thus, \begin{equation*} \dfrac{n}{7} = 2\left(\dfrac{n}{14}\right) =
2\left(a + \dfrac{10}{14}\right) = 2a + 2\left(\dfrac{10}{14}\right) =
2a + \dfrac{10}{7} = 2a + \dfrac{7}{7} + \dfrac{3}{7} = (2a+1) +
\dfrac{3}{7}\text{.} \end{equation*} So, since $a$ is an integer, $2a
+ 1$ is an integer, which is our $b$ from the original equation. Thus, $r = 3$.
To her, this method was not very intuitive. She did understand the explanation. Are there any suggestions for how I can explain this in another way?
|
It might help to go back to the basic definition of multiplication as repeated addition:
$$2 \cdot 5 = 2 + 2 + 2 + 2 + 2 = 10$$
and that division is the opposite of multiplication:
$$\frac{10}{2} \mapsto 10 \underbrace{{}- 2 - 2 - 2 - 2 - 2}_{\text{5 times}} = 0.$$
So $\frac{10}{2}=5$, remainder $0$
Also, for $\frac{11}{3} = \rightarrow 11 \underbrace{{}- 3 - 3 - 3}_{\text{3 times}} = 2$
So $\frac{11}{3}=3$, remainder $2=3+\frac{2}{3}$
If $\frac{n}{14}=a+\frac{10}{14}$, that means $n\underbrace{{}-14-14-14-\cdots}_{a\text{ times}}=10$
Here's the simple/intuitive part:
Since $2(7)=14$, division by 7 will also have a "remainder" of 10
But 10 > 7, so we can subtract out one more 7: $10 - 7 = 3$.
Therefore, the remainder ($r$) is 3: $r=3$
(Also note, $b=2a+1$. Because $14=2(7)$, we double $a$.
Because we subtracted one more 7, we add 1.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/915238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
}
|
Value of $\psi\left(\frac{1}{2}\right)$ I apologise if this is a dumb question, but I have trouble deriving $\displaystyle\psi\left(\frac{1}{2}\right)=-\gamma-2\ln{2}$. I have tried the following.
\begin{align}
\psi\left(\frac{1}{2}\right)
&=\lim_{N\to\infty}\left[-\gamma-2+\sum^N_{k=1}\left(\frac{1}{k}-\frac{1}{k+1/2}\right)\right]\\
&=\lim_{N\to\infty}\left[-\gamma-1+\sum^{\lfloor \frac{N-1}{2} \rfloor}_{k=1}\frac{1}{2k}+\sum^{\lfloor \frac{N-1}{2} \rfloor}_{k=1}\frac{1}{2k+1}-2\sum^N_{k=1}\frac{1}{2k+1}\right]\\
&=\lim_{N\to\infty}\left[-\gamma-1+\sum^{\lfloor \frac{N-1}{2} \rfloor}_{k=1}\left(\frac{1}{2k}-\frac{1}{2k+1}\right)-2\sum^N_{k=1+\lfloor \frac{N-1}{2} \rfloor}\frac{1}{2k+1}\right]\\
&=-\gamma-\ln{2}-\lim_{N\to\infty}\sum^N_{k=1+\lfloor \frac{N-1}{2} \rfloor}\frac{2}{2k+1}
\end{align}
I have trouble showing that the last sum $\to \ln{2}$ as $N \to \infty$, and would like to seek your assistance in evaluating the limit. Thank you in advance.
|
One could evaluate the limit as follows. Set $N=2 n$:
$$\lim_{N\to \infty} \sum_{k=1+\lfloor{\frac{N-1}{2}}\rfloor}^{N} \frac{2}{2k+1}=\lim_{n\to \infty} \sum_{k=n}^{2n} \frac{2}{2k+1}=\lim_{n\to \infty} 2\cdot\left(\sum_{k=0}^{2n} \frac{1}{2k+1}-\sum_{k=0}^{n-1} \frac{1}{2k+1}\right)=2\cdot\lim_{n\to \infty} \left(\left(\sum_{k=1}^{4n+1} \frac{1}{k}-\sum_{k=1}^{2n} \frac{1}{2k}\right)-\left(\sum_{k=1}^{2n-1} \frac{1}{k}-\sum_{k=1}^{n-1} \frac{1}{2k}\right)\right)=2\cdot\lim_{n\to \infty}{H_{4n+1}-\frac{1}{2}H_{2n}-H_{2n-1}+\frac{1}{2}H_{n-1}}=2\cdot\lim_{n\to \infty}{(H_{4n+1}-\ln(4n+1))+\ln(4n+1)-\frac{1}{2}(H_{2n}-\ln(2n))-\frac{1}{2}\ln(2n)-(H_{2n-1}-\ln(2n-1))-\ln(2n-1)+\frac{1}{2}(H_{n-1}-\ln(n-1))+\frac{1}{2}\ln(n-1)}=2\cdot\lim_{n\to \infty}{(H_{4n+1}-\ln(4n+1))-\frac{1}{2}(H_{2n}-\ln(2n))-(H_{2n-1}-\ln(2n-1))+\frac{1}{2}(H_{n-1}-\ln(n-1))}+2\cdot\lim_{n\to \infty}{\ln(4n+1)-\frac{1}{2}\ln(2n)-\ln(2n-1)+\frac{1}{2}\ln(n-1)}=2\cdot(\gamma-\frac{1}{2}\gamma-\gamma+\frac{1}{2}\gamma)+2\cdot\lim_{n\to \infty}{\ln(\frac{4n+1}{2n-1})-\frac{1}{2}\ln(\frac{2n}{n-1})}=2\cdot(\ln(2)-\frac{1}{2}\ln(2))=\ln(2)$$
Here I used that:
$$\lim_{n\to \infty}H_n-\ln(n)=\gamma$$
Where $\gamma$ is Euler's constant.
Note that there is a faster way to evaluate $\psi\left(\frac{1}{2}\right)$:
$$\psi\left(\frac{1}{2}\right)=-\gamma-2+\sum_{k=1}^{\infty} \left(\frac{1}{k}-\frac{1}{k+\frac{1}{2}}\right)=-\gamma+\sum_{k=1}^{\infty} \left(\frac{1}{k}-\frac{1}{k-\frac{1}{2}}\right)=-\gamma+\sum_{k=1}^{\infty} \left(\frac{2}{2k}-\frac{2}{2k-1}\right)=-\gamma-2\cdot\sum_{k=1}^{\infty} \left(\frac{1}{2k-1}-\frac{1}{2k}\right)=-\gamma-2\cdot\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=-\gamma-2\ln(2)$$
I hope, this is helpful and I apologize for all mistakes in my English, I'm not a native speaker. :)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Closed form for $1 + 3 + 5 + \cdots +(2n-1)$ What is the closed summation form for $1 + 3 + 5 + \cdots + (2n-1)$ ?
I know that the closed form for $1 + 2 + 3+\cdots + n = n(n+1)/2$ and I tried plugging in $(2n-1)$ for $n$ in that expression, but it didn't produce a correct result:
$(2n-1)((2n-1)+1)/2$
plug in 3
$(2n-1)((2n-1)+1)/2 = 6 != 1 + 2 + 3 = 9$
|
$$M=1+3+5+7+...+(2n-3)+(2n-1)\\M=(2n-1)+(2n-3)+...+7+5+3+1\\M+M=(1+2n-1)+(3+2n-3)+(5+2n-5)+....(2n-3+3)+(2n-1+1)\\n-term\\M+M=n(2n)\\2M=2n^2\\M=n^2
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
}
|
Find a solution to Laplace's equation that satisfies polar coordinates and show that any solution produces perpendicular lines. 7a. Find a solution of Laplace's equations $u_{xx}+u_{yy}=0$ of the form $u(x,y)=Ax^2+Bxy+Cy^2 (A^2+B^2+C^2 \neq 0)$ which satisfies the boundary condition $u(cos ( \theta),sin( \theta))=cos(2\theta)+sin(2\theta)$ for all points $(cos(\theta),sin(\theta))$ on the circle $x^2+y^2=1$
My Attempt: We need to find a solution to Laplace's equation in the form of $u(x,y)=Ax^2+Bxy+Cy^2 (A^2+B^2+C^2 \neq 0)$ that satisfies the polar coordinate conditions.
Suppose our solution to Laplace's equation is $u(x,y) = \frac{1}{2}x^2+xy-\frac{1}{2}y^2$
By taking the $u_x$, $u_{xx}$ $u_y $ and $u_{yy}$, we have
$u_x = x + y$
$u_{xx} = 1$
$u_y = x - y$
$u_{yy} = -1$
Therefore, $u_{xx}+u_{yy} = 0 \rightarrow 1-1 = 0$ and we have satisfied Laplace's equation.
7b. Show that the graph of any solution $u(x,y)$ of Laplace's equation of the form in $(a)$ intersections the $xy$-plane in a pair of perpendicular lines through $(0,0)$
I have some equations that I've found in that particular form which are the following:
$2x^2+4xy-2y^2$
$x^2+2xy-y^2$
$3x^2+6xy-3y^2$
$\frac{-1}{2}x^2+xy+\frac{1}{2}y^2$
So what I need to do is to graph these equations and show that they are intersecting the xy-plane with perpendicular lines. I know that a perpendicular line has a negative slope and on top of that it intersects whereas parallel lines are just two lines that don't cross in the graph.
It must also follow the condition that I must be in a circle of radius 1 because
$x^2+y^2 =1$ and we know that $x^2+y^2 =r^2$ from vector calculus, so $r^2 =1$ and the $r = 1$... well a +1 and a -1 to be exact because we need to add those signs after we take the square root. SO, the solution that should not only satisfy the equation for Laplace must also not leave the circle which is really small. So it could be that the numerical values of A B and C must be either 1 as in on the circle or $\frac{1}{2}$ meaning inside the circle.
So, maybe another solution might be $\frac{1}{4}x^2+\frac{1}{2}xy-\frac{1}{4}y^2$
$u_x = \frac{1}{2}x+\frac{1}{2}$
$u_{xx} = \frac{1}{2}$
$u_y = \frac{1}{2}x-\frac{1}{2}y$
$u_{yy} = -\frac{1}{2}$
$u_{xx}+u_{yy} = \frac{1}{2}-\frac{1}{2} = 0$
My question is how do I graph these solutions? Do I graph as normal by putting some values of x to obtain my y or some values of y to obtain my x and then plot the points? It's been a long time. The last time I graph was in vector calculus with the x,y,z.
I want to use these equations
$u(x,y) = \frac{1}{2}x^2+xy-\frac{1}{2}y^2$ [original from 7a]
$\frac{1}{4}x^2+\frac{1}{2}xy-\frac{1}{4}y^2$
$x^2+2xy-y^2$
|
The graph of the function $u$ is the subset $\{(x,y,u(x,y)):x,y\in\Bbb R\}\subset\Bbb R^3$. The points on the $xy$-plane are precisely those with $z$-coordinate $0$. Hence the intersection with our graph with the $xy$-plane is the set of all points $(x,y)$ with $u(x,y)=0$. You have $u(x,y)=x^2+2xy-y^2=0$. Can you solve this? As a hint, complete the square on the first two terms, ignoring the third, and then proceed from there.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Inequality $\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$ with weird condition I want to prove the following inequality:
$$\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$$
Where $a,b,c$ are positive reals and with the horrible condition:
$$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2$$
It was a problem of the MEMO 2011. What I tried so far is the following:
Make the substitution $x=\frac{a}{1+a}$, $y=\frac{b}{1+b}$, $z=\frac{c}{1+c}$ so $a=\frac{x}{1-x}$, $b=\frac{y}{1-y}$, $c=\frac{z}{1-z}$. The conditions then change to $x+y+z=2$ and $0<x,y,z<1$ since $\frac{a}{1+a}<1$ and so the other two terms. The original inequality changes to:
$$\frac{\sqrt \frac{x}{1-x}+\sqrt \frac{y}{1-y}+\sqrt \frac{z}{1-z}}{2}\ge\sqrt \frac{1-x}{x}+\sqrt \frac{1-y}{y}+\sqrt \frac{1-z}{z}$$
Now we have $y+z=2-x>1>x$ and similarly $x+y>z$, $z+x>y$. This is equivalent to $x=u+v$, $y=v+w$ and $z=w+u$. So with this change of variables, the condition $0<x,y,z<1$ disappears. The other condition becomes $(u+v)+(v+w)+(w+u)=2 \iff u+v+w=1$. The inequality changes to:
$$\frac{\sqrt \frac{u+v}{1-u-v}+\sqrt \frac{v+w}{1-v-w}+\sqrt \frac{w+u}{1-w-u}}{2}\ge\sqrt \frac{1-u-v}{u+v}+\sqrt \frac{1-v-w}{v+w}+\sqrt \frac{1-w-u}{w+u} \iff \frac{\sqrt \frac{u+v}{w}+\sqrt \frac{v+w}{u}+\sqrt \frac{w+u}{v}}{2}\ge\sqrt \frac{w}{u+v}+\sqrt \frac{u}{v+w}+\sqrt \frac{v}{w+u} \iff \sqrt \frac{u+v}{2w}+\sqrt \frac{v+w}{2u}+\sqrt \frac{w+u}{2v}\ge\sqrt \frac{2w}{u+v}+\sqrt \frac{2u}{v+w}+\sqrt \frac{2v}{w+u}$$
Am I right so far? The last inequality seems kind of "beauty" since each term on the righthand side has its reciprocal on the lefthand side. But I wasn't able to find a solution from there, so any help is highly appreciated! Thanks.
|
The condition $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2 \iff abc = a+b+c+2$, so as you deduced through two successive substitutions, we can have
$$a = \frac{u+v}w, \, b = \frac{v+w}u, \, c = \frac{w+u}v$$ to get
$$\sum_{cyc} \sqrt{\frac{u+v}w} \ge 2 \sum_{cyc} \sqrt{\frac{w}{u+v}} \iff \sum_{cyc} \frac{u+v-2w}{\sqrt{w(u+v)}}\ge 0$$
Now note that $u+v-2w$ and $w(u+v)$ are arranged oppositely and apply Chebyshev's inequality. Use $\sum_{cyc} (u+v-2w)=0$ to conclude.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor series form.
I mean:
$$\int _0 \frac{d}{dx}\sum_{n=1}^\infty (n^2+n^3)x^{n-1}dx=\int_0 \sum_{n=1}^\infty \frac{n^2+n^3}{n-1}x^{n-2}dx$$
But it doesn't seem to work in this exercise.
|
If $|x|<1$ :
The series converges by the ratio test
$\displaystyle\sum_{n=1}^\infty (n^2+n^3)x^{n-1}=\frac{x+1}{(1-x)^3}+\frac{x^2+4x+1}{(1-x)^2(x^2-2x+1)}=\frac{4 x + 2}{x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1}$
If $|x|>1$ :
The series diverges by the ratio test
$\displaystyle\sum_{n=1}^\infty (n^2+n^3)x^{n-1}=\sum_{n=1}^\infty\frac{n^2x^n}x+\sum_{n=1}^\infty\frac{n^3x^n}x=\sum_{n=1}^{\infty} n^{2} x^{n - 1} \left(n + 1\right)$ is the best I can come with
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
Solving an equation over the reals: $ x^3 + 1 = 2\sqrt[3]{{2x - 1}}$ Solve the following equation over the reals:$$
x^3 + 1 = 2\sqrt[3]{{2x - 1}}
$$
I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by $(x-1)$.. But I can't see the solution, how do I go from here?
|
We have $$x^3+1=2(2x-1)^{1/3}\iff x^3=2(2x-1)^{1/3} -1.$$
Here, setting $y=(2x-1)^{1/3}$ gives us
$$y^3=2x-1 \ \ \text{and}\ \ x^3=2y-1.$$
Hence, we have
$$\begin{align}y^3-x^3=(2x-1)-(2y-1)&\Rightarrow (y-x)(y^2+yx+x^2)=2(x-y)\\&\Rightarrow (y-x)(y^2+yx+x^2+2)=0\\&\Rightarrow (y-x)\{(x+(y/2))^2 + (3/4)y^2+2\}=0\\&\Rightarrow y=x.\end{align}$$
Hence, we have
$$x^3=2x-1\iff (x-1)(x^2+x-1)=0\iff x=1,\frac{-1\pm\sqrt 5}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/929053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Calculus Limits Problem L'Hopital's Rule is not allowed.
Question 1:
$$\lim_{x\to -2} \frac{\sqrt{6+x}-2}{\sqrt{3+x}-1} = \ ?$$
I tried to cross multiply $\frac{\sqrt{6+x}-2}{\sqrt{3+x}-1}$with $\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ and I got $x+2$ on both LHS and RHS thus I conclude
$\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ is equal to $\frac {\sqrt{6+x}-2)}{\sqrt{3+x}-1}$.
Thus when I sub $x = -2$ into $\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ I got $0.5$.
Question 2:
$$\lim_{x\to \pi} \sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = \ ?$$
I said that $\sin\frac{x-\pi}{x+\pi}$ tends to $0$.
For $\sin\frac{x+\pi}{x-\pi}$, I use squeeze theorem and said that it can lies between $1$ and $-1$.
Thus, $\sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = 0$.
|
Answer 1:
Set $x=-2+y$
$$\lim_{x\to -2} \frac{\sqrt{6+x}-2}{\sqrt{3+x}-1} = \lim_{y\to 0} \frac{\sqrt{4+y}-2}{\sqrt{1+y}-1}$$
$$=\lim_{y\to 0} \frac{(y/4)+O(y^2)}{(y/2)+O(y^2)}=\lim_{y\to 0} \frac{y/4}{y/2}=(1/2)$$
Answer 2:
Set $x=\pi+y$
$$\lim_{x\to \pi} \sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = \lim_{y\to 0} \sin\frac{y+2\pi}{y}\sin\frac{y}{y+2\pi} =I $$
$$|I|=\lim_{y\to 0} \left|\sin\frac{y+2\pi}{y}\right|\left|\sin\frac{y}{y+2\pi}\right|\le \lim_{y\to 0} 1\left|\sin\frac{y}{2\pi}\right|=0$$
So $I=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/930498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Proof by induction that $3^n \geq 2n^2 + 3n$ for $n \ge 4$ Problem:
If $n$ is a natural number and $n\geq4$, then $3^n \geq 2n^2 + 3n$. (Prove by Induction.)
Attempt at solution:
1) Given: $n$ is a natural number, $n \geq 4$.
2) Let $P(n)$ be the statement "$3^n \geq 2n^2 + 3n$."
3) $P(4) = 3^4 > 2(4)^2 + 3(4)$, i.e. $81 > 44$, thus the base case of $P(4)$ is true.
4) Assume $P(k) = 3^k \geq 2k^2 + 3k$ is true, where $k$ is a natural number and $k \geq 4$.
5) $P(k + 1) = 3^k(3) = 3^{k+1} > (2k^2 + 3k)(3) = 6k^2 + 9k$.
And this is where I am stuck. I know that I am trying to get $P(k+1)$ into the form $3^{k+1} > 2(k+1)^2 + 3(k + 1)$, and I have tried all sorts of algebraic manipulation, but I still am nowhere close to arriving at the correct form. How should I proceed?
|
Try expanding $2(k+1)^2 + 3(k+1)$ in order to compare it with $6k^2 + 9k$ (I.e., what you need to ensure that $6k^2 + 9k \geq 2(k+1)^2 + 3(k+1))$:
$$\begin{align} 2(k+1)^2 + 3(k+1) & = 2(k^2 + 2k + 1) + 3k + 3 \\ &= 2k^2 + 4k + 2 + 3k + 3 \\ &= 2k^2 + 7k +5\end{align}$$
Now, all that remains (while keeping in mind that $\bf k\geq 4)$ is to show that $$\begin{align} 6k^2 + 9k & \geq 6k^2 + 7k + 5 \\
& \geq 2k^2 + 7k + 5 \\
& = 2(k+1)^2 + 3(k+1)
\end{align}$$
So indeed, with the base case, and the fact that $P(k) \implies P(k+1)$, we have established that $P(n)$ holds for all $n \geq 4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/932348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
}
|
Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ and $x < - \sqrt{6}$
But the solution says that $x > -\sqrt{6}$, what am I not getting?
|
$$|x^2 - 4| < 2 \implies -2 < x^2 - 4 < 2$$
$$\implies 2 < x^2 < 6$$
$$ \implies \sqrt{2} < |x| < \sqrt{6}$$
So
$\sqrt{2} < -x < \sqrt{6} \implies -\sqrt{2} > x > -\sqrt{6}$
or
$ \implies \sqrt{2} < x < \sqrt{6}$ and its done...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/932930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Equation with two unknowns I have this equation to solve and I solved it but don't know if the result is correct.
$\begin{cases}2\pi r_1+2\pi r_2=24 & (1) \\ \pi r_1^2+\pi r_2^2=20 & (2)\end{cases}$
Equation $(1)$ gives $r_1=\frac{12-\pi r_2}{\pi}$
Plug that into $(2)$, we have $\pi\big(\frac{12-\pi r_2}{\pi}\big)^2+\pi r_2^2=20 \Rightarrow 2\pi r_2^2-24r_2+\frac{144}{\pi}-20=0$.
$\frac {2\pi r_2^2-24r_2+\frac{144}{\pi}-20}{2\pi}=0$
So to solve the unknowns I use the reduced quadratic equation
$r_2^2 - 3.819r_2 + 4.11 = 0$
$r = - \frac {- 3.819}{2} {+ \choose -} \sqrt{\frac {- 3.819^2}{2^2}-4.11}$
$r = 1.9 {+ \choose -} \sqrt{-0.463}$
So because I don't have complex numbers in my course I think that this equation does not have an solution.
I'm I correct in my assumption or is my assumption wrong?
Thanks!!
|
Let's consider the system
\begin{cases}
x+y=a\\
x^2+y^2=b
\end{cases}
which is the same as yours with $a=12/\pi$ and $b=20/\pi$. We can rewrite the second equation as
$$
(x+y)^2-2xy=b
$$
so, taking into account the first equation, it becomes
$$
2xy=a^2-b
$$
Now the problem is reduced to finding two numbers of which we know the sum and the product: they are the roots of
$$
z^2-az+\frac{a^2-b}{2}=0
$$
The discriminant is
$$
a^2-4\frac{a^2-b}{2}=a^2-2a^2+2b=2b-a^2
$$
so we have real solutions if and only if
$$
2b-a^2\ge0
$$
In your case
$$
2\frac{20}{\pi}-\frac{144}{\pi^2}=\frac{40\pi-144}{\pi^2}
$$
and $144/40=3.6>\pi$, so the discriminant is negative.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/935485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{1}{4}\int\frac{du}{\sqrt{u}}$$
$$\frac{1}{2}\sqrt{u}+C$$
$$\frac{1}{2}\sqrt{1+2x^2}+C$$
I am confused why I get two different answers.
|
Why not directly? Since
$$\int\frac1{\sqrt x}dx=\int x^{-1/2}dx=\frac{x^{1/2}}{1/2}+C=2\sqrt x+C$$
we get that for any differentiable (and positive) function $\;f\;$:
$$\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$
In our case,
$$f(x)=1+2x^2\;,\;\;f'(x)=4x\implies\int\frac x{\sqrt{1+2x^2}}dx=\frac14\int\frac{(1+2x^2)'}{\sqrt{1+2x^2}}dx=$$
$$\frac14(2)\sqrt{1+2x^2}+C=\frac12\sqrt{1+2x^2}+C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/936324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
}
|
Calculate the binomial sum $ I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i} $ I need any hint with calculating of the sum
$$
I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i}.
$$
Maple give the strange unsimplified result
$$
I_n={\frac {1/12\,i\sqrt {3} \left( - \left( \left( 1+i\sqrt {3} \right)
^{2\,{\it n}+2} \right) ^{2}+16\, \left( -1 \right) ^{2\,{\it n}}
\left( {2}^{2\,{\it n}} \right) ^{2} \right) }{{2}^{2\,{\it n}}
\left( 1+i\sqrt {3} \right) ^{2\,{\it n}+2}}},
$$
Сalculation for small $n$ are as follows $I_1=-1, I_2=0,I_3=1, I_4=-1, \ldots$
and leads to a hypothese:
$
I_n= -1 \text{ for } n=3k+1, =0, \text{ for } n=3k+2,=1, \text{ for } n=3k.
$
But how to prove it?
|
Since $ \dbinom{n}{r} = 0 $ for $ r > n $, we can rewrite the sum as
$$ \text{S} = \sum_{r=0}^{\infty} (-1)^r \dbinom{2n+1-r}{r} $$
From the Binomial Theorem, we see that the sum is the coefficient of $x^n$ in
$$f(x) = x^n (1-x)^{2n+1} + x^{n-1} (1-x)^{2n} + \ldots $$
$$ = \dfrac{(1-x)^{n+1}}{x^2-x+1} $$
Note that,
$$ \sum_{n=0}^{\infty}{{U}_{r}(a) {x}^{r}} = \dfrac{1}{x^2 -2ax+1} $$
where $ U_{r} (x) $ is the Chebyshev Polynomial of the second kind.
Putting $a = \dfrac{1}{2}$, we see that,
$$ f(x) = (1-x)^{n+1} \sum_{n=0}^{\infty}{{U}_{r} \left( \dfrac{1}{2} \right) {x}^{r}} \quad (*) $$
To calculate the coefficient of $x^n$ in $ (*) $, we see that coefficient of $x^{n-r}$ for fixed $r$ in $(1-x)^{n+1}$ is $ (-1)^{n-r} \dbinom{n+1}{n-r} $, so coefficient of $x^n$ is,
$$ \sum_{r=0}^{n} (-1)^{n-r} \dbinom{n+1}{n-r} U_{r} \left(\dfrac{1}{2}\right) $$
Substituting $ r \mapsto r-1 $, we have,
$$ \text{S} = \sum_{r=1}^{n+1} (-1)^{n+1-r} \dbinom{n+1}{n+1-r} U_{r-1} \left(\dfrac{1}{2}\right) $$
$$ = (-1)^{n+1} \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} U_{r-1} \left(\dfrac{1}{2}\right) \quad \quad \left( \because \dbinom{n}{n-r} = \dbinom{n}{r} \right) $$
$$ = (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because U_{r-1} \left(\dfrac{1}{2}\right) = \sin \left( \dfrac{r \pi}{3} \right) \right) $$
$$ (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because \sin 0 = 0 \right) $$
Now, $$\displaystyle \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) = \Im \left( \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} e^{\frac{i r \pi}{3}} \right) = (-1)^{n+1} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) $$
$$ \implies \text{S} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \quad \square $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/936864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
}
|
Does the series converge or diverge? I want to check, whether $$\sum\limits_{n=0}^{\infty }{\frac{n!}{(a+1)(a+2)...(a+n)}}$$
converges or diverges.
$a$ is a constant number
Ratio test
$$\begin{align}
& \frac{a_{n}}{a_{n-1}}=\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}\cdot \frac{(a+1)(a+2)...(a+(n-1))}{(n-1)!}=\frac{n}{a+n} \\
& \underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{a_{n-1}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{a+n}=1 \\
\end{align}$$
We know nothing
Root test
$$\sqrt[n]{a_{n}}=\sqrt[n]{\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}}=$$ I can't :(
I can only resolve the serie if "a" is a integer number
$$\begin{align}
& \text{if }a\in \mathbb{Z} \\
& (a+1)(a+2)...(a+(n-1))(a+n)=(n+a)(a+(n-1))...(a+2)(a+1)= \\
& (n+a)(n+a-1)(n+a-2)...(a+2)(a+1)=\frac{(n+a)!}{a!} \\
& \Rightarrow a_{n}=\frac{a!n!}{(n+a)!} \\
& \text{if }a<0\Rightarrow \text{(}n+a)<n,a\in \mathbb{Z} \\
& \underset{n\to \infty }{\mathop{\lim }}\,a_{n}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a!n!}{(n+a)!}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a!n(n-1)...(n+a)!}{(n+a)!}= \\
& \underset{n\to \infty }{\mathop{\lim }}\,a!n(n-1)...(n+a-1)=\infty \\
& \Rightarrow \sum\limits_{n=0}^{\infty }{a_{n}=\infty } \\
& \text{if }a>1,a\in \mathbb{Z} \\
& a_{n}=\frac{a!n!}{(n+a)!}=\frac{a!n!}{(n+a)(n-1+a)...\underbrace{(n-a+a)!}_{n!}}=\frac{a!}{(n+a)(n-1+a)...(n+1)} \\
& \Rightarrow \\
& \frac{a!}{\underbrace{(n+a)(n-1+a)...(n+1)}_{a\text{ times}}}<\frac{a!}{(n+1)^{a}}\Rightarrow \\
& \sum\limits_{n=0}^{\infty }{\frac{a!}{(n+a)(n-1+a)...(n+1)}<\sum\limits_{n=0}^{\infty }{\underbrace{\frac{a!}{(n+1)^{a}}<\infty }_{a>1}}} \\
& \text{if }a=1 \\
& \sum\limits_{n=0}^{\infty }{\frac{a!n!}{(n+a)!}}=\sum\limits_{n=0}^{\infty }{\frac{a!n!}{(n+1)!}}=\sum\limits_{n=0}^{\infty }{\frac{a!}{n+1}}=\infty \\
\end{align}$$
But how i can resolve if "a" is not a integer number?
|
I don't know much about Gamma Function and summation under the integral so here is a more elementary proof.
*
*If $a>0$ we have $\ln a_n=-(\ln (1+\frac{a}{1})+\ln (1+\frac{a}{2})+\dots+\ln (1+\frac{a}{n}))$ and using
$$x\ge\ln(1+x)\ge x-\frac{x^2}{2}$$
(for $x>0$ proven by using monotonic function) we have
$$-a\sum_{i=1}^n\frac{1}{n}\le\ln a_n\le-a\sum_{i=1}^n\frac{1}{n}+\frac{a^2}{2}\sum_{i=1}^n\frac{1}{n^2}\le-a\sum_{i=1}^n\frac{1}{n}+C$$
(where $C=\frac{a^2}{2}\sum_{i=1}^\infty\frac{1}{n^2}$) Then using the fact that:
$$\ln(n+1)=\int_1^{n+1}\frac{1}{x}dx\leq\sum_{i=1}^n\frac{1}{n}\leq1+\int_1^{n}\frac{1}{x} dx=1+\ln(n)$$
we have
$$e^{-a(1+\ln n)}\le a_n\le e^{C-a(\ln(n+1))}$$
or
$$e^{-a}n^{-a}\le a_n\le e^C(n+1)^{-a}$$
Since $a_n$ is positive, the series converges if and only if $a>1$
*
*If $a\le 0$ then from $n\ge |a|$ we have $|a_n|$ increases and doesn't change sign so $\lim_{n\rightarrow\infty}\sup |a_n|\ne 0$ then the sum diverges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/938791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal. Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal.
Part A:
$$T(x,y,z)=\begin{pmatrix} 1 & 1 & 0 \\ 1 & -1 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z\\ \end{pmatrix}$$
\begin{equation*}
\begin{split}
det(T(x,y,z)-I_n \lambda ) & =\begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix} \\
& =(1-\lambda)\begin{vmatrix} -1-\lambda & 0 \\ 0 & 1-\lambda \\ \end{vmatrix}-\begin{vmatrix} 1 & 0 \\ 1 & 1-\lambda \\ \end{vmatrix}=(1-\lambda)^2(-1-\lambda)-(1-\lambda) \\
& =\begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix} \\
& =-(1-\lambda)[(1-\lambda)(1+\lambda)+1] \\
& =-(1-\lambda)[1-\lambda^2+1] \\
& =(\lambda-1)[2-\lambda^2]=0 \\
\end{split}
\end{equation*}
Hence $\lambda=1, \pm \sqrt{2}$.
When $\lambda=1$:
$\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,1,0)^T, (0,0,1)^T \}$.
When $\lambda=\sqrt{2}$:
$\begin{pmatrix} 1-\sqrt{2} & 1 & 0 \\ 1 & -1-\sqrt{2} & 0 \\ 1 & 0 & 1-\sqrt{2} \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 1-\sqrt{2} \\ 0 & 1-\sqrt{2} & 1-\sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,\sqrt{2}-1,0)^T, (1-\sqrt{2},1+\sqrt{2},0)^T \}$.
When $\lambda=-\sqrt{2}$:
$\begin{pmatrix} 1+\sqrt{2} & 1 & 0 \\ 1 & -1+\sqrt{2} & 0 \\ 1 & 0 & 1+\sqrt{2} \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 1+\sqrt{2} \\ 0 & -1+\sqrt{2} & 1+\sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,1-\sqrt{2},0)^T, (1+\sqrt{2},1-\sqrt{2},0)^T \}$.
Are my eigenvectors correct? Is there a better way to find see if the two vectors are orthogonal other than using the cross-product method?
|
Transformation T(x,y,z) is defined as in the question posted $$
T(x,y,z) =
\left[ {\begin{array}{cc}
1 & 1 & 0\\
1 &-1 & 0\\
1 & 0 & 1\\
\end{array} } \right]\left[ {\begin{array}{cc}
v_1\\
v_2\\
v_3\\
\end{array} } \right]
$$
which is of the form $T(x,y,z)=Ax.$ Such that $A=\left[ {\begin{array}{cc}
1 & 1 & 0\\
1 &-1 & 0\\
1 & 0 & 1\\
\end{array} } \right]$ and the vector $x =\left[ {\begin{array}{cc}
v_1\\
v_2\\
v_3\\
\end{array} } \right] $ respectively. For a matrix A of 3*3 order,there are three Eigen values that can be determined from it. For each Eigen value, a vector is determined which is named as Eigen vector, hence three vectors are determined from three Eigen values. Meaning of Eigen values and Eigen vectors can be well-described as below: As mentioned above, Eigen values of A can be determined by the following the equation $$ Ax = \lambda I_{n} x$$ where $I_n$ is the unit matrix with n as the order of square matrix involved. Here, $n =3.$ From this equation, It can be understood that the Matrix $A$ operates on a vector $x$ so as to result in a vector which is parallel to the original vector $x.$ And $\lambda$ is any scalar that operates on the vector $x.$ It can be rephrased as: what are vectors (here, say three values for vectors $x$) for which Matrix $A$ operates on each of them to give a vector parallel to it. Upon simplifying the above equation, we get the equation (i), $$(A-\lambda I_3)x=0$$ $$ det\left|A-\lambda x\right|=0$$ $$\left| {\begin{array}{cc}
1-\lambda & 1 & 0\\
1 &-1-\lambda & 0\\
1 & 0 & 1-\lambda\\
\end{array} } \right| =0 $$
solving this equation results in third order polynomial of the following form (calculations procedure shown in the question is sufficient),$$(\lambda -1)(2-\lambda^{2})=0.$$ obviously, $\lambda=1,\sqrt{2},-\sqrt{2}.$ Case 1: $\lambda=1$$$\left| {\begin{array}{cc}
0 & 1 & 0\\
1 &-2 & 0\\
1 & 0 & 0\\
\end{array} } \right|\left[ {\begin{array}{cc}
v_1\\
v_2\\
v_3\\
\end{array} } \right] =\left[ {\begin{array}{cc}
0\\
0\\
0\\
\end{array} } \right]$$ On solving this equation, we get vector $x$ is of the form $\left[ {\begin{array}{cc}
0\\
0\\
v_3\\
\end{array} } \right] $ such that $v_3$ is independent of both $v_1$ and $v_2$
.By choosing $v_3=1$, results in Eigen vector $(\epsilon_1)$ $\left[ {\begin{array}{cc}
0\\
0\\
1\\
\end{array} } \right]$ corresponding to Eigen value $\lambda_1=1.$ By substituting other two Eigen values, corresponding Eigen vectors are obtained. Case 2: $\lambda=\sqrt{2}$ Eigen vector, $\epsilon_2$ is $\left[ {\begin{array}{cc}
1\\
\sqrt{2}-1\\
\sqrt{2}+1\\
\end{array} } \right].$
Case 3: $\lambda=-\sqrt{2}$ Eigen vector, $\epsilon_3$ is $\left[ {\begin{array}{cc}
1\\
-\sqrt{2}-1\\
1-\sqrt{2}\\
\end{array} } \right].$ There are three vectors $\epsilon_1$,$\epsilon_2$ and $\epsilon_3$ obtained as Eigen vectors corresponding to Eigen values $\lambda=1,\sqrt{2} and -\sqrt{2}.$ To verify, these eigen vectors are orthogonal, the necessary condition for orthogonality, based on the notation used here is,$$\epsilon_i .\epsilon_j =0$$ here dot denotes dot product or inner product and further i and j can take values from 1 to 3, such that $i\neq j. $ i.e, $\epsilon_1.\epsilon_2 =0$ shows $\epsilon_1$ is orthogonal to $\epsilon_2$ and $\epsilon_1.\epsilon_3 = 0$ shows $\epsilon_1$ is orthogonal to $\epsilon_3.$ and $\epsilon_2.\epsilon_3 = 0$ shows $\epsilon_2$ is orthogonal to $\epsilon_3.$ If $$\epsilon_1.\epsilon_2 =\epsilon_1.\epsilon_3 =\epsilon_2.\epsilon_3 = 0.$$ then, $\epsilon_1$,$\epsilon_2$ and$\epsilon_3$ are mutually orthogonal to each other. From the vectors obtained, it is found that they are not orthogonal to each other.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/939393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Square in Interval of Primes Denote by $a_n$ the sum of the first $n$ primes. Prove that there is a perfect square between $a_n$ and $a_{n+1}$, inclusive, for all $n$.
The first few sums of primes are $2$, $5$, $10$, $17$, $28$, $41$, $58$, $75$. It seems there is a perfect square between each pair of successive sums. In addition, we can put a bound on $a_n$, namely $a_n \le 2+3+5+7+9+11+...+(2n+1)=n^2+2n+5$.
|
Let $p$ be the $(n+1)$st prime. Then $a_n\le (-8)+(1+3+5+7+9+\ldots +p-2)=(\frac{p-1}{2})^2-8$ provided $p\ge 11$ (the smaller cases can be dealt with by checking manually). So with $b_n:=\lfloor \sqrt {a_n}\rfloor $ we have $b_n< \frac{p-1}{2}$. Then $a_n<(b_n+1)^2=b_n^2+2b_n+1< a_n+p=a_{n+1}$ as was to be shown.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/939747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
What is the limit of this specific function? Please evaluate the following limit for me:
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$
I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the denominator so it always stay in the indeterminate form.
|
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} = \lim_{x \to -1} \frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)} \\ \\ = \lim_{x \to -1} \frac{x^2+8-9}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x^2-1}{(x+1)(\sqrt{x^2+8}+3)}=\\ \\ \lim_{x \to -1} \frac{(x-1)(x+1)}{(x+1)(\sqrt{x^2+8}+3)}=\lim_{x \to -1} \frac{x-1}{\sqrt{x^2+8}+3}=\frac{-2}{6}=-\frac{1}{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/940354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
}
|
Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
|
Notice that
$$(a+b+c+d)^2=\underbrace{a^2+b^2+c^2+d^2}_{\text{the sum of square of all terms}}+\underbrace{2ab+2ac+2ad+2bc+2bd+2cd}_{\text{the sum of the double products of the terms taken 2 by 2 }}$$
so we find
$$\left(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 ..\right)^2=\underbrace{ 1^2+(2\times 1\times\frac12 x)}_{=1+x}+\underbrace{(\frac12 x)^2-2\times \frac18x^2}_{=0}\\\underbrace{-2\times \frac12x\frac18x^2+2\times\frac1{16}x^3}_{=0}+\cdots$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/940758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
}
|
Proving the general formula [nx] where [.] is the floor function. I've been trying to solve a exercise that asks me to prove the following generalization for the floor function:
$$\lfloor nx\rfloor = \sum_{k=0}^{n-1} {\lfloor x + \frac kn \rfloor}$$
I've already proven the special cases where $n = 1,2,3$. I proved it as follows:
$Case \;(i) \;x \in Z: $
$ If\;x \in Z,\; 2x \in Z.\;$
$$\lfloor2x\rfloor = 2x = x + x = \lfloor x\rfloor+\lfloor x + \frac 12 \rfloor$$
$Case \;(ii)\;x \in R - Z:$
$Let \;y = \lfloor x\rfloor.\;$
$$y \lt x \lt y + 1 \\\Rightarrow 2y \lt 2x \lt 2y + 2 \\\Rightarrow 0 \lt 2x - 2y \lt 2$$
$Subcase \;(a)\; 0 \lt 2x - 2y \lt 1 \Rightarrow 0 \lt x - y \lt \frac 12:
Since\; x - y < 0.5,\; \lfloor x\rfloor = \lfloor x +\frac12\rfloor\;given\;that\;y=\lfloor x\rfloor.$
$$0 \lt 2x - 2\lfloor x\rfloor\lt 1\\\Rightarrow 2\lfloor x\rfloor \lt 2x \lt 2\lfloor x\rfloor+1\\\Rightarrow\lfloor 2x\rfloor = 2\lfloor x\rfloor = \lfloor x\rfloor+\lfloor x\rfloor = \lfloor x\rfloor + \lfloor x + \frac 12\rfloor$$
$Subcase \;(b) \;1 \lt 2x - 2y \lt 2 \Rightarrow \frac 12 \lt x - y \lt 1 \Rightarrow \lfloor x\rfloor+1 = \lfloor x +\frac12\rfloor\;given\;that\;y=\lfloor x\rfloor.$
$$1 \lt 2x - 2\lfloor x\rfloor\lt 2\\\Rightarrow 2\lfloor x\rfloor +1\lt 2x \lt 2\lfloor x\rfloor+2\\\Rightarrow\lfloor 2x\rfloor = 2\lfloor x\rfloor +1= \lfloor x\rfloor+\lfloor x\rfloor +1= \lfloor x\rfloor + \lfloor x + \frac 12\rfloor$$
$QED:\;\lfloor 2x \rfloor = \lfloor x\rfloor + \lfloor x + \frac 12\rfloor$
I've proven for n = 3 similarly. Can somebody suggest a way to generalise it for all n, where n is a positive integer? This is not a homework question, but I would like if you only guide me in the right direction, instead just telling me the answer.
|
Here is a classical neat solution:
Let
$$f(x)= \lfloor nx\rfloor - \sum_{k=0}^{n-1} \lfloor x + \frac kn \rfloor$$
Then the following are immediate:
*
*$f(x+\frac{1}{n})=f(x)$.
*$f(x)=0$ for all $x \in [0, \frac{1}{n})$.
Moreover, any function satisfying these two must be identically 0.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/941098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following
$u = \sqrt{x^2+4 }$ , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$ , $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck here
$\int \dfrac{1}{4 x^4} \dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck here please help
|
Use Substitution instead. Let $u^2=x^2+4$. Then $u\,du=x\,dx$ and we end up integrating $(u^2-4)(u^2)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/942385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$
I rewrote the function to the form
$$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
and figured that the answer would be $0$, but apparently this is wrong.
The correct answer is $\displaystyle{{1 \over 2}\left(\,a - b\,\right)}$.
|
You correctly started with $$x^2\Big(\sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} - \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}}\Big)$$ Now consider that, for small values of $\epsilon $, $$\sqrt{1+\epsilon}=1+\frac{\epsilon }{2}-\frac{\epsilon ^2}{8}+\frac{\epsilon ^3}{16}+O\left(\epsilon
^4\right)$$ and replace first $\epsilon$ by $\dfrac{a}{x^2} +\dfrac{1}{x^4}$ and then by $\dfrac{b}{x^2} +\dfrac{1}{x^4}$. You then arrive to $$x^2\Big(\sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} - \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}}\Big)=\frac{a-b}{2}+\frac{b^2-a^2}{8 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/942514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Finding the cube root of a complex number $z$ $\text{Let }z = -2-2i \text{ where }i \text{ is imaginary. Find in Modulus-Argument form the cube roots of }z$
So far I've done this
$$r = \sqrt8 = 2 \sqrt2 \\
\alpha = \frac{-\pi + \frac{\pi}{4}}{3} = \frac{-\pi}{4}
$$
Which then leads me to believe that the first answer in Mod-Arg form is
$$
z_0 = 2\sqrt2 (\cos\frac{-\pi}{4} + i\sin\frac{-\pi}{4})
$$
Then I increases the value of K by $1$ (Using De Moivre's Theorem for finding nth roots)
$$
r = 2\sqrt2 \\
\alpha = \frac{(-\pi + \frac{\pi}{4})+2\pi(1)}{3} = \frac{5\pi}{12} \\
\text{Thus } z_1 = 2\sqrt2(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12})
$$
and finally, where $K = n-1 =2$
$$
r = 2\sqrt2 \\
\alpha = \frac{(-\pi + \frac{\pi}{4})+2\pi(2)}{3} = \frac{13\pi}{12} \\
$$
But $\frac{13\pi}{12} > \pi$, so $\frac{13\pi}{12} -2\pi = \frac{-11\pi}{12}$, so
$$
z_2 = 2\sqrt2(\cos\frac{-11\pi}{12} + i\sin\frac{-11\pi}{12})
$$
The actual answers have exactly the same arguments, except the modulus is only $\sqrt2$, why so? Did I miss something in my working out? Thanks in advance.
|
Seems like you forgot to take the cube root of $2\sqrt{2}$, after which, you get the answer.
If you look take a complex number $re^{i\theta}$ to the $n$th power, you get $r^n e^{in\theta}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/943967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Question about $\lim_{x \to -\infty}\frac{\sqrt{10+11x^2}}{12+13x}$
$\lim_{x \to -\infty}\dfrac{\sqrt{10+11x^2}}{12+13x}$
= multiply top and bottom by $\dfrac{1}{x}=-\dfrac{1}{\sqrt{x^2}}$
My question is, why is the negative sign in front so crucial, I don't get it:
$\lim_{x \to -\infty}-\dfrac{\sqrt{10/x^2+11x^2/x^2}}{12/x+13x/x}=-\dfrac{\sqrt{11}}{13}$
Why wouldn't the answer be $\dfrac{\sqrt{11}}{13},$ the limit as $x$ goes to positive infinity, $x \rightarrow \infty$?
Thank you.
|
Maybe this is easier to visualize:
\begin{align*}
\lim_{x \to -\infty} \frac{\sqrt{10+11x^2}}{12+13x} & = \lim_{x \to -\infty} \frac{\sqrt{x^2\big(\frac{10}{x^2}+11\big)}}{x\big(\frac{12}{x}+13\big)} \\
& = \lim_{x \to -\infty} \frac{\sqrt{x^2} \sqrt{\frac{10}{x^2}+11}}{x\big(\frac{12}{x}+13\big)} \\
& = \lim_{x \to -\infty} \frac{|x| \sqrt{\frac{10}{x^2}+11}}{x\big(\frac{12}{x}+13\big)} \\
& = \lim_{x \to -\infty} -\frac{\sqrt{\frac{10}{x^2}+11}}{\frac{12}{x}+13} \quad\text{because $\frac{|x|}{x}=-1$ when $x < 0$}\\ & =-\frac{\sqrt{11}}{13}
\end{align*}
But intuitively, if the limit exists, then it must be negative because the numerator is always positive while the denominator is negative whenever $x < -1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/945239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find $\cos(x+y)$ and $\sin(x+y)$ given that $\cos x + \cos y = a$ and $\sin x + \sin y = b$ If $\cos x + \cos y = a$ and $\sin x + \sin y = b$.
Find $\cos(x+y)$ and $\sin(x+y)$.
I only need some hints to start as I am not able to get any way to go forward to.
|
Using Prosthaphaeresis Formulas
$$2\sin\frac{x+y}2\cos\frac{x-y}2=a$$
and $$2\cos\frac{x+y}2\cos\frac{x-y}2=b$$
Divide to find $\tan\dfrac{x+y}2$ assuming $ab\cos\dfrac{x-y}2\ne0$
Now apply Weierstrass substitution
Alternatively find $a^2+b^2,a^2-b^2,ab$ and use Prosthaphaeresis Formulas
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/948150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Missing root of equation $\tan(2x)=2\sin(x)$ I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong?
$$
\tan(2x)=2\sin x
$$
$$
\frac{(\sin 2x)}{(\cos 2x)}=2\sin x
$$
$$
\frac{(2\sin x \cos x)}{((\cos x)^2-(\sin x)^2)}=2\sin x
$$
$$
2\sin x\cos x=2\sin x\cos^2 x-2\sin^3 x
$$
$$
\sin x\cos x=\sin x(\cos^2 x-\sin^2x)
$$
$$
\cos x=\cos^2 x-\sin^2 x
$$
$$
\cos x=\cos^2 x+\cos^2x-1
$$
$$
2\cos^2x-\cos x-1=0
$$
Solving this quadratic equation gives $\cos x=1$ or $\cos x=-\frac 12$
$$
\cos x=1=\cos \theta
$$
$$
x=n2\pi
$$
$$
\cos x=-\frac 12=\cos(\frac{2\pi}3)
$$
$$
x=\pm\frac{2\pi}3+n2\pi
$$
|
Classic mistake occurs at the line where you go to cancel a $\sin x$ from each side...that only holds if $\sin x$ isn't 0. You also get answers whenever $\sin x=0$, which occurs at increments of $n\pi$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/949519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Integrate by partial fraction decomposition $$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx$$
Here's what I have so far...
$$\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)} = \frac{\mathrm A}{x+1}+\frac{\mathrm Bx+\mathrm C}{x^2+2x+5}\\$$
$$5x^2 + 9x + 16 = \mathrm A(x^2+2x+5) + (\mathrm Bx+\mathrm C)(x+1)=\\$$
$$\mathrm A(x^2+2x+5) + \mathrm B(x^2+x)+\mathrm C(x+1)=\\$$
$$(\mathrm A+\mathrm B)x^2 + (2\mathrm A + \mathrm B + \mathrm C)x + (5\mathrm A+\mathrm C)\\$$
$$\mathrm A=-3,\;\mathrm B=8,\;\mathrm C = 31$$
$$$$
$$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx = \int\bigg(-\frac{3}{x+1}+\frac{8x+31}{x^2+2x+5}\bigg)dx\Rightarrow$$
$$\int-\frac{3}{x+1}dx +\int\frac{8}{x^2+2x+5}dx+\int\frac{31}{x^2+2x+5}dx $$
Hopefully I've got it correct until this point (if not, someone point it out please!). I can do the first integration by moving the -3 out and using $u=x+1$ to get
$$-3 \ln(x+1)$$
but I'm stuck on the next two.
|
PS:
Thomas Andrews says the values of $A,B,C$ are in error. I haven't checked those, but the technique outlined below still works if different numbers are involved.
end of PS
$$
\int\frac{8x+31}{x^2+2x+5}\,dx
$$
First let $w=x^2+2x+5$ so that $dw=(2x+2)\,dx$. Then we have
$$
\int\frac{8x+31}{x^2+2x+5}\,dx = 4\int\frac{2x+2}{x^2+2x+5}\,dx+ \int\frac{23\,dx}{x^2+2x+5}
$$
Use the substitution to handle the first integral. Then
$$
\overbrace{\int\frac{23\,dx}{x^2+2x+5} = \int\frac{23\,dx}{(x+1)^2+2^2}}^{\text{completing the square}} = \frac{23}2\int\frac{dx/2}{\left(\frac{x+1}{2}\right)^2+1} = \frac{23}2 \int\frac{du}{u^2+1}
$$
and get an arctangent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/950462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Inverting a Characteristic Function for half-cubic Student's T entailing a Modified Bessel of 2nd kind The Characteristic function of the Student's T with $\alpha$ degrees of freedom,
$C(t)=\frac{2^{1-\frac{\alpha }{2}} \alpha ^{\alpha /4} \left| t\right| ^{\alpha /2}
K_{\frac{\alpha }{2}}\left(\sqrt{\alpha } \left| t\right| \right)}{\Gamma
\left(\frac{\alpha }{2}\right)}$ entails a modified Bessel function of the second kind
$K_{\alpha/2}\left(\sqrt{\alpha } \left| t\right| \right)$. To invert the Fourier to get the probability density of the $n$-summed variable when $\alpha$ is not an integer poses problem as the equation below seems integrable otherwise. Of particular interest is the distribution for $\alpha= 3/2$ ("halfcubic"). With $n$ an integer ( $n >2$):
$$f_n(x)= \left(\frac{3^{3/8}}{\sqrt[8]{2} \,\Gamma \left(\frac{3}{4}\right)}\right)^n \int_{-\infty }^{\infty } e^{-i\, t x} \left| t\right| ^{\frac{3 n}{4}} K_{\frac{3}{4}}\left(\sqrt{\frac{3}{2}} \left| t\right| \right)^n \, dt$$
I tried all manner of expansions and reexpressions of the Bessel into other functions (Hypergeometric, Gamma) to no avail. One good news is that $n=2$ works on Mathematica because the Wolfram library has the square of a Bessel function. It would be great to get the solution for at least $n=3$.
|
Well, according to Mathematica
FourierTransform[
Abs[t]^(9/4)*BesselK[3/4, (\[Sqrt](3/2)*Abs[t])^3], t, w],
which translates to
$$\mathcal{F}_t\left[\left| t\right| ^{9/4} K_{3/4}\left(\left(\sqrt{\frac{3}{2}} \left| t\right| \right)^3\right)\right](w)$$
is equal to
$$\frac{4\ 2^{3/8} \pi \, _1F_4\left(\frac{11}{12};\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{5}{6};-\frac{w^6}{39366}\right)}{9\ 3^{5/8} \Gamma \left(\frac{5}{12}\right)}+\frac{2\ 2^{13/24} \sqrt{\pi } w^4 \Gamma \left(\frac{19}{12}\right) \, _1F_4\left(\frac{19}{12};\frac{7}{6},\frac{4}{3},\frac{3}{2},\frac{5}{3};-\frac{w^6}{39366}\right)}{729\ 3^{5/8} \Gamma \left(\frac{7}{6}\right)}-\frac{w^2 \Gamma \left(\frac{1}{4}\right) \, _1F_4\left(\frac{5}{4};\frac{2}{3},\frac{5}{6},\frac{7}{6},\frac{4}{3};-\frac{w^6}{39366}\right)}{27 \sqrt[8]{2} 3^{5/8}}.$$
The Fourier transform in Mathematica is
$$\dfrac{1}{\sqrt{2 \pi }}\int _{-\infty }^{\infty } d t\, f(t)\, e^{i t \omega }.$$
To find that result by hand is... hard, I guess.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/951177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Partial Fraction Decomposition Problem I am having trouble with this problem.
I need to integrate: $$\frac1{T^4}\times \frac1{K-T}$$ with respect to $T$.
If I do PFD:
$$\frac{A}{T^4} + \frac{B}{T^3} + \frac{C}{T^2} +\frac{D}{T} + \frac{E}{K-T}$$
Does this look right?
From here, I am having trouble solving for the coeffs. Can someone help me figure out the method for solving for these coeffs? If I multiply left side denominator across, I just get $1 =$ "a bunch of stuff"
Any help?
|
\begin{align*} \frac1{T^4(K-T)} &= \frac{A}{T^4} + \frac{B}{T^3} + \frac{C}{T^2} +\frac{D}{T} + \frac{E}{K-T} \\ 1 &= (K-T)A + T(K-T)B + T^2(K-T)C + T^3(K-T)D + T^4E \\ \text{Now equate coefficients} \\ \text{(Unity)} \qquad 1 &= KA \implies A=\tfrac1K \\ (T) \qquad 0 &=-A+KB \implies B = \tfrac1{K^2} \\ (T^2) \qquad 0 &= -B + KC \implies C= \tfrac1{K^3} \\ (T^3) \qquad0 &= -C + KD \implies D = \tfrac1{K^4} \\ (T^4) \qquad0 &= -D + E \implies E = \tfrac1{K^4}\\[12pt] \frac1{T^4(K-T)} &= \frac{1}{KT^4} + \frac{1}{K^2T^3} + \frac{1}{K^3T^2} +\frac{1}{K^4T} + \frac{1}{K^4(K-T)} \\\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/953529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Confusion about a certain series expansion While reading some old notes on contour integration, I noticed the author uses series expansion:
$$\frac{\sinh sx}{\sinh sa}=\frac{sx(1+\frac{1}{6}s^2x^2+\cdots)}{sa(1+\frac{1}{6}s^2a^2+\cdots)}=\frac{x}{a}(1+\frac{1}{6}s^2(x^2-a^2)+\cdots)$$
for small $s$
This may be a silly question, but could someone tell me the justification behind this?
|
The first part is an application of the Taylor series of $\sinh(z)$, which is
$$\sinh(z)=z+\frac{z^3}{3!}+\frac{z^5}{5!}+\frac{z^7}{7!}...$$
Collecting $z$ we get
$$\sinh(z)=z(1+\frac{z^2}{3!}+\frac{z^4}{5!}+\frac{z^6}{7!}...)$$
which corresponds to the first expression provided in the question.
The second expression, where the ratio of the two series $1+\frac{x^2}{3!}+\frac{x^4}{5!}+\frac{x^6}{7!}...$ and $1+\frac{a^2}{3!}+\frac{a^4}{5!}+\frac{a^6}{7!}...$ is substituted by $1$ plus their difference, is explained by a rough approximation. This is based on the fact that, if we have two numbers $1+p$ and $1+q$ with $p<<1$ and $q<<1$, their ratio $1+r$ can be estimated as
$$\frac{(1+p)}{(1+q)}=(1+r)$$
$$(1+p)=(1+r)(1+q) \\ =1+q+r+qr$$
Neglecting the "small" term $qr$, we get $1+p=1+q+r$, and then
$$1+r=1+p-q$$
In the equation shown in the OP, this approximation can be applied if $s$ is sufficiently small.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/955810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Discrete math. Finding a perfect square. The problem is: Find all natural numbers $n$ for which $2^n + 1$ is a perfect square?
I am having a bit of trouble finding a generic way of finding these numbers. Of course the first obvious solution is $n = 3.$ For which we have $8 + 1 = 3^2.$
Anyone has any smart ideas?
|
Method $\#1:$
If $2^n+1=m^2\iff2^n=(m+1)(m-1)$
We can easily test for $n\le2$
For $n>2,$ clearly $m$ is odd
and we have $$2^{n-2}=\frac{m-1}2\cdot\frac{m+1}2$$
But as $\displaystyle\frac{m+1}2-\frac{m-1}2=1,\left(\frac{m-1}2,\frac{m+1}2\right)=1,$ at least one of them is odd
But each divides $2^{n-2},$ the odd must be $\pm1$
If $\displaystyle\frac{m-1}2=1, m=3$ which is a legitimate solution
If $\displaystyle\frac{m+1}2=-1, m=-3$ which is also a valid solution
But, $\displaystyle\frac{m-1}2=-1$ and $\displaystyle\frac{m+1}2=1$ makes $2^n=0$
Method $\#2:$
If $n\ge1,2^n+1$ is odd, for the existence of a square we need $2^n+1=(2a+1)^2\iff2^n=4a(a+1),4$ must divide $2^n\implies n\ge2$
So, we have $2^{n-2}=a(a+1)$ and again, $(a+1,a)=(a+1-a,a)=1$
As exactly, one of $a,a+1$ is odd
Now any odd divisor of $2^{n-2}$ must be $\pm1$
Case $\#1:$ If $a$ is odd
Case $\#2:$ If $a+1$ is odd
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/955907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
}
|
Difference in derivative between $\frac{6}{3x^2+1}$ and $\frac{6}{3x^2}$ Is there a difference in derivative between $\frac{6}{3x^2+1}$ and $\frac{6}{3x^2}$ I thought there wouldn't be and I've asked several people all with different results?
|
Let $$f(x) = \frac{6}{3x^2+1}$$
Then we have
$$\tag{1}f'(x) = \frac{d/dx(6)\cdot(3x^2+1)-6\cdot d/dx(3x^2+1)}{(3x^2+1)^2} = \frac{-6\cdot6x}{(3x^2+1)^2}$$
Since $\frac{d}{dx} ~3x^2+1 = \frac{d}{dx}~ 3x^2$ there is no difference upstairs in $(1)$, but only in the denominator.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/955980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
What am I doing wrong? Finding a limit as $x$ approaches $0$ $$\lim_{x\to 0} {a-\sqrt{a^2-x^2}\over x^2} =$$
$${a-\sqrt{a^2-x^2}\over x^2}\cdot{a+\sqrt{a^2-x^2}\over a+\sqrt{a^2-x^2}} = $$
$$a^{2} - a^{2} - x^{2}\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}} $$
$$-x^{2}\over ax^{2}+x^{2}\sqrt{a^{2}-x^{2}}$$
$$-1\over a+\sqrt{a^{2}-x^{2}}$$
Can anyone tell me where I'm wrong?
|
Third equality $$\left (a - \sqrt{a^2-x^2}\right )\left (a+\sqrt{a^2+x^2}\right ) \color{blue}{=x^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/960015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find and sketch the line $x=1$ under the mapping $f(z)=1/z$ Find and sketch the image of the vertical line $x=1$ under the mapping $f(z)=\frac 1z$
I started by using $u(x,y) + iv(x,y)=f(z)=\frac 1z = \frac 1{x+iy}$ From here I multiplied $f(z)$ by the conjugate to get $\frac x {x^2 + y^2} - \frac {iy} {x^2 + y^2}$ Then I took $u(x,y)=\frac {x} {x^2 + y^2}$ and $v(x,y)=- \frac {iy} {x^2 + y^2}$ Now, plugging in $x=1$ for $U(1,y)$ and $v(1,y)$ to get $u(1,y)=\frac {1} {1 + y^2}$n and $v(1,y)=\frac {-y} {1 + y^2}$ and then Im stuck. Im not sure how to go from here to a graph.
|
Using $u_y=u(1,y)$ and $v_y=v(1,y)$, you obtain the equation
$(u_y-\frac{1}{2})^{2} +v_y^{2} =(\frac{1}{2})^{2}$ for any $y\in\mathbb{R}$.
Conversely you have for any point $z=x+iy$ in the circle with center $(\frac{1}{2},0)$ and radius $\frac{1}{2}$
\begin{equation}
(x-\frac{1}{2})^{2} +y^{2} =(\frac{1}{2})^{2}\\
\Rightarrow x^2-x+y^2=0
\end{equation}
so
$\frac{1}{z}=\frac{x-iy}{x^2+y^2}=1-\frac{iy}{x^2+y^2}$.
Therefore $f(z)=\frac{1}{z}$ is a bijection of the vertical line $x=1$ and the circle $(x-\frac{1}{2})^2 +y^2=(\frac{1}{2})^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/960162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Proof of an identity involving binomial coefficients I have found numerically that the following identity holds:
\begin{equation}
\sum_{n=0}^{\frac{t-x}{2}} n 2^{t-2n-x}\frac{\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}{\binom{2t}{t+x}} = \frac{x^2+t^2-t}{2t-1},
\end{equation}
where $n$, $t$, and $x$ are positive integers ($x \leq t$). To make it more visible, values of $n$ range from $0$ to $\frac{t-x}{2}$.
Any clue about how to prove it?
Thanks, Antonio
|
The right-hand side is not correct, but we can show for non-negative integers $0\leq x\leq t$:
\begin{align*}
\sum_{n=0}^{\frac{t-x}{2}} n 2^{t-2n-x}\frac{\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}{\binom{2t}{t+x}} = \frac{1}{2}\left(\frac{x^2+t^2-t}{2t-1}-x\right)\tag{1}
\end{align*}
It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n\tag{2}
\end{align*}
We start with the left-hand side of (1). Multiplication with $\binom{2t}{t+x}$ gives
\begin{align*}
\color{blue}{\sum_{n=0}^{\frac{t-x}{2}}}&\color{blue}{n2^{t-2n-x}\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}\\
&=2^{t-x}\sum_{n=1}^{\frac{t-x}{2}}n\binom{t}{n}\binom{t-n}{n+x}\frac{1}{4^n}\tag{3}\\
&=t2^{t-x}\sum_{n=1}^{t}\binom{t-1}{n-1}\binom{t-n}{n+x}\frac{1}{4^n}\tag{4}\\
&=t2^{t-x}\sum_{n=0}^{t-1}\binom{t-1}{n}\binom{t-1-n}{n+x+1}\frac{1}{4^{n+1}}\tag{5}\\
&=t2^{t-x}\sum_{n=0}^{t-1}\binom{t-1}{n}[z^{n+x+1}](1+z)^{t-1-n}\frac{1}{4^{n+1}}\tag{6}\\
&=\frac{t}{4}2^{t-x}[z^{x+1}](1+z)^{t-1}\sum_{n=0}^{t-1}\binom{t-1}{n}\left(\frac{1}{4z(1+z)}\right)^n\tag{7}\\
&=\frac{t}{4}2^{t-x}[z^{x+1}](1+z)^{t-1}\left(1+\frac{1}{4z(1+z)}\right)^{t-1}\tag{8}\\
&=\frac{t}{4^t}2^{t-x}[z^{t+x}](1+2z)^{2t-2}\tag{9}\\
&=\frac{t}{4^t}2^{t-x}\binom{2t-2}{t+x}2^{t+x}\tag{10}\\
&=t\binom{2t-2}{t+x}\\
&=t\binom{2t}{t+x}\frac{(t-x)(t-x-1)}{2t(2t-1)}\tag{11}\\
&=\frac{1}{2}\binom{2t}{t+x}\frac{t^2-2tx+x^2-t+x}{2t-1}\\
&\,\,\color{blue}{=\frac{1}{2}\binom{2t}{t+x}\left(\frac{x^2+t^2-t}{2t-1}-x\right)}
\end{align*}
and the claim (1) follows.
Comment:
*
*In (3) we use the binomial identity $\binom{n}{t}=\frac{n}{t}\binom{n-1}{t-1}$. We also set the lower limit to $n=1$ skipping a zero term.
*In (4) we use the binomial identity
\begin{align*}
\binom{t}{n+x}\binom{t-n-x}{t-2n-x}&=\frac{t!}{(n+x)!(t-n-x)!}\cdot\frac{(t-n-x)!}{(t-2n-x)!n!}\\
&=\frac{t!}{n!(t-n)!}\cdot\frac{(t-n)!}{(n+x)!(t-2n-x)!}\\
&=\binom{t}{n}\binom{t-n}{n+x}
\end{align*}. We also set the upper limit to $n=t$ without changing anything, since we are adding zeros only.
*In (5) we shift the index to start with $n=0$.
*In (6) we apply (2) to the right-hand binomial coefficient.
*In (7) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (8) we apply the binomial theorem.
*In (9) we do some simplifications and apply the rule as in (7).
*In (10) we select the coefficient of $z^{t+x}$.
*In (11) we apply the binomial identity $\binom{p}{q}=\frac{p}{p-q}\binom{p-1}{q}=\frac{p(p-1)}{(p-q)(p-q-1)}\binom{p-2}{q}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/964008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Evaluating $ \sum\frac{1}{1+n^2+n^4} $ How to evaluate following expression?
$$ \sum_{n=1}^{\infty}\frac{1}{1+n^2+n^4}$$
I doubt it is a telescopic Sum.
|
Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then:
$$ n^4+n^2+1 = (n^2-\omega)(n^2-\omega^2), $$
so:
$$\frac{1}{1+n^2+n^4}=\frac{1}{i\sqrt{3}}\left(\frac{1}{n^2-\omega}-\frac{1}{n^2-\omega^2}\right)$$
and:
$$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{\sqrt{3}}\Im\sum_{n=1}^{+\infty}\frac{1}{n^2-\omega}$$
can be computed through the identity:
$$\sum_{n=1}^{+\infty}\frac{1}{n^2+a}=\frac{-1+\pi\sqrt{a}\coth(\pi\sqrt{a})}{2a}$$
that follows from considering the logarithmic derivative of the Weierstrass product for the $\sinh$ function. By putting all together, we have:
$$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{6}\left(-3+\pi\sqrt{3}\tanh\frac{\pi\sqrt{3}}{2}\right)$$
as stated by WA.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/966942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
}
|
Finding $ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $ $$ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $$
If I pick $ x = 0$ I get:
$$ \lim_{(x,y) \to (0,0)} \frac{0}{2y^2} = 0$$
So if the limit exists it must be $0$
Now for ${(x,y) \to (0,0)}$ I have $xy \to 0$
So I can use the Taylor series of $sin(t) = t + o(t)$ where $t \to 0$
$$0 \leq \frac{\sin^2(xy)}{3x^2+2y^2} = \frac{x^2y^2 + 2o(x^2y^2) + o (x^2y^2)}{3x^2 + 2y^2} =$$
$$\frac{x^2y^2 + o (x^2y^2)}{3x^2 + 2y^2} = $$
Now I can use the polar coordinates:
$$\frac{\rho^4 \cos^2(\theta)\sin^2(\theta) + o (\rho^4 \cos^2(\theta)\sin^2(\theta))}{3\rho^2 \cos^2(\theta) + 2\rho^2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 \cos^2(\theta) + 2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 - 3 \sin^2(\theta) + 2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 - \sin^2(\theta)}\leq$$
$\frac{\rho^2}{2} \to 0$ for $\rho \to 0$
I would like to know if it is solved in the right way
|
We have $\;3x^2+2y^2\geq 2\sqrt{6}|xy|\;$ by AM-GM, thus we have
$\dfrac{\sin^2(xy)}{2\sqrt{6}|xy|}=\left(\dfrac{\sin(xy)}{xy}\right)^2\dfrac{|xy|}{2\sqrt{6}}\to0\;$ for $\;(x,y)\to(0,0)\;.$
I think your proof is ok. You can read also my proof if you want, it is a little bit shorter.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/967716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Use row reduction to show that the determinant is equal to this variable. Show
determinant of:
\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}
is equal to $(b - a)(c - a)(c - b)$
I'm not sure if you can use squares or square roots hmmm.. please help me. I'm sure it's a simple question. Much appreciated.
|
Subtracting $\DeclareMathOperator{Col}{Col}\Col_1$ from $\Col_2$ and $\Col_3$ gives
$$
\begin{pmatrix}
1 & 0 & 0 \\
a & b-a & c-a \\
a^2 & b^2-a^2 & c^2-a^2 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 \\
a & b-a & c-a \\
a^2 & (b+a)(b-a) & (c+a)(c-a)
\end{pmatrix}
$$
The determinant is then
\begin{align*}
(b-a)(c+a)(c-a)-(c-a)(b+a)(b-a)
&=(b-a)(c-a)(c+a-b-a) \\
&=(b-a)(c-a)(c-b)
\end{align*}
Of course, if you'd rather use row operations then you could consider the transpose and use the fact that $\det(A^\top)=\det(A)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/969304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Proof by induction that $f(n) = 1-2^{2^n}$, where $f(0) = 3$ and $f(n) = 2 f(n-1) - (f(n-1))^2$ I am doing a textbook question which state that a function $f:\mathbb{N}\to\mathbb{Z}$ is a recursively defined as shown bellow
$f(0) =3$,
$f(n) = 2\cdot f(n-1) -(f(n-1))^2 $ if $n\ge1$.
Prove that $f(n) = 1-2^{2^n}$ for all integer $n\ge1$.
I am trying to prove this by induction, so I started by using my base step :
Let $n=1 : f(1) = -3 $
then I did the inductive step for $n = k-1$
which gave me $f(k-1) = 1-2^{2^{k-1}}$ but I don't know how to proceed from here.. can one explain. thanks
|
$$\begin{align}f(k)&=2f(k-1)-(f(k-1))^2\\&=2 \left(1-2^{2^{k-1}}\right)-\left({1-2^{2^{k-1}}}\right)^2\\&=2-2^{2^{k-1}+1}-\left(1-2^{2^{k-1}+1}+2^{2^k}\right)\\&=1-2^{2^{k}}.\end{align}$$
P.S.
$$2\left(1-2^{2^{k-1}}\right)=2-2^1\cdot 2^{2^{k-1}}=2-2^{1+2^{k-1}}=2-2^{2^{k-1}+1}.$$
$$\left(1-2^{2^{{k-1}}}\right)^2=1^2-2\cdot 1\cdot 2^{2^{k-1}}+\left(2^{2^{k-1}}\right)^2=1-2^{2^{k-1}+1}+2^{2^{k-1}\times 2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/969923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
When simplifying $\sin(\arctan(x))$, why is negative $x$ not considered? Let $u = \arctan(x)$, hence $x = \tan(u)$ for $u$ belongs in $(-\frac\pi2, \frac\pi2)$. Since $u$ belongs in $(-\frac\pi2, \frac\pi2)$, we consider $\sin(u)$ where $u$ belongs in $(-\frac\pi2, \frac\pi2)$.
I used the unit circle to determine that the hypotenuse is $\sqrt{x^2 + 1}$ and got an answer $\sin(u) = \frac{x}{\sqrt{x^2 + 1}}$ when I consider than the angle $u$ lies between $(0, \frac\pi2)$.
That's what my textbook says too. However, why don't we also consider when $x$ is negative, and the angle $u$ lies between $(-\frac\pi2, 0)$ ?
|
To add some additional information regarding user46234's comment, we can also understand the omission of a $\pm$ sign as a simple consequence of algebra.
Here are the relevant pieces of information to conclude this:
*
*By definition, $\sqrt{x^2}=|x|$
*$\arctan$ is a strictly increasing function bounded between $(-\frac{\pi}{2},\frac{\pi}{2})$
*$\arctan(0)=0$
*$\sin(x) \geq 0$ if $x \in [0,\frac{\pi}{2})$ and $\sin(x) \lt 0$ if $x \in (-\frac{\pi}{2},0)$
*$\cos(x) \gt 0$ if $x \in (-\frac{\pi}{2},\frac{\pi}{2})$
Working forward from the above bulleted points:
Let $y=\arctan(x)$.We know that $y \in (-\frac{\pi}{2},\frac{\pi}{2})$. By definition, $\tan(y)=\tan\circ \arctan(x)=x \implies \frac{\sin(y)}{\cos(y)}=x$
To further simplify, we use the Pythagorean identity $\cos^2(y)+\sin^2(y)=1 \implies \cos^2(y)=1-\sin^2(y)$. Taking the square root of both sides yields: $|\cos(y)|=\sqrt{1-\sin^2(y)}$. Note that $y \in (-\frac{\pi}{2},\frac{\pi}{2})$, which means that $\cos(y) \gt 0$...and therefore $|\cos(y)|=\cos(y)$, so that we have $\cos(y)=\sqrt{1-\sin^2(y)}$.
We then have that $x=\frac{\sin(y)}{\sqrt{1-\sin^2(y)}}$. Rearranging and then squaring gives us $\big(1-\sin^2(y)\big)\cdot x^2=\sin^2(y)$. Because $x=\tan(y)$, we can use the tangent expression of the Pythagorean identity to further simplify: $1+\frac{\sin^2(y)}{\cos^2(y)}=\frac{1}{\cos^2(y)} \implies1+\tan^2(y)=\frac{1}{1-\sin^2(y)} \implies \frac{1}{1+\tan^2(y)}=1-\sin^2(y) \implies\frac{1}{1+x^2}=1-\sin^2(y)$. Note that in order to apply this substitution, we need to make sure that $\cos(y) \neq 0$...however, given that $y \in (-\frac{\pi}{2},\frac{\pi}{2})$, we know that $\cos(y) \neq 0$. We can then apply the relevant substitution that gives us:
$$\frac{x^2}{1+x^2}=\sin^2(y)$$
Taking the square root of this expression gives us:
$$\frac{|x|}{\sqrt{1+x^2}}=|\sin(y)| \quad (\dagger_1)$$
Now, suppose $x \geq 0$. Then $y \in [0,\frac{\pi}{2})$. We then know that $\sin(y) \geq 0$, so we must have that $|x|=x$ and $|\sin(y)|=\sin(y)$. Under these circumstances, $(\dagger_1)$ is just $\frac{x}{\sqrt{1+x^2}}=\sin(y)$.
Next, suppose $x \lt 0$. Then $y \in (-\frac{\pi}{2},0)$. We then know that $\sin(y) \lt 0$, so we must have that $|x|=-x$ and $|\sin(y)|=-\sin(y)$. Under these circumstances, $(\dagger_1)$ is $\frac{-x}{\sqrt{1+x^2}}=-\sin(y)$, which can of course be rewritten as $\frac{x}{\sqrt{1+x^2}}=\sin(y)$.
Thus, across all relevant cases, we must have that $\sin(\arctan(x))=\sin(y)=\frac{x}{\sqrt{1+x^2}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/970505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
}
|
The common tangents to the circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ form an equilateral triangle Problem :
Show that the common tangents to circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ form an equilateral triangle.
Solution :
Let $C_1 : x^2+y^2+2x=0$
here centre of the circle is $(-1,0) $ and radius 1 unit.
$C_2:x^2+y^2-6x=0$
here centre of the circle is $(3,0) $ and radius 3 units.
But how to proceed to prove that the tangents form equilateral triangle please suggest thanks.
|
More systematic and direct method!
Tangent line to $y=f(x)$ at the point $(x_0,y_0)$ is:
$$y=y_0+y'(x_0)(x-x_0).$$
Let $(x_1,y_1)$ and $(x_2,y_2)$ be the tangent points of the common increasing tangent line to the circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$, respectively.
Equate the slopes and intercepts of the tangent line:
$$\begin{cases}y'(x_1)=y'(x_2)\\ y_1-x_1y'(x_1)=y_2-x_2y'(x_2)\end{cases} \Rightarrow \begin{cases}\frac{-x_1-1}{\sqrt{-x_1^2-2x_1}}=\frac{-x_2+3}{\sqrt{-x_2^2+6x_2}}\\ \frac{-x_1}{\sqrt{-x_1^2-2x_1}}=\frac{3x_2}{\sqrt{-x_2^2+6x_2}}\end{cases}$$
Divide $(1)$ by $(2)$ to find:
$$x_1=\frac{3x_2}{3-4x_2}$$
Substitute it to $(2)$:
$$\frac{x_1^2}{-x_1^2-2x_1}=\frac{9x_2^2}{-x_2^2+6x_2} \Rightarrow \frac{9x_2^2}{15x_2^2-18x_2}=\frac{9x_2^2}{-x_2^2+6x_2} \Rightarrow x_2=1.5 \Rightarrow x_1=-1.5.$$
Hence, the tangent line is:
$$y=\frac1{\sqrt{3}}x+\sqrt{3} \Rightarrow \tan \alpha=\frac1{\sqrt{3}} \Rightarrow \alpha=30^\circ.$$
Note: Similarly, you can find the decreasing tangent line.
Refer to the diagram (for verifying other given answers and finding more methods):
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/972825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
}
|
Using only addition, subtraction and multiplication I have the numbers 6, 30, 8, 8, 3, 7, 1, 2, and 5. Using only addition, subtraction, and multiplication, can you use those numbers to make 60, 54, and 52?
|
$$6+30+8+8+3+7+1+2\color{red}{-}5=60.$$
$$6+30+8+8\color{red}{-}3+7+1+2\color{red}{-}5=54.$$
$$6+30+8+8\color{red}{-}3+7\color{red}{-}1+2\color{red}{-}5=52.$$
P.S.
$$30\times 2=60.$$
$$6\times (8+1)=54.$$
$$(8+5)\times (7-3)=52.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/973161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proving AM-GM for the special case $n=3$ I know the AM-GM inequality and its proof which is relatively complex, though the case for $n=2$ is quite simple. However, I don't know of any special easier proof for the case $n=3$, specifically:
$$\frac{a+b+c}3\ge \sqrt[3]{abc}$$
What is the most elegant proof for this? :)
|
The case for $n=3$ can be proved by using the cases for $n=2,4$.
For $p,q\gt 0$, we have$$(\sqrt p-\sqrt q)^2\ge0\iff \frac{p+q}{2}\ge\sqrt{pq}.$$
So, we have for $s,t,u,v\gt 0,$$$s+t\ge 2\sqrt{st},\ \ \ u+v\ge 2\sqrt{uv}.$$
Hence, we have
$$s+t+u+v\ge 2\sqrt{st}+2\sqrt{uv}\ge 2\sqrt{2\sqrt{st}\cdot 2\sqrt{uv}}=4(stuv)^{1/4}.$$
Here, setting $s=a,t=b,u=c,v=\frac{a+b+c}{3}$ gives us
$$a+b+c+\frac{a+b+c}{3}\ge 4\left(\frac{abc(a+b+c)}{3}\right)^{1/4}\iff \frac{a+b+c}{3}\ge\sqrt[3]{abc}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/973679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
}
|
Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\sqrt{6}-5)$$
Could you show me the steps to get the final result?
Thank you.
|
We have
$$x(5+2\sqrt{6})=289$$
$$\Rightarrow x=\frac{289}{5+2\sqrt{6}}$$
$$\Rightarrow x=\frac{289}{5+2\sqrt{6}}\frac{5-2\sqrt{6}}{5-2\sqrt{6}}$$
$$\Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}$$
$$\Rightarrow x=289(5-2\sqrt{6})$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/977429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.
"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.
|
$$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b-(c-a)}{2(b+c)}=$$
$$=\sum_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(a+c)}\right)=\sum_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/980751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
}
|
Asymptotic approximation of the arctangent? That is, I am looking for an algebraic function $f(x)$ that approximates $\arctan x$ for large values of $x$.
The approximation could be reasonably modest -- perhaps something like
$$\tan (f(x)) = \frac{\pi}{4} + O\left(\frac{1}{x^2}\right).$$
|
Use the following
$$\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} \Leftrightarrow$$
$$\arctan(x) = \frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)$$
The series for $\arctan(x)$ is
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$
Now substitute $\frac1x$ to get the result
$$\arctan(x)=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^3}-\frac{1}{5x^5} +\ldots$$
So for large values of $x$
$$\arctan(x)\sim\frac{\pi}{2}$$
which is logical, because $\displaystyle \lim_{x\to\infty}\arctan(x)=\frac{\pi}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/982838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
Find $\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$ How can we find the integral:
$$\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$$
I tried to find and got it to be $\cfrac{\pi}{\sqrt2}$. Am I correct? Please help me with an appropriate method. I tried to use sum of residue.
|
First, use the change of variable $x=\frac{1}{u} => dx = -\frac{1}{u^2}du $
$I = \int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{u^2} {1 + {u^4}}} \;{\mathrm{d}}u $
=> $2*I = 2*\int_{ 0 }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}} \;{\mathrm{d}}x => I= \int_{ 0 }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}} \;{\mathrm{d}}x $
Now: Let: $t= x-\frac{1}{x}$ => $dt= \frac{1+x^2}{x^2}dx $ x->0, t->$-\infty$ ; x->$+\infty$ , t->$+\infty$
=> $ I = \int_{ - \infty }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}*\frac{x^2}{1+x^2}} \;{\mathrm{d}}t = \int_{ - \infty }^{ + \infty } {\frac{1} {x^{-2} + {x^2}}} \;{\mathrm{d}}t$
$ x^2 + x^{-2} = (x-\frac{1}{x})^2 +2 = t^2 +2 $
=> $ I = \int_{ - \infty }^{ + \infty } {\frac{1} {2 + {t^2}}} \;{\mathrm{d}}t $
Let : t = $\sqrt{2}*v$
$I = \frac{1}{2}*\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {(\frac{t}{\sqrt{2}})^2}}} \;{\mathrm{d}}t = \frac{1}{\sqrt{2}}*\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {v^2}}} \;{\mathrm{d}}v = \frac{\pi}{\sqrt{2}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/985837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
}
|
Continued fraction of the golden ratio It is known, that the continued fraction of $\phi = \frac{1+\sqrt{5}}{2}$ is $[\bar{1}]$. This can be shown via the equation $x^2-x-1=0$:
$$ x^2-x-1=0 \Rightarrow x = 1+\frac{1}{x} = 1+ \frac{1}{1+\frac{1}{x}} = \cdots $$
As far as I can see, the only thing that has been used here is that $\phi$ is a root of the polynomial $x^2-x-1$. My question:
This polynomial has 2 roots. Why do we get the continued fraction of $\phi$ and not those of the other root? What has to be done to get the continued fraction of the other root with this method?
|
The golden ratio
$$
x : 1 = 1 + x : x
$$
leads to the equation
$$
x^2 - x - 1 = 0 \quad (\#)
$$
It can be transformed to two different equations of the form
$$
x = F(x)
$$
which then can be used to substitute the $x$ on the right hand side by $F(x)$
$$
x = F(F(x)) = F(F(F(x))) = \cdots
$$
Your transformed version of $(\#)$ was this equation:
$$
x = 1 + \frac{1}{x} \quad (*)
$$
The repeated substitution of the RHS $x$ in $(*)$ with the RHS term
$$
x \mapsto 1 + \frac{1}{x}
$$
results in the expansion
\begin{align}
x
&= 1 + \frac{1}{x} \quad (\mbox{EX}1.1) \\
&= 1 + \frac{1}{1 + \frac{1}{x}} \quad (\mbox{EX}1.2) \\
&= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{x}}} \quad (\mbox{EX}1.3) \\
&= \cdots
\end{align}
and leads to the continued fraction
$$
x = 1 + \frac{1\rvert}{\lvert 1} + \frac{1\rvert}{\lvert1} + \cdots \quad (\mbox{CF}1)
$$
This continued fraction is positive (consisting only of additions and divisions of positive numbers) and will converge to the positive solution of $(\#)$.
Note that each of the steps $(\mbox{EX}1.n)$ is an equation equivalent
to equation $(\#)$, having two solutions for $x$, while $(\mbox{CF}1)$
converges only to the positive root.
This is because the continued fraction is the limit of these finite fractions:
\begin{align}
c_0 &= 1 \\
c_1 &= 1 + \frac{1}{1} = 2 \\
c_2 &= 1 + \frac{1}{1 + \frac{1}{1}} = 1.5 \\
c_3 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}} = 1.\overline{6} \\
c_4 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}} = 1.6 \\
\end{align}
Note that compared to the expansions $(\mbox{EX}1.n)$ the right hand side $1/x$ is dropped, which makes the difference.
To get the other root, the golden ratio equation $(\#)$
needs to be transformed into
$$
x = \frac{1}{-1 + x} \quad (**)
$$
Repeated substitution of the RHS $x$ in $(**)$ with the RHS term
$$
x \mapsto \frac{1}{-1 + x}
$$
results in the expansion
\begin{align}
x
&= 0 + \frac{1}{-1 + x} \quad (\mbox{EX}2.1) \\
&= 0 + \frac{1}{-1 + \frac{1}{-1 + x}} \quad (\mbox{EX}2.2) \\
&= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1 + x}}} \quad (\mbox{EX}2.3) \\
&= \cdots
\end{align}
leads to the continued fraction
$$
x = 0 + \frac{1\rvert}{\lvert-1} + \frac{1\rvert}{\lvert-1} + \cdots \quad (\mbox{CF}2)
$$
for the negative root of $(\#)$.
Note that all equations $(\mbox{EX}2.m)$ are equivalent to equation $(\#)$ and the equations $(\mbox{EX}1.n)$, thus have two solutions for $x$.
However the derived continued fraction $(\mbox{CF}2)$ is the limit of these finite fractions:
\begin{align}
d_0 &= 0 \\
d_1 &= 0 + \frac{1}{-1} = -1 \\
d_2 &= 0 + \frac{1}{-1 + \frac{1}{-1}} = -0.5 \\
d_3 &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1}}} = -0.\overline{6} \\
d_4 &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1}}}} = -0.6\\
\end{align}
They result from dropping the right hand side $1/(-1+x)$ sub term.
Note that the negative root is nicely approached from above by the even numbered terms and from below by the odd numbered terms.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/985996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find out the primitive polynomial GF(3) 1.) $x^2 + 2x$
2.) $x^2 + 1$
3.) $x^2 + 2$
4.) $x^2 + 2x$
5.) $x^2 + 2x + 1$
6.) $x^2 + 2x + 2$
7.) $x^2 $
8.) $x^2 + x + 2$
9.) $x^2 + x + 1$
Can any one help me in listing out primitive polynomials and tell me why is it a primitive polynomial please.
|
$$x^2 + 1,
x^2 + 2x + 2,
x^2 + x + 2,$$
are primitive polynomials over $gf(3)$.
Because they are irreducible and monic polynomials
Monic means coefficient of highest exponent of $x$ is $1$.
And irreducible means can not be further factorise into small polynomials.
Let's take first case $x^ 2 + 2x$ is further factorise as $x(x + 2)$.
Third case $x^2 + 2$ is monic but it is factorise as $(x-1)(x+1)$.
Because in $gf(3), 2 = -1$ and hence $x^2+2 = x^2 -1 = (x-1)(x+1)$.
And you can also check it by taking $x = 1$.
$1^2 + 2 = 3$ which equal to $0$ ( because $3 \pmod3 = 0$) so $x-1$ is factor of $x^2 + 2$ in $gf(3)$.
Now check other polynomials by yourself and get answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/986920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Consider the lists of length six made with the symbols $P, R, O, F, S$ where repetition is allowed. Consider the lists of length six made with the symbols $P, R, O, F, S$ where
repetition is allowed. (For example, the following is such a list: $(P,R,O,O,F,S)$.)
How many such lists can be made if the list must end in an $S$ and the symbol
$O$ is used more than once?
Here's my attempt at counting the lists:
$S$ must be the last entry of the list, so there is $1$ possible value. $O$ must occur more than once, so I arranged $1$'s in the other five entries to represent possible combinations of $O$ in the list.
$\begin{align}1\cdot1\cdot5\cdot5\cdot5\cdot1 + 1\cdot5\cdot 1 \cdot5\cdot5\cdot1 + 1 \cdot 5 \cdot 5 \cdot 1 \cdot 5 \cdot 1+ 1 \cdot 5 \cdot 5 \cdot 5 \cdot 1 \cdot 1 + 5\cdot1\cdot1\cdot5\cdot5\cdot 1 + 5 \cdot 1 \cdot 5\cdot1\cdot5\cdot1 + 5\cdot1\cdot5\cdot5\cdot1\cdot1+ \cdots + 5 \cdot 5 \cdot 5 \cdot1\cdot1\cdot1 \end{align}$
Also: Is there an easier way to count this?
|
We want to count $5$ letter words with at least $2$ O letters.
*
*The number of arbitrary $5$ letter words is $5^5$.
*The number of $5$ letter words with exactly one "O" is $5 \cdot 4^4$ (fix the location of the O, and the rest can consist of the four remaining letters).
*The number of $5$ letter words without the letter O is $4^5$.
Since (2) and (3) are disjoint, the number of $5$ letter words with at least $2$ O letters is
$$5^5 - 5 \cdot 4^4 - 4^5 = 821$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/988782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How to find the roots of $(\frac{z-1}{z})^5=1$
Write down the fifth roots of unity in the form $\cos \theta + i \sin \theta$ where $ 0 \leq \theta \leq 2\pi$
Hence, or otherwise, find the fifth roots of i in a similar form
By writing the equation $(z-1)^5=z^5$ in the form :
$${\left(\frac{z-1}{z}\right)}^{5}=1$$
show that its roots are:
$$\frac{1}{2}\left(1- i \cot \frac{1}{4} \pi k\right)$$ where $k=0,1,2,3,4$
I know how to do the first 2 parts :
First part:
$$z^5=1= \cos 2\pi + i \sin 2 \pi $$
$$z= \left( \cos \frac{2\pi k}{5} + i \sin \frac{2\pi k}{5}\right)$$
where $k=0,1,2,3,4$
Second Part:
$$z^5=i= \cos \pi + i sin \pi $$
$$z= \left(\cos \frac{\pi k}{5} + i \sin \frac {\pi k}{5}\right)$$
where $k=0,1,2,3,4$
I have no clue how to do the third part, Please help.
How to do the third part?
Please note: I am a beginner to the complex world
|
Solve$$\displaystyle \left(\frac{z-1}{z}\right)^5=1\rightarrow \frac{z-1}{z}=e^{\Large \frac{i\cdot{2k}\pi}{5}}$$ for $k=0,1,2,3,4$.
Hence $$\displaystyle 1-\frac{1}{z}=e^{\Large \frac{i\cdot{2k}\pi}{5}}\rightarrow 1-e^{\Large \frac{i\cdot{2k}\pi}{5}}=\frac{1}{z}\rightarrow z=\frac{1}{1-e^{\Large \frac{i\cdot{2k}\pi}{5}}}=\frac{1}{\left(1-\cos\left(\frac{2k\pi}{5}\right)\right)-i\sin\left(\frac{2k\pi}{5}\right)}$$ Now, $$\frac{\left(1-\cos\left(\frac{2k\pi}{5}\right)\right)+i\sin\left(\frac{2k\pi}{5}\right)}{\left(1-\cos\left(\frac{2k\pi}{5}\right)\right)^2+\sin^2\left(\frac{2k\pi}{5}\right)}=\frac{\left(1-\cos\left(\frac{2k\pi}{5}\right)\right)+i\sin\left(\frac{2k\pi}{5}\right)}{2-2\cos\left(\frac{2k\pi}{5}\right)}=\\=\frac{2\sin^2\left(\frac{k\pi}{5}\right)+i\cdot 2\sin\left(\frac{k\pi}{5}\right)\cos\left(\frac{k\pi}{5}\right)}{4\sin^2\left(\frac{k\pi}{5}\right)}=\frac{\sin\left(\frac{k\pi}{5}\right)+i\cos\left(\frac{k\pi}{5}\right)}{2\sin\left(\frac{k\pi}{5}\right)}=\\=\frac{1}{2}\left(1+i\cdot \cot\left(\frac{k\pi}{5}\right)\right)$$
So the question is incorrect.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/990314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Inequality relating coefficients and roots of a complex polynomial While going through some olympiad handouts I stumbled upon a problem related to an upper bound for the Mahler measure, which stated that
Given a polynomial $f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0 \in \mathbb{C}[x]$ that has roots $z_1, z_2, ... , z_n \in \mathbb{C},$ prove that:
$\prod_{k = 1}^n$max$(1,|z_k|) \leq \sqrt{1+|a_{n-1}|^2 + ... + |a_{0}|^2}.$
I am unsure of how to relate the sum of the squares of the magnitudes of the coefficients to the product of the roots. A hint revolved around using the product $f(x)\bar{f}(\frac{1}{x}),$ where $\bar{f}$is the polynomial with coefficients conjugate to $f,$ but I did not get anywhere. What other approaches are best to prove the statement?
|
Suppose the roots of polynomial $f(z)$ are $z_1,z_2,\cdots,z_n$,
where, $|z_1| \ge |z_2| \ge \cdots \ge |z_m| > 1 \ge |z_{m+1}| \ge \cdots \ge |z_n|$
Let, $g(z) = z^nf\left(\frac{1}{z}\right) = 1 + a_{n-1}z + \dots + a_0z^n$
Then, $\{1/z_k\}_{1\le k \le m}$ are the zeros of $g$ in the disk $|z| \le r = 1-\epsilon < 1$, where, $\epsilon$ is chosen such that $g(re^{i\theta}) \neq 0$ for $\theta \in [0,2\pi]$.
Then, using Jensen's Formula (proof) we have: $\displaystyle \log|r^m z_1\cdots z_m| = \frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,d\theta$
i.e.,$$|r^m z_1\cdots z_m| = \exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,d\theta\right)$$
Applying Jensen Inequality, $\displaystyle \exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,d\theta\right) \le \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,d\theta$
followed by Cauchy-Schwarz Inequality yields,
$\displaystyle \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,d\theta \le \frac{1}{2\pi}\left(\int_0^{2\pi}\,d\theta\int_0^{2\pi} |g(re^{i\theta})|^2\,d\theta\right)^{1/2} = \sqrt{1+|a_{n-1}|^2 + ... + |a_{0}|^2}$
Therefore, $\displaystyle |r^m z_1\cdots z_m| \le \sqrt{1+|a_{n-1}|^2 + ... + |a_{0}|^2}$, and letting $\epsilon \to 0$, $r \to 1$ gives the desired result.
A Second Approach
We can also use induction.
The 2-norm of a polynomial $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_0$ is defined as,
$\lVert f \rVert_2 := \sqrt{|a_n|^2+|a_{n-1}|^2 + ... + |a_{0}|^2}$
and, we have:
Lemma: $\displaystyle \lVert (x-z)f(x) \rVert_2 = \lVert (\overline{z}x-1)f(x) \rVert_2 $
$\displaystyle \begin{align}\lVert (x-z)f(x) \rVert_2^2 &= \sum\limits_{j=0}^{n+1} |a_{j-1} - za_j|^2 \textrm{ ,where, $a_{-1} = a_{n+1} = 0$ in notation} \\ &= \sum\limits_{j=0}^{n+1} (a_{j-1} - za_j)\overline{(a_{j-1} - za_j)} \\ &= (1+|z|^2)\lVert f(x) \rVert_2^2 - \sum\limits_{j=0}^{n+1} (a_{j-1}\overline{za_j} + za_j\overline{a_{j-1}}) \\ &= (1+|z|^2)\lVert f(x) \rVert_2^2 - \sum\limits_{j=0}^{n+1} (z\overline{a_{j-1}}a_{j} + \overline{z}a_{j-1}\overline{a_{j}}) \\ &= \sum\limits_{j=0}^{n+1}(\overline{z}a_{j-1} - a_j)(z\overline{a_{j-1}} - \overline{a_j}) \\ &= \sum\limits_{j=0}^{n+1}|\overline{z}a_{j-1} - a_j|^2 = \lVert (\overline{z}x-1)f(x) \rVert_2^2 \end{align}$
as before let us write the roots of the polynomial in the order, $|z_1| \ge |z_2| \ge \cdots \ge |z_m| > 1 \ge |z_{m+1}| \ge \cdots \ge |z_n|$ .
Consider the polynomial: $\displaystyle g(x) = \prod\limits_{j=1}^m (\overline{z_j}x-1)\prod\limits_{j=m+1}^n (x - z_j) = \sum\limits_{j=0}^{n}b_jx^j$,
where, $|b_n| = |z_1z_2\cdots z_m|$
Now,
$\begin{align}|z_1z_2\cdots z_m| = |b_n| &\le \sqrt{|b_n|^2+\cdots+|b_0|^2} \\ & = \lVert g(x) \rVert_2 \\ &= \lVert \frac{g(x)}{\overline{z_1}x - 1}(x-z_1) \rVert_2 \\ &= \lVert \frac{g(x)}{\prod\limits_{j=1}^m (\overline{z_j}x-1)}\prod\limits_{j=1}^m(x-z_1) \rVert_2 [\textrm{ ,by repeated application of lemma}] \\ &= \lVert f \rVert_2 \end{align}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/990925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
}
|
Prove this equality $$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
This is how I did it:
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4 \;\;\; |^{3}$$
$$2+11i+3\sqrt[3]{(2+11i)^2(2-11i)}+3\sqrt[3]{(2+11i)(2-11i)^2}+2-11i=64$$
$$4+3\sqrt[3]{(2+11i)(2-11i)} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=64$$
$$3\sqrt[3]{4+121} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=60$$
$$15 \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=60$$
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
Is this correct?
|
You can begin with
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=x$$
and find $x$.
Alternatively, let $z=2+11i$. You have to show that $\sqrt[3]z+\sqrt[3]{\bar z}=4$. Since $\sqrt[3]z$ and $\sqrt[3]{\bar z}$ are conjugate, it suffices to show that $\Re\sqrt[3] z=2$.
If you try to solve for $b$
$$(2+ib)^3=2+11i$$
you get
$$8-6b^2=2$$
and
$$12b-b^3=11$$
and both equations are satisfied if $b=1$.
Now, simply compute $(2+i)^3$ to check that $\sqrt[3]{2+11i}=2+i$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/991516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet? Suppose $a$, $b$ and $c$ are three prime numbers.
How to prove that $a^2 + b^2 \neq c^2$?
|
From $a^2+b^2=c^2$ we get $a^2=c^2-b^2=(c+b)(c-b)$, i.e. a factorization of $a^2$ into two distinct factors $c+b>c-b$. The only such factorizations for the square of a prime is $a^2\cdot 1$, i.e. we conclude $c-b=1$, hence $b=2$, $c=3$. But then $a^2=5$, qea.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/991947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 6
}
|
Closed form of $\int_0^{\infty} \frac{\log(x)}{\cosh(x) \sec(x)- \tan(x)} \ dx$ What real analysis tools would you recommend me for getting the closed form of the integral below?
$$\int_0^{\infty} \frac{\log(x)}{\cosh(x) \sec(x)- \tan(x)} \ dx$$
|
The place I would start is the nifty result, proven here, that
$$\frac{\sin{x}}{\cosh{t} - \cos{x}} = 2 \sum_{k=1}^{\infty} e^{-k t} \sin{k x} $$
Of course, the integral actually looks like
$$\int_0^{\infty} dx \frac{\cos{x}}{\cosh{x} - \sin{x}} \log{x} $$
so we need to map $x \mapsto \pi/2 - x$ and we have that the integral is actually
$$2 \sum_{m=0}^{\infty} (-1)^m \int_0^{\infty} dx \, e^{-(2 m+1) x} \cos{(2 m+1) x}\, \log{x} + 2 \sum_{m=1}^{\infty} (-1)^{m+1} \int_0^{\infty} dx \, e^{-2 m x} \sin{2 m x}\, \log{x} $$
We then note that
$$-k \int_0^{\infty} dx \, e^{-(1-i) k x} \log{x} = \frac{\log (k)}{2}+\frac{\gamma }{2}+\frac{\log (2)}{4}+\frac{\pi }{8}+i \left(\frac{\log (k)}{2}+\frac{\gamma }{2}+\frac{\log (2)}{4}-\frac{\pi }{8}\right)$$
So the integral is
$$-\sum_{m=0}^{\infty} (-1)^m \frac{\log{(2 m+1)}}{2 m+1} - \sum_{m=1}^{\infty} (-1)^{m+1} \frac{\log{(2 m)}}{2 m}\\ - \left (\gamma + \frac12 \log{2} + \frac{\pi}{4} \right )\sum_{m=0}^{\infty} \frac{(-1)^m}{2 m+1} - \left (\gamma + \frac12 \log{2} - \frac{\pi}{4} \right )\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{2 m}$$
The third and fourth sums are well known. The second is less-well known, but may be shown to be
$$\frac{1}{4} \left(3 \log ^2(2)-2 \gamma \log (2)\right)$$
(See, for example, Williams & Hardy, The Red Book of Mathematical Problems, problem 71.)
The first sum, however, is less well known. It may be evaluated by considering the derivative of a generalized zeta function (or a Lerch transcendent). The result is
$$\sum_{m=0}^{\infty} (-1)^m \frac{\log{(2 m+1)}}{2 m+1} = \frac12 \log{2} \left [\psi{\left ( \frac{3}{4} \right )} - \psi{\left ( \frac{1}{4} \right )} \right ] - \frac14 \left [ \gamma_1{\left ( \frac{3}{4} \right )}-\gamma_1{\left ( \frac{1}{4} \right )}\right ]$$
where $\gamma_1(a)$ is a generalized Stieltjes constant.
Put this all together and the result agrees with a numerical evaluation performed with Mathematica ($ \approx -1.35775$).
ADDENUDUM
Apparently I forgot that I had encountered that first sum some time ago here, and the result may be expressed in simpler terms:
$$ \sum_{m=1}^{\infty} (-1)^{m} \frac{\log{(2 m+1)}}{2 m+1} = -\frac{\pi}{4} \gamma - \frac{\pi}{4} \log{\frac{\Gamma{\left ( \frac{3}{4} \right )}^4}{\pi}} $$
so we may now put this all together as follows:
$$\frac{\pi}{4} \gamma + \frac{\pi}{4} \log{\frac{\Gamma{\left ( \frac{3}{4} \right )}^4}{\pi}} - \frac{3}{4} \log^2{2} + \frac12 \gamma \log{2} - \left (\gamma + \frac12 \log{2} + \frac{\pi}{4} \right )\frac{\pi}{4} - \left (\gamma + \frac12 \log{2} - \frac{\pi}{4} \right )\frac12 \log{2}$$
or
$$\int_0^{\infty} dx \frac{\cos{x}}{\cosh{x} - \sin{x}} \log{x} = \frac{\pi}{4} \log{\frac{\Gamma{\left ( \frac{3}{4} \right )}^4}{\pi}} - \log^2{2} - \frac{\pi^2}{16}$$
ADDENDUM II
The methodology outlined above may be applied to other, similar integrals. For example, it is a simpler matter to show that
$$\int_0^{\infty} dx \frac{x \sin{x}}{\cosh{x}-\cos{x}} \log{x} = \frac{1}{6} \pi ^2 \left(-12 \log{A}+1+\frac{\log{2}}{2}+\log{\pi} \right) $$
where $A$ is Glaisher's constant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/992140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
}
|
A triangle ABC with vertex $C(4,3)$. The bisector and the median line equation drawn from the same vertex are given. Find the vertices A & B. A triangle $\triangle ABC$ one of his vertex is the point $C(4,3)$.
The bisector line equation is $x+2y-5=0$ and the median line equation is $4x+3y-10=0$ drawn from the same vertex.
Find the coordinates of the vertices A and B.
What i have done:
I started using the fact that the two equations starts on the same vertex, but is not $C$.
So, let $A$ be that vertex, i calculate the intersection point by the system:
$$y_A=-1/2\cdot x_A+5/2$$
$$y_A=-4/3\cdot x_A+10/3$$
Which implies that $A(1,2)$
But i dont know how to find $B$.
|
The segment connecting $A$ in $(1,2)$ with point $C$ in $(4,3)$ lies on the line $y=\frac{x}{3}+\frac{5}{3}$ and has length $\sqrt{3^2+1^2}=\sqrt{10}$.
Note that the bisector crosses the $x$-axis in point $(5,0)$. Let us call $D$ this point. The segment $CD$ lies on the line $y=-3x+15$ and has length $\sqrt{3^2+1^2}=\sqrt{10}$.
Because segments $AC$ and $CD$ have equal length and lie on lines whose slopes are one the negative inverse of the other, we get that $ACD$ is an isosceles right triangle. Thus, the bisector $AD$ forms with $AC$ a $45^o$ degree angle, which means that the angle $CAB$ is right. As a result, the leg $AB$ of the right triangle $ABC$ has slope equal to $CD$, and then lies on the line $y=-3x+5$.
Now let us call $s$ the slope of the segment $BC$ and $E$ its middle point. This segment lies on the line $y=sx+3-4s$, which crosses the median $y=\frac{10-4x}{3}$ in $(\frac{12s+1}{3s+4}, \frac{12-6s}{3s+4})$. So the length of segment $CE$ is
$$\sqrt{(4-\frac{12s+1}{3s+4})^2+(3-\frac{12-6s}{3s+4})^2} \\ = \frac{\sqrt{15^2+(15s)^2}}{3s+4}= \frac{15\sqrt{s^2+1}}{3s+4}$$
On the other hand, the segment $BC$ crosses the leg $AB$ in $(\frac{4s+2}{s+3}, \frac{9-7s}{s+3})$. So the length of segment $BE$ is
$$\sqrt{( \frac{4s+2}{s+3} -\frac{12s+1}{3s+4})^2+( \frac{9-7s}{s+3} -\frac{12-6s}{3s+4})^2 } \\ = \frac{\sqrt{(5-15s)^2+(5s-15s^2)^2}}{(s+3)(3s+4)}= \frac{5(3s-1)\sqrt{s^2+1}}{(s+3)(3s+4)}$$
Since $BE=CE$ we get
$$\frac{5(3s-1)\sqrt{s^2+1}}{(s+3)(3s+4)}=\frac{15\sqrt{s^2+1}}{3s+4})$$
which reduces to
$$\frac{3s-1}{s+3}=3$$
The last equation has no real solutions, except for $s \rightarrow \infty$. This means that the segment $BC$ is vertical, i.e., it lies on a line of the form $x=k$. Since it includes point $C$, we get that segment $BC$ lies on the line $x=4$. Knowing this, it is not difficult to find its intersection with line $AB$, allowing to conclude that point $B$ is in $(4,-7)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/993697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find $C$ such that $x^2 - 47x - C = 0$ has integer roots, and further conditions Have been working on this for years. Need a system which proves that there exists a number $C$ which has certain properties. I will give a specific example, but am looking for a system which could possibly be generalized.
Find (or prove there exists) $C$, such that the quadratic $x^2 - 47x - C = 0$ has integer roots, and furthermore, $C$ must have ALL OF $2$, $3$ and $5$ as its only prime factors (though each of these can be to any positive integer power).
|
Say
$$
x^2 -47x - c = (x-a)(x-b)
$$
for integers $a$ and $b$.
Then
$$
a + b = 47 \\
-ab = C.
$$
So $a$ and $b$ must have only $2,3,5$ in their prime factorization.
Say $b$ is divisible by $5$ (one of them must be). So say $b = 5n$. So then $47 - a$ is divisible by $5$. And $a$ must be $-13 , -8 , -3 , 2, 7, 12, 17, \dots$. So $a = 2 + 5m$. Pick out those that are divisible by only $2$ or $3$.
By trial and error
$$
a = -3, b = 50
$$
would work.
Another solution is
$$
a = 72, b = - 25.
$$
Note, for example that $a=12$ would give $b = 35$, but then $b$ is divisible by $7$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 3
}
|
How to integrate $\int\frac{\ln x\,dx}{x^2+2x+4}$
$$K=\int\frac{\ln x\,dx}{x^2+2x+4}$$
I did this $x^2+2x+4=(x+\alpha)(x+\beta)$, then used partial fraction, I am then unsure how to integrate $\int\frac{\ln x}{x+c}\,dx$.
I tried Integration by parts also taking first function as both of them which ended nowhere
Also I need help how to evaluate this, with or without(which I think might be easier) using the above:
$$I=\int_0^{\infty}\frac{\ln x\,dx}{x^2+2x+4}$$
I did $x\mapsto 1/x$and then added those to get:
$$I=\frac32\int_0^{\infty}\frac{(x^2-1)\ln x \,dx}{(x^2+2x+4)(4x^2+2x+1)}=?$$
|
One way to attack improper integrals of the form
$$\int_0^{\infty} dx \, f(x) $$
is to use the residue theorem, i.e. contour integration in the complex plane. To do this, one considers the contour integral
$$\oint_C dz \, f(z) \log{z} $$
where $C$ is a keyhole contour about the positive real axis, of inner radius $\epsilon$ and outer radius $R$. We consider the limits as $\epsilon \to 0$ and $R \to \infty$. In most cases, the integrals about the inner and outer circles vanish in these limits (I will not prove here).
In this case, our statement of the residue theorem takes the form
$$\int_0^{\infty} dx \frac{\log^2{x} - (\log{x}+i 2 \pi)^2}{x^2+2 x+4} = i 2 \pi \sum_{k=1}^2\frac{\log^2{z_k}}{2 z_k+2}$$
where $z_{1,2}= -1 \pm i \sqrt{3}$, or, in other words, $z_1 = 2 e^{i 2 \pi/3}$ and $z_2=2 e^{i 4 \pi/3}$. Note that the arguments of the $z_k$ are between $[0,2 \pi]$ necessarily because of how $C$ was defined. Thus,
$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2+2 x+4} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2+2 x+4} = i 2 \pi \left [\frac{(\log{2} + i 2 \pi/3)^2}{i 2 \sqrt{3}} + \frac{(\log{2} + i 4 \pi/3)^2}{-i 2 \sqrt{3}} \right ]$$
Equating real and imaginary parts, we find that
$$\int_0^{\infty} dx \frac{1}{x^2+2 x+4} = \frac{\pi}{\sqrt{3}} $$
$$\int_0^{\infty} dx \frac{\log{x}}{x^2+2 x+4} = \frac{\pi}{ 3 \sqrt{3}} \log{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
}
|
Converting Summation to Expression How does the summation break down from
$$\displaystyle\sum_{n \geq 0} (x + x^2) ^ n$$
to
$$\frac1{1 - x - x^2} $$
per this answer?
|
Assuming $$|x+x^2|\lt 1$$
$$s=\displaystyle\sum_{n \geq 0} (x + x^2) ^ n$$
Notice that
$$\quad \\\\\\\ \ \ \quad s=1+(x+x^2)+(x+x^2)^2+(x+x^2)^3+\cdots$$
$$s(x+x^2)=(x+x^2)+(x+x^2)^2+(x+x^2)^3+\cdots$$
$$s-s(x+x^2)=s(1-(x+x^2))=1$$
$$\Rightarrow s=\frac{1}{1-(x+x^2)}=\frac{1}{1-x-x^2}$$
$$\Rightarrow \displaystyle\sum_{n \geq 0} (x + x^2) ^ n=\frac{1}{1-x-x^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find sufficient and necessary conditions in which $f(x)$ is a natural number Let us consider the function $$f(x)=(-√3+2)^{2^{x-2}}+(√3+2)^{2^{x-2}}$$ where $x≥3$.
We know for example that if $x$ is an integer, then $f(x)$ is also an integer.
My question is: Find sufficient and necessary conditions in which $f(x)$ is a natural number.
|
1st claim :
Let $k \in \mathbb{N}$ with $k \geq 2$. Then :
$$ \Big( 2\cosh(x) = k \Big) \, \Leftrightarrow \, x = \ln \Big( \frac{k+\sqrt{k^2-4}}{2} \Big). $$
Let $k \in \mathbb{N}$ with $k \geq 2$. We have :
$$ \begin{align*}
2\cosh(x) = k &\Leftrightarrow {} e^{2x} - ke^{x} + 1 = 0 \\
\end{align*}
$$
Let $u=e^{x} > 0$. We solve $u^2 - k u + 1 =0$ and find that $u=e^{x}=k+\sqrt{k^2-4}$.
2nd claim :
For all $x \in \mathbb{R}$ with $x \geq 3$,
$$ f(x) = 2 \cosh \big( 2^{x-2} \ln(\sqrt{3}+2) \big). $$
Just note that : $\ln(-\sqrt{3}+2) = - \ln(\sqrt{3}+2)$.
3rd claim :
Let $k \in \mathbb{N}$ with $k \geq 2$. Then :
$$f(x) =k \; \Leftrightarrow \; x = \frac{1}{\ln(2)} \Bigg( (\ln \circ \ln) \Big( \frac{k+\sqrt{k^2-4}}{2} \Big) - (\ln \circ \ln)(\sqrt{3}+2) \Bigg) + 2. $$
We have :
$$\begin{align*}
f(x) = k &\Leftrightarrow {} 2^{x-2}\ln(\sqrt{3}+2) = \underbrace{\ln\Big( \frac{k+\sqrt{k^2-4}}{2} \Big)}_{>0} \\[2mm]
&\Leftrightarrow (x-2)\ln(2) + (\ln \circ \ln)(\sqrt{3}+2) = (\ln \circ \ln)\Big( \frac{k+\sqrt{k^2-4}}{2} \Big) \\[2mm]
\end{align*}
$$
Which gives the expected value for $x$.
Then, you can verify that if $x$ is given as in the third claim, then $f(x)$ is an integer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Help with epsilon delta proof $x^3$ is near $27$ when $x$ is near $3$ but $x$ is not equal to $3$.
So I have
$$0<|x-3|<\delta \implies 3-\delta<x<3+\delta$$
$$|x^3-27|<\epsilon \implies|(x-3)(x^2+3x+3^2)|<\epsilon$$
$$=(x^2+3x+9)|(x-3)|<\epsilon \implies28|(x-3)|<\epsilon$$
$$=|(x-3)|<\epsilon/28$$
How do I prove $x^2+3x+9$ is less than $28$?
My attempt
$|x-3|<\delta$
so $$|x^2+3x+9|=|x^2+3x-18+27|\leq|x^2+3x-18|+27 \\ =|(x-3)(x+6)|+27
=|(x-3)(x-3+9)|+27$$
I'm not really sure where to go from here.
|
There's a safer way to think. Let $\epsilon > 0$, and suppose that $0 < |x - 3| < \delta$. let's find $\delta$. Notice: $$|x| - 3 < |x-3| < \delta \implies |x| < \delta + 3.$$
Once we find $\delta$, any other value $\delta'$ less than this $\delta$ will also work, so we impose that $\delta \leq 1$. If the value of $\delta$ we find is greater than $1$, if we choose anything less than $1$, it will also work. We have: $$|x^3 - 27| = |(x - 3)(x^2+3x+9)| = |x^2+3x+9||x-3|.$$
We use the triangle inequality, and the hypothesis $|x-3| < \delta $ to get: $$|x^3-27| \leq (x^2+3|x|+9)\delta.$$
Now, we use that $|x| < 3+\delta < 3+1 = 4$. Hence: $$|x^3-27| < (16+12+9)\delta = 37\delta$$
I want $37\delta < \epsilon$, so $\delta < \epsilon/37$.
The crux here is that this argument only holds if $\delta < 1$ too. So, our final answer is that: given $\epsilon > 0$, if $\delta = \min\{1, \epsilon/37\}$, then: $$0<|x-3|<\delta \implies |x^3 - 27| < \epsilon.$$
You might find my answer here helpful. I give the general strategy for dealing with epsilon-delta proofs for limits with polynomials.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Integration of $ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $ How can I integrate this by changing variable?
$$ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $$
Thanks.
|
Notice that in this case by doing some rearrangement
$$\frac{p(x)}{q(x)}=A\frac{q(x)}{q(x)}+B\frac{q'(x)}{q(x)}$$
$$\begin{align}
\int\frac{2\sin x + \cos x}{\sin x + 2\cos x}dx&=-\frac{3}{5}\int\frac{\cos x-2\sin x}{\sin x + 2\cos x}dx+\frac{4}{5}\int\frac{\sin x+2\cos x}{\sin x + 2\cos x}dx\\
\end{align}$$
$$\int\frac{2\sin x + \cos x}{\sin x + 2\cos x}dx=\frac{1}{5}\left(4x-3\log(\sin x+2\cos x)\right)+C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1004318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
}
|
$\lim_{x\rightarrow\pm\infty} \frac{2x}{2x-\sqrt{4x^2-2x}}$ $$\lim_{x\rightarrow\pm\infty} \frac{2x}{2x-\sqrt{4x^2-2x}}$$
What I did
I multiplied by the denominator's conjugate and got the following $$2x+\sqrt{4x^2-2x}$$
My question is, what would I now do to evaluate the limit? The positive infinity I would do as follows:
$$\lim_{x\rightarrow\infty}2x+\sqrt{4x^2-2x} = 2\cdot\infty + \sqrt{4\cdot(\infty)^2 - 2\cdot\infty} = \infty???$$
|
When $x \to +\infty$
$$2x+\sqrt{4x^2-2x} = 2x\left(1 +\sqrt{1- \dfrac{1}{2x}}\right) \sim 4x\to +\infty$$
When $x \to -\infty$
$$2x+\sqrt{4x^2-2x} = 2x\left(1 -\sqrt{1- \dfrac{1}{2x}}\right) = \dfrac{1}{1 +\sqrt{1- \dfrac{1}{2x}}} \to \dfrac{1}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1004620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Why is $2x-1=7$ not $x=-4 \text{ or } x=4$ How would you explain why $3(2x-1)^2=147$, is $2x-1=7 \text{ or } 2x-1=-7$. But not $2x=8 \text{ so } x=4 \text{ or } x=-4$?
|
$$ 3(2x-1)^2=147 $$
$$ (2x-1)^2=\frac{147}{3} $$
$$ (2x-1)^2=49 $$
$$ (2x-1)^2=(\pm 7)^2 $$
$$ \sqrt{(2x-1)^2}=\sqrt{(\pm 7)^2} $$
$$ 2x-1=\pm 7$$
$$ 2x=1\pm 7$$
$$ x=\frac{1\pm 7}{2}$$
Therefore
$$ x_1=\frac{1+ 7}{2}=\frac82=4$$
$$ x_2=\frac{1- 7}{2}=-\frac62=-3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1005733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Specific question on proving continuity of this function If $c\in\mathbb{R}$, I want to show that $f(x)=\dfrac{cx}{x^2+c^2}$ is continuous using $\varepsilon-\delta$. Unfortunately from an algebraic standpoint I seem to be missing the strategy. I let $p\in\mathbb{R}$ ($p\neq 0)$, and I want $\forall \varepsilon>0$ $\exists \delta>0$ such that $\left|x-p\right|<\delta \Rightarrow \left|f(x)-f(p)\right|<\varepsilon$.
$\left|f(x)-f(p)\right|=\left|\dfrac{cx}{x^2+c^2}-\dfrac{cp}{p^2+c^2}\right|=|c|\left|\dfrac{x}{x^2+c^2}-\dfrac{p}{p^2+c^2}\right|,$ but I do not see how to proceed.
|
Choose $\delta < |p|/2$ such that when $|x-p| < \delta$,
$$||x|-|p|| \leq |x-p|< |p|/2$$
and
$$|p|/2 <|x| < 3|p|/2$$
Note that
$$|f(x) - f(p)| < \frac{|c||p||x|}{|x^2+c^2||p^2+c^2|}|x-p|+\frac{|c|^3}{|x^2+c^2||p^2+c^2|}|x-p| \\ < \frac{3|c|p^2}{2|p^2/2+c^2||p^2+c^2|}|x-p|+\frac{|c|^3}{|p^2/2+c^2||p^2+c^2|}|x-p|\\ = \left[\frac{3|c|p^2}{2|p^2/2+c^2||p^2+c^2|}+\frac{|c|^3}{|p^2/2+c^2||p^2+c^2|}\right] |x-p|$$
Choose $\delta'$ such that the RHS is less than $\epsilon$ if $|x-p|< \delta'$:
$$\delta' = \epsilon \left[\frac{3|c|p^2}{2|p^2/2+c^2||p^2+c^2|}+\frac{|c|^3}{|p^2/2+c^2||p^2+c^2|}\right]^{-1}$$
Therefore, if $|x-p| < \min(\delta, \delta')$ then $|f(x) - f(p)| < \epsilon$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1007032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Better substitution calculating integral? I'm calculating
$$ \iint\limits_S \, \left(\frac{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}{1+\frac{x^2}{a^2}+\frac{y^2}{b^2}} \right)^\frac{1}{2} \, dA$$ with $$S =\left\{ (x, \, y) \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\right\}.$$
I take
$$x = ar\cos \theta$$
$$y= br\sin \theta$$
and the integral becomes
$$ ab\int_0^{2\pi}d\theta\int_0^1\, \left(\frac{1-r^2}{1+r^2} \right)^\frac{1}{2} r \, dr$$
What is better substitution to calculate inner integral?
I tried with $r= \sin \vartheta$, a friend mine told me $u=1+r^2$. Thanks for any suggestions and helping ideas.
|
One way to do this is to substitute $u^2 = 1 + r^2$, so $2u \, du = 2r\, dr$, or $r\, dr = u\, du$
$$\int_0^1 \sqrt{\frac{1-r^2}{1+r^2}}\,r\,dr = \int_0^1 \sqrt{\frac{2-(1+r^2)}{1+r^2}}\,r\,dr\\
= \int_1^\sqrt{2} \sqrt{\frac{2-u^2}{u^2}}\,u\,du = \int_1^\sqrt{2} \sqrt{2- u^2} \, du $$
Now substitute $u = \sqrt{2} \sin \phi$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1007587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer:
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?
|
To avoid that nasty quotient rule
(and throwing away that "$2$"),
$\begin{array}\\
\left(\frac{1+4\cos x}{ \sqrt{x+4\sin x}}\right)'
&=\left((1+4\cos x)(x+4\sin x)^{-1/2}\right)'\\
&=(1+4\cos x)\left((x+4\sin x)^{-1/2}\right)'
+(1+4\cos x)'(x+4\sin x)^{-1/2}\\
&=(1+4\cos x)\left((-1/2)((x+4\sin x)'(x+4\sin x)^{-3/2}\right)
+(-4\sin x)(x+4\sin x)^{-1/2}\\
&=(1+4\cos x)\left((-1/2)((1+4\cos x)(x+4\sin x)^{-3/2}\right)
+(-4\sin x)(x+4\sin x)^{-1/2}\\
&=\frac{(-1/2)(1+4\cos x)^2
+(-4\sin x)(x+4\sin x)}{(x+4\sin x)^{3/2}}\\
&=-\frac{(1+4\cos x)^2
+(8\sin x)(x+4\sin x)}{2(x+4\sin x)^{3/2}}\\
&=-\frac{1+8\cos x+16\cos^2 x
+8x\sin x+32\sin^2 x}{2(x+4\sin x)^{3/2}}\\
&=-\frac{1+8\cos x+16
+8x\sin x+16\sin^2 x}{2(x+4\sin x)^{3/2}}\\
\end{array}
$
Or,
looking at
Przemysław Scherwentke's
marvelous suggestion,
since
$(x+4\sin x)'
=1+4\cos x
$
and
$(f'f^{-1/2})'
=f'' f^{-1/2}+f'(-1/2)f'f^{-3/2}
=\dfrac{f'' f+(-1/2)f'^2}{f^{-3/2}}
$,
if
$f(x)
=x+4\sin x
$,
$f'(x)
=1+4\cos x
$
and
$f''(x)
=-4\sin x
$
we get
$\begin{array}\\
(f'f^{-1/2})'
&=f'' f^{-1/2}+f'(-1/2)f'f^{-3/2}\\
&=\dfrac{(-4\sin x) (x+4\sin x)+(-1/2)(1+4\cos x)^2}{(x+4\sin x)^{3/2}}\\
&=\dfrac{-4x\sin x-16\sin^2 x)-(1/2)(1+8\cos x+16\cos^2x}{(x+4\sin x)^{3/2}}\\
&=-\dfrac{8x\sin x+32\sin^2 x+1+8\cos x+16\cos^2x}{(x+4\sin x)^{3/2}}\\
\end{array}
$.
By a miracle,
the two results are the same.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1008087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Determine all values of n such that this quadratic Determine all values of $n^2 + 19n + 99$ is a perfect square. I tried setting some square $b^2$ equal to the following, and then factoring as a Diophantine equation with $2$ variables... Didn't work.
|
$$
\begin{array}{c}
n^2 + 19n+99 = r^2 \\
(4n^2+76n) = 4r^2 - 396 \\
(2n+19)^2 -361 = 4r^2 - 396 \\
r^2 - (2n+19)^2 = 35 \\
(r+ (2n+19)) (r-(2n+19)) = 35
\end{array}
$$
Then if: $$
\begin{array}{ccc}
(r+ (2n+19)) =35, & (r-(2n+19)) = 1 & \longrightarrow n = -1 \\
(r+ (2n+19)) =7, & (r-(2n+19)) = 5 & \longrightarrow n = -9 \\
(r+ (2n+19)) =5, & (r-(2n+19)) = 7 & \longrightarrow n = -10 \\
(r+ (2n+19)) =1, & (r-(2n+19)) = 35 & \longrightarrow n = -18 \\
(r+ (2n+19)) =-35, & (r-(2n+19)) = -1 & \longrightarrow n = -18 \\
(r+ (2n+19)) =-7, & (r-(2n+19)) = -5 & \longrightarrow n = -10 \\
(r+ (2n+19)) =-5, & (r-(2n+19)) = -7 & \longrightarrow n = -9 \\
(r+ (2n+19)) =-1, & (r-(2n+19)) = -5 & \longrightarrow n = -1 \\
\end{array}
$$
So only those four integer solutions work. $n$ must be negative, by the way.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1009662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
proving that $(n-1)^n>n^{n-1}$ I want to prove that $(n-1)^n>n^{n-1}$, for $n>4$, $n$ is an integer.
So I divided by $n^n$ and got:
$(1-\frac{1}{n})^{n}>\frac{1}{n}$
I know that $(1-\frac{1}{n+1})^{n+1}$>$(1-\frac{1}{n})^{n+1}=(1-\frac{1}{n})^n(\frac{n-1}{n})>\frac{1}{n}(\frac{n-1}{n})$.
How can I continue?
Thanks.
|
Solution using induction:
Base case - $n=4$, so $(4-1)^4>4^{4-1}$ which is true.
Inductive step - assume that the statement holds for some k and show that it holds for k+1, hence we want to show that $k^{k+1}>(k+1)^k$ for $k \ge 3$.
Our assumption is $(k-1)^k>k^{k-1}$ or alternatively $\displaystyle k-1>\left(1+\frac{1}{k-1}\right)^{k-1}$.
Dividing both sides by $k^k$ we get that we need to show that $\displaystyle k>\left(1+\frac{1}{k}\right)^k$ for $k \ge 3$.
It is obvious that $\displaystyle 1+\frac{1}{k}<1+\frac{1}{k-1} \rightarrow \left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{k-1}\right)^k$, so it is sufficient to show that $\displaystyle k>\left(1+\frac{1}{k-1}\right)^k$.
We can write: $$\displaystyle \left(1+\frac{1}{k-1}\right)^k=\left(1+\frac{1}{k-1}\right)\cdot{\left(1+\frac{1}{k-1}\right)^{k-1}}\underbrace{<}_{using \ assumption}\left(1+\frac{1}{k-1}\right)(k-1)=k$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1011035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
Proof $(1+1/n)^n$ is an increasing sequence I need help proving $a_n=\left(\dfrac{n+1}{n}\right)^n$ is increasing sequence on the positive integers.
An exercise in the analysis book by Mattuck asks to prove $a_n=\left(\dfrac{2^n+1}{2^n}\right)^{2^n}$ is increasing.
But this is easy since $\left(\dfrac{2^n+1}{2^n}\right)^{2^n}=\left(\left(\dfrac{2^n+1}{2^n}\right)^{2}\right)^{2^{n-1}} =\left(\dfrac{2^{2n}+2\left(2^n\right)+1}{2^{2n}}\right)^{2^{n-1}}=$
$\left(\dfrac{2^{n-1}+1+\frac{1}{2^{n+1}}}{2^{n-1}}\right)^{2^{n-1}}>\left(\dfrac{2^{n-1}+1}{2^{n-1}}\right)^{2^{n-1}}$ as desired.
I am having difficulty in the general case:
I must prove $\left(\frac{n+2}{n+1}\right)^{n+1}>\left(\frac{n+1}{n}\right)^n$.
This is equivalent to $(n+2)^{n+1}n^n>(n+1)^{2n+1}$ which is the same as proving $\frac{n+2}{n+1}>\left(\frac{(n+2)(n)}{(n+1)^2}\right)^n$ which is the same as $\frac{n+2}{n+1}>\left(1-\frac{1}{(n+1)^2}\right)^n$
I'm stumped. Thanks in advance. Oh, and I don't want to use the derivative.
|
Use the Bernoulli inequality,
$$1+(n+1)(a-1)\leq a^{n+1}$$ with
$$a=\frac{n(n+1)}{(n+1)^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1011414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Show sequance is monotonic Let $x>0$ (fixed) and $n$ be natural. Show that $$\displaystyle (x^n+x^{n-1}+...+1)^{\frac{1}{n}}$$ is monotonic.
I tried by induction but didn't work but intuition tells me it's decreasing.
|
The first couple are decreasing at least:
$(x+1)^2-(x^2+x+1)^1=x$
$(x^2+x+1)^3-(x^3+x^2+x+1)^2=x^5+3x^4+3x^3+3x^2+x\\=x(x+1)(x^3+x^2+x+1)+x^2(x^2+x+1)$
Next is $(x^3+x^2+x+1)^4-(x^4+x^3+x^2+x+1)^3$, which is
$$x^{11}+4x^{10}+10x^9+16x^8+22x^7+25x^6+22x^5+16x^4+10x^3+4x^2+x$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1012071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Another quick induction question for a recursively defined sequence (with closed form formula given) I was given:
A sequence is defined recursively by $a_0 = 1$, $a_1 = 4$, and for $n\ge2$, $a_n = 5a_{n-1} - 6a_{n-2}$.
Use induction to prove that the closed form formula for $a_n$ is $a_n = 2\cdot3^n-2^n, n\ge0$.
So far for my proof, I have the following:
Basis: $a_0=1; 1=2\cdot3^0-2^0; 1=2\cdot1-1; 1=1$ and $a_1=4; 4=2\cdot3^1-2^1; 4=2\cdot3^-2; 4=4$
Inductive Step: Assume $a_n = 2\cdot3^n-2^n$.
Therefore, $a_{n+1}=5a_{n+1-1}-6a_{n+1-2};$
$a_{n+1}=5a_{n}-6a_{n-1}$
$a_{n+1}=5(2\cdot3^n-2^n)-6a_{n-1}$
What do I do next?
|
Substitute $a_{n-1}=2\cdot3^{n-1}-2^{n-1}$:
$$\begin{align}
a_{n+1}&=5(2\cdot3^n-2^n)-6a_{n-1}\\
&=5(2\cdot3^n-2^n)-6(2\cdot3^{n-1}-2^{n-1})\\
&=30\cdot3^{n-1}-10\cdot2^{n-1}-12\cdot3^{n-1}+6\cdot2^{n-1}\\
\end{align}$$
Can you finish?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1015576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Inequality with five variables Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$
Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is true
and for even $n$ and positives $a_i$ the following inequality is true.
$$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$
|
Here is a full proof.
Let us start the discussion for general $n$. Denote $S = \sum_{i=1}^n a_i$.
Since by AM-GM, $S \geq n \sqrt[n]{a_1a_2...a_n}$, we have
$$1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} \geq \frac{S}{S - (n-2)\sqrt[n]{a_1a_2...a_n}}$$
Hence a tighter claim is (simultaneously defining $L$ and $R$):
$$L = \sum_{cyc}\frac{a_i}{a_i+a_{i+1}}\geq 1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} = R$$
and it suffices to show that one.
We write $2 L \geq 2 R$ or $L \geq 2 R- L$ and add on both sides a term $$\sum_{cyc}\frac{a_{i+1}}{a_i+a_{i+1}}$$ which leaves us to show
$$n = \sum_{cyc}\frac{a_i + a_{i+1}}{a_i+a_{i+1}}\geq 2+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{S} + \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}}$$
or, in our final equivalent reformulation of the L-R claim above,
$$ \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} \leq (n - 2) (1- \frac{n \sqrt[n]{a_1a_2...a_n}}{S} )$$
For general odd $n$ see the remark at the bottom. Here
the task is to show $n=5$.
Before doing so, we will first prove the following Lemma (required below), which is the above L-R-inequality for 3 variables (which is tighter than the original formulation, hence we cannot apply the proof for $n=3$ given above by Michael Rozenberg for the original formulation):
$$ \frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c} \leq (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )$$
This Lemma is, from the above discussion, just a re-formulation of the claim in $L$ and $R$ above, for 3 variables, i.e.
$$ \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq 1+\frac{3\sqrt[3]{a \, b \ c}}{2(a+b+c)}$$
By homogeneity, we can demand $abc=1$ and prove, under that restriction, $$ \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq 1+\frac{3}{2(a+b+c)}$$
This reformulates into
$$ \frac{a\; c}{a +b} + \frac{b\; a}{b +c} + \frac{c\; b}{c +a} \geq \frac{3}{2}$$
or equivalently, due to $abc=1$,
$$ \frac{1}{b(a +b)} + \frac{1}{c(b +c)} + \frac{1}{a(c +a)} \geq \frac{3}{2}$$
which is known (2008 International Zhautykov Olympiad), for some proofs see here: http://artofproblemsolving.com/community/c6h183916p1010959
Hence the Lemma holds.
For $n=5$, we rewrite the LHS of our above final reformulation by adding and subtracting terms:
$$ \frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{d-c}{d+c} + \frac{e-d}{e+d} + \frac{a-e}{a+e} = \\
(\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + (\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e})
$$
This also holds for any cyclic shift in (abcde), so we can write
$$ 5 (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{d-c}{d+c} + \frac{e-d}{e+d} + \frac{a-e}{a+e}) = \\
\sum_{cyc (abcde)} (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + \sum_{cyc (abcde)}(\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + \sum_{cyc (abcde)} (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e})
$$
Using our Lemma, it suffices to show (with $S = a +b+c+d+e$)
$$
\sum_{cyc (abcde)} (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )
+ \sum_{cyc (abcde)}(1- \frac{3 \sqrt[3]{a\, c \, d}}{a + c+ d} )
+ \sum_{cyc (abcde)}(1- \frac{3 \sqrt[3]{a\, d \, e}}{a + d+ e} )
\leq 15 (1- \frac{5 \sqrt[5]{a b c d e }}{S} )
$$
which is
$$
\sum_{cyc (abcde)} (\frac{\sqrt[3]{a\, b \, c}}{a + b+ c} + \frac{\sqrt[3]{a\, c \, d}}{a + c+ d} + \frac{\sqrt[3]{a\, d \, e}}{a + d+ e} )
\geq 25 \frac{\sqrt[5]{a b c d e }}{S}
$$
Using Cauchy-Schwarz leaves us with showing
$$
\frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2}{\sum_{cyc (abcde)}(a + b+ c)}
+
\frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, c \, d})^2}{\sum_{cyc (abcde)}(a + c+ d)}
+
\frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, d \, e})^2}{\sum_{cyc (abcde)}(a + d+ e)}
\geq 25 \frac{\sqrt[5]{a b c d e }}{S}
$$
The denominators all equal $3S$, so this becomes
$$
(\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2
+
(\sum_{cyc (abcde)} \sqrt[6]{a\, c \, d})^2
+
(\sum_{cyc (abcde)} \sqrt[6]{a\, d \, e})^2
\geq 75 \sqrt[5]{a b c d e }
$$
Using AM-GM gives for the first term
$$
(\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2 \geq ( 5 (\prod_{cyc (abcde)} \sqrt[6]{a\, b \, c} )^{1/5})^2
= 25 (\prod_{cyc (abcde)} ({a\, b \, c} ) )^{1/15} = 25 \sqrt[5]{a b c d e }
$$
By the same procedure, the second and the third term on the LHS are likewise greater or equal than $25 \sqrt[5]{a b c d e }$. This concludes the proof.
Remarks:
*
*the tighter $L-R$-claim used here is - for general $n$ - asked for in the problem given at Cyclic Inequality in n (at least 4) variables
*For general odd $n$, the above reformulation can be used again. For odd $n>5$, take the method of adding and subtracting terms to form smaller sub-sums which are cyclically closed in a smaller number of variables, and apply previous results for smaller $n$ recursively.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1017110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 0
}
|
Minimum without using of differential calculus Find minimum of $$x + y^5$$ where $x>0,y>0 $ $xy=1$ without using of differential calculus.
|
$x+y^5=\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+y^5$. By inequality between arithmetic mean and geometric mean we get $$\frac{x+y^5}{6}\geq \sqrt[6]{\frac{x^5}{5^5}y^5}=5^{-5/6}\Rightarrow x+y^5\geq 6\cdot 5^{-5/6}$$ and equality holds iff $\frac{x}{5}=y^5$, so corresponding values of arguments are $x=\sqrt[6]{5},y=\sqrt[6]{\frac{1}{5}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1020164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
summation of $\sum^n_{k=0} (n-k)^2$ I'm trying to find the recurrence of
$$ T(n) = T (n-1) + n^2$$
After following the steps,
$$T (n) = T (n-1) + n^2 = T (n-2) + (n-1)^2 + n^2 $$
$$T (n) = T (n-2) + (n-1)^2 + n^2 = T(n-3) + (n-3)^2 + (n-1)^2 + n^2 $$
$$T (n) = T (n-3) + (n-3)^2 + (n-1)^2 + n^2 =T (n-4) + (n-4)^2 + (n-3)^2 + (n-1)^2 + n^2 $$
i generalize recurrence relation at the kth step of the recursion, which is
$$ T(n) = T (n-k) + \sum^n_{k=0} (n-k)^2$$
Just wondering,
What is the summation of
$$\sum^n_{k=0} (n-k)^2$$
is it the same as $$ n(n+1)(2n+1)/6$$
?
|
$$\sum^n_{k=0} (n-k)^2=n^2+(n-1)^2+\cdots + 0^2= 0^2+1^2+\cdots +n^2 =\sum^n_{k=0} k^2=\frac16n(n+1)(2n+1).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1020993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Inhomogeneus recurrence relation $a_{n+1} = 2a_n+3^n+4^n$ So this was given in class and the teacher weren't able to solve it, and I was wondering how a solution can be given?
$a_{n+1} = 2a_n+3^n+4^n, \enspace a_0 = 1$
Usually we'd consider the solution $a_n$ to be of the form $a_n = a_n^{(h)}+a_n^{(p)}$, where $a_n^{(h)}$ is a solution to the homogeneous recurrence relation and $a_n^{(p)}$ is a particular solution to the recurrence relation. Thus $a_n^{(h)} = c_1\times 2^n$ and $a_n^{(p)} = c_2(3^n+4^n)$, but this strategy didn't work, as solving for $c_2$ in
$c_2(3^{n+1}+4^{n+1})+c_2(3^n+4^n) = 3^n+4^n$
there doesn't seem to be a way to cancel out the $n$'s this equation. Is there maybe another strategy that would make this possible to solve?
Thanks in advance.
|
$$\dfrac{a_{n+1}}{2^{n+1}} = \dfrac{a_n}{2^n} + \dfrac{1}{2}(\dfrac{3}{2})^n + \dfrac{1}{2}2^n$$
Let $b_{n+1} = \dfrac{a_{n+1}}{2^{n+1}}$, we have $$b_{n+1} = b_n + \dfrac{1}{2}(\dfrac{3}{2})^n + \dfrac{1}{2}2^n $$
so $b_n = b_0 + \sum_{k=1}^n\left(\dfrac{1}{2}(\dfrac{3}{2})^k + \dfrac{1}{2}2^k\right)$
Can you go from this?(In general for $a\neq 1$, we have $\sum_{k=1}^n a^k = \dfrac{a-a^{n+1}}{1-a}$)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1021595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Exact closed form expression of $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$ Exact closed form of this expression $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$
I assume this means there is just one $2^0$ and one $2^{2n}$ and a double of all the terms in between?
|
Your series is
$1+2+\cdots+2^n$
$\,\,\,\,\,\,\,\,\,2+\cdots+2^n+2^{n+1}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2^n+2^{n+1}+\cdots+2^{2n}$
And Sum can be written as
$$\begin{align}s&=1+2(2+4+\cdots+2^{n})+2^{n+1}+2^n+2^{n+1}+2^{n+2}+\cdots2^{2n}\\
&=1+2(2+4+\cdots+2^{n})+2^{n+1}+(2^n+2^{n+1}+2^{n+2}+\cdots2^{2n})\\
\end{align}$$
So, answer is
$$1+2\cdot\frac{2^{n+1}-2}{2-1}+2^{n+1}+\frac{2^{2n+1}-2^{n}}{2-1}=2^{2n+1}+3\cdot2^{n+1}-2^n-3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1022497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.