Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$f(x)$=$\left\lfloor { x }^{ 2 } \right\rfloor -\left\lfloor x \right\rfloor ^{ 2 }$ is discontinuous for all integer values of x except only at x=1 How to prove that $f(x)$=$\left\lfloor { x }^{ 2 } \right\rfloor -\left\lfloor x \right\rfloor ^{ 2 }$ is discontinuous for all integer values of x except only at $x=1$ ?
Ya,even I used intuition at the first go taking some trial values and checking.Then I plotted the graph on wolfram alpha.But if you can think of some other rigorous proof for this one,let me know.Thanks!
| We could deal all at once with all integers $a$. However, it is helpful to develop intuition by first looking at positive $a$.
Let $a$ be a positive integer. If $x$ is close enough
to $a$ but smaller than $a$, then $a^2-1\lt x^2 \lt a^2$, so $\lfloor x^2 \rfloor=a^2-1$. It follows that $f(x)=(a^2-1)-(a-1)^2=2a-2$. Thus the limit of $f(x)$ as $x$ approaches $a$ from the left is $2a-2$. Since $f(a)=0$, We conclude that $f$ is not continuous at $a$ if $2a-2\ne 0$, that is, if $a\ne 1$.
Let $a=1$. Then the limit of $f(x)$ as $x$ approaches $a$ from the left is $2a-2$, which is $0$. It is not hard to see that the limit of $f(x)$ as $x$ approaches $a$ from the right is $0$. Also, $f(0)=0$, so $f$ is continuous at $a=0$.
Next we deal with integers $a\le 0$. Again, $f(a)=0$. If $x$ is close enough to $a$ but larger than $a$, then $\lfloor x^2\rfloor=a^2-1$, while $(\lfloor x\rfloor)^2=(a-1)^2$. So $f(x)=2a-2$. Thus the limit of $f(x)$ as $x$ approaches $a$ from the right is $2a-2$. However, $f(a)=0$, and $2a-2$ cannot be $0$ at $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Results for values of the residues in the Residue Theorem
*
*If the sum of the residue is $0$, what can I conclude: that the value of the integral is $0$ or that the integral diverges?
*If the residue tends to $\infty$, should I conclude that the integral diverges?
| If you REALLY want to use the Laurent series expansion, this is how it goes:
\begin{align}
\dfrac{1}{1+z} (z e^{1/z}) &= (1-z+z^2-z^3+\dots)\cdot z (1+\dfrac{1}{z}+\dfrac{1}{2!z^2} + \dfrac{1}{3!z^3}+\dots)\\
& = z^{-1}\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dfrac{1}{5!}+\dots\right)+\text{other terms we don't care about}
\end{align}
Now, if you look at the coefficient,
$$\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dfrac{1}{5!}+\dots\right)$$ is exactly
\begin{align}
e^z & = 1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!}+\dfrac{z^4}{4!}+\dfrac{z^5}{5!}+\dots\\
e^{-1}&=1+(-1)+\dfrac{(-1)^2}{2!} + \dfrac{(-1)^3}{3!}+\dfrac{(-1)^4}{4!}+\dfrac{(-1)^5}{5!}+\dots\\
e^{-1}&=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dfrac{1}{5!}+\dots
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Limits and cube roots I'm rather stumped at the moment. I can graph the following equation but I'm having trouble solving it algebraically.
$$
\lim_{x\to 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1}
$$
Where do I start?
| This solution uses what is called: The conjugate expression. Using
\begin{equation*}
(a-b)(a+b)=a^{2}-b^{2}
\end{equation*}
with $a=\sqrt{x}$ and $b=1,$ you get
\begin{equation*}
\left( \sqrt{x}-1\right) (\sqrt{x}+1)=x-1.
\end{equation*}
[The conjugate of $(\sqrt{x}-1)$ is $(\sqrt{x}+1)$)]. The same way, using
\begin{equation*}
(a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3}
\end{equation*}
with $a=\sqrt[3]{x}$ and $b=1,$ you get
\begin{equation*}
(\sqrt[3]{x}-1)(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)=x-1.
\end{equation*}
[The conjugate of $(\sqrt[3]{x}-1)$ is $(\left( \sqrt[3]{x}\right) ^{2}+\sqrt%
[3]{x}+1)$]. Therefore
\begin{eqnarray*}
\frac{(\sqrt[3]{x}-1)}{\left( \sqrt{x}-1\right) } &=&\frac{(\sqrt[3]{x}-1)}{%
\left( \sqrt{x}-1\right) }\frac{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}%
+1)}{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)}\frac{(\sqrt{x}+1)}{(%
\sqrt{x}+1)} \\
&=&\frac{(\sqrt[3]{x}-1)\cdot (\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)%
}{\left( \sqrt{x}-1\right) \cdot (\sqrt{x}+1)}\cdot \frac{(\sqrt{x}+1)}{%
(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)} \\
&=&\frac{x-1}{x-1}\cdot \frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+%
\sqrt[3]{x}+1)} \\
&=&\frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)}
\end{eqnarray*}
and the limit follows
\begin{equation*}
\lim_{x\rightarrow 1}\frac{(\sqrt[3]{x}-1)}{\left( \sqrt{x}-1\right) }%
=\lim_{x\rightarrow 1}\frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+%
\sqrt[3]{x}+1)}=\frac{(\sqrt{1}+1)}{(\left( \sqrt[3]{1}\right) ^{2}+\sqrt[3]{%
1}+1)}=\frac{2}{3}.
\end{equation*}
${\bf NOTE:}$ You can take it as a Definition for further uses: The conjugate
expression of $A$ (which may contains radicals) is the expression $B$ (which
may contains radicals), if $AB$ contains no radicals! For example, the
conjugate of $\sqrt[4]{x}-1$ is obtained from the formula
\begin{equation*}
(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})=a^{4}-b^{4}
\end{equation*}
by taking $a=\sqrt[4]{x}$ and $b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coefficient of $x^9$ in the Product Find the Coefficient of $x^9$ in $$G(x)=(1+x)(1+x^2)(1+x^3)(1+x^4)\cdots(1+x^{100})$$
My Try:
$$G(x)=P(x)(1+x^{10})(1+x^{11})\cdots(1+x^{100})=P(x)(1+O(x^{10}))$$ Hence Coefficient of $x^9$ in $G(x)$ is Coefficient of $x^9$ in $P(x)$ where $$P(x)=(1+x)(1+x^2)(1+x^3)\cdots(1+x^9)=R(x)+x^9R(x)$$ So Coefficient of $x^9$ in $P(x)$ is $1$+coeff of $x^9$ in $ R(x)$ where $$R(x)=(1+x)(1+x^2)\cdots(1+x^8)$$
But the method is becoming lengthy. Can i have any good approach
| HINT: The desired coefficient is the number of ways to express $9$ as a sum of distinct positive integers. Since $1+2+3+4>9$, you need only consider sums of at most three integers.
| {
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Proof coutinuity $f:=\left\{\begin{matrix}
\frac{xy}{\sqrt{x}+y^2}, x,y\neq 0\ begin & \\
0, x,y=0 &
\end{matrix}\right.$
Is f continuous in (0,0)?
My idea is:
$\left | f(x,y) \right | = \left | \frac{x*y}{\sqrt{x}+y^2} \right |\leq \left | \frac{x*y}{\sqrt{x}+\sqrt{y}} \right|$
we know:
$\sqrt{x}+\sqrt{y}>\sqrt{x+y} \implies \left | \frac{x*y}{\sqrt{x}+\sqrt{y}} \right| \leq\left | \frac{x*y}{\sqrt{x+y}} \right|$
we know:
$\sqrt{x+y} \geq 2*\sqrt{x}\sqrt{y} \implies \left | \frac{x*y}{\sqrt{x+y}} \right| \leq \left | \frac{x*y}{(x*y)^{\frac{1}{4}}} \right|=(x*y)^{\frac{3}{4}}$
with $(x,y) \implies 0$ we get $(x*y)^{\frac{3}{4}}$ =0
Can someone please check my suggestion?
| We assume that $x\ge 0$. Now, note that
$$\sqrt{x}+y^2\ge\sqrt{x} \implies \frac{1}{\sqrt{x}+y^2}\le \frac{1}{\sqrt{x}}\implies \frac{xy}{\sqrt{x}+y^2}\le \sqrt{x}y \to 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x^3-9x-5=0$, then what is $x^4-18x^3-81x^2-12$ If we have $x^3-9x-5=0$, then what $x^4-18x^3-81x^2-12$ equals to?
This is a multiple choices question.
A)$5$ B)$25$ C)$42$ D)$67$ E)$81$.
My attempt,
By long division of polynomials, we have $x^4-18x^3-81x^2-12=(x-18)(x^3-9x-5)+(-72x^2-157x-102)$
Since $x^3-9x-5=0$, whatever $x$ could be, $x^4-18x^3-81x^2-12=-72x^2-157x-102$
How to proceed?
P/S. Conclusion, NONE of the options can be right.
| Simply use synthetic substitution to solve for $x$ in the first equation, then plug that value into the second.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How can one simplify the following expression? I have the expression:
$$-\left(\frac{A}{3a} + \frac{B}{3b} + \frac{C}{3c}\right) \pm \frac{8}{3} \sqrt{b^{2}c^{2}A^{2} + a^{2}c^{2}B^{2} + a^{2}b^{2}C^{2} - abc^{2}AB - ab^{2}cAC - a^{2}bcBC}$$
and I would like to be able to simplify it, if it is at all possible.
Thank you.
| $$-\left(\frac{A}{3a} + \frac{B}{3b} + \frac{C}{3c}\right) \pm \frac{8}{3} \sqrt{{a^2b^2c^2}\left(\frac{A^2}{a^2} +\frac{B^2}{b^2} + \frac{C^2}{c^2} - \frac{AB}{ab} - \frac{AC}{ac} - \frac{BC}{bc}\right)},$$
$$-\left(\frac{A}{3a} + \frac{B}{3b} + \frac{C}{3c}\right) \pm \frac{8|abc|}{3} \sqrt{\frac{A^2}{a^2} +\frac{B^2}{b^2} + \frac{C^2}{c^2} - \frac{AB}{ab} - \frac{AC}{ac} - \frac{BC}{bc}},$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration consquence How does it follow from $$\int_0^\infty \left(\frac{1}{a^2+x^2} \right) dx= \frac{\pi}{2a}$$ that $$\int_0^\infty \left(\frac{1}{\left({a^2+x^2}\right) ^2 } \right) dx= \frac{\pi}{4a^3}\text{ ?}$$
| From $\displaystyle\int_0^\infty \left(\frac{1}{a^2+x^2} \right) dx= \frac{\pi}{2a}$, we can differentiate both sides with respect to $a$:
$\displaystyle\int_0^\infty \dfrac{\partial}{\partial a}\left(\frac{1}{a^2+x^2} \right) dx= \dfrac{\partial}{\partial a}\frac{\pi}{2a}$
$\displaystyle\int_0^\infty \left(-\frac{2a}{(a^2+x^2)^2}\right) dx= -\frac{\pi}{2a^2}$
Do you see how to finish? (Note you should also verify for yourself that the conditions for differentiating under the integral sign are met.)
EDIT: The other way to handle the second integral is to use trigonometric substitution.
Specifically, let $x = a\tan \theta$. Then, $dx = a\sec^2\theta\,d\theta$ and $a^2+x^2 = a^2\sec^2\theta$. Hence,
$\displaystyle\int_{0}^{\infty}\dfrac{dx}{(a^2+x^2)^2} = \int_{0}^{\pi/2}\dfrac{a\sec^2\theta\,d\theta}{(a^2\sec^2\theta)^2} = \dfrac{1}{a^3}\int_{0}^{\pi/2}\cos^2\theta\,d\theta$, which easily evaluates to $\dfrac{\pi}{4a^3}$.
Of course, this doesn't "follow" from $\displaystyle\int_0^\infty \left(\frac{1}{a^2+x^2} \right) dx= \frac{\pi}{2a}$, which is why everyone suggested to differentiate with respect to $a$.
| {
"language": "en",
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"source": "stackexchange",
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Find error in integration of $\int \frac {\sin 2x}{\sin^4 x + \cos^4 x} \, dx$? Find error in integration of $\int \frac {\sin 2x}{\sin^4 x + \cos^4 x}dx$?
The answer is supposed to be ($\arctan \tan^2 x + C$), but I obtained ($-\arctan \cos2x + C$) as follows. Please identify the error.
$$\int \frac {\sin 2x}{\sin^4 x + \cos^4 x}dx$$
$$= \int \frac {\sin 2x}{(\sin^2 x+ \cos^ 2x)^2 - 2\cos^2 x\sin^2 x}dx$$
$$= \int \frac {\sin 2x}{1- \frac{\sin^2 2x}{2}}dx$$
$$= \int \frac {2\sin 2x}{2- \sin^2 2x}dx$$
$$= \int \frac {2\sin 2x}{1 + \cos^2 2x}dx$$
$1 + \cos^2 2x = t ; dt = 2(\cos 2x)(-\sin 2x)(2)dx; 2\sin 2x.dx= \frac{-dt}{2\cos 2x} = \frac{-dt}{2(\sqrt{t-1})};$
$$\int \frac{\frac{-dt}{2\sqrt {t-1}}}{t} = -1/2\int \frac{dt}{t.\sqrt{t-1}}$$
$\sqrt{t-1} = u; du=\frac{dt}{2\sqrt{t-1}}; 2u.du = dt$;
$$= -1\int \frac{du}{u^2+1} = -\arctan{u} = -\arctan \sqrt{t-1} = -\arctan \sqrt{1+\cos^2 2x-1} = -\arctan \cos2x$$
Also, is there an easier method to this problem?
| Since
$$
\begin{align}
1
&=(\sin^2(x)+\cos^2(x))^2\\[6pt]
&=\sin^4(x)+\cos^4(x)+2\sin^2(x)\cos^2(x)\\[2pt]
&=\sin^4(x)+\cos^4(x)+\frac12\sin^2(2x)
\end{align}
$$
we can proceed with partial fractions
$$
\begin{align}
&\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm{d}x\\[6pt]
&=\int\frac{\sin(2x)}{1-\frac12\sin^2(2x)}\,\mathrm{d}x\\
&=\frac1{\sqrt2}\int\left(\frac1{1-\frac1{\sqrt2}\sin(2x)}-\frac1{1+\frac1{\sqrt2}\sin(2x)}\right)\,\mathrm{d}x\\
&=\frac1{\sqrt2}\int\left(\frac1{1-\frac{\sqrt2\tan(x)}{1+\tan^2(x)}}-\frac1{1+\frac{\sqrt2\tan(x)}{1+\tan^2(x)}}\right)\frac{\mathrm{d}\tan(x)}{1+\tan^2(x)}\\
&=\frac1{\sqrt2}\int\left(\frac1{\tan^2(x)-\sqrt2\tan(x)+1}-\frac1{\tan^2(x)+\sqrt2\tan(x)+1}\right)\,\mathrm{d}\tan(x)\\
&=\int\left(\frac1{(\sqrt2\tan(x)-1)^2+1}-\frac1{(\sqrt2\tan(x)+1)^2+1}\right)\,\mathrm{d}\sqrt2\tan(x)\\[12pt]
&=\arctan(\sqrt2\tan(x)-1)-\arctan(\sqrt2\tan(x)+1)+C+\tfrac\pi2\\[9pt]
&=\arctan\left(\frac{-1}{\tan^2(x)}\right)+C+\tfrac\pi2\\[9pt]
&=\bbox[5px,border:2px solid #C0A000]{\arctan\left(\tan^2(x)\right)+C}
\end{align}
$$
We can also proceed by writing everything in terms of $\cos(2x)$
$$
\begin{align}
\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}\,\mathrm{d}x
&=\int\frac{\sin(2x)}{1-\frac12\sin^2(2x)}\,\mathrm{d}x\\
&=-\int\frac1{1+\cos^2(2x)}\,\mathrm{d}\cos(2x)\\[9pt]
&=\bbox[5px,border:2px solid #C0A000]{-\arctan(\cos(2x))+C}
\end{align}
$$
These two answers are copacetic since
$$
\begin{align}
\tan\left(\arctan\left(\tan^2(x)\right)+\arctan(\cos(2x))\right)
&=\frac{\tan^2(x)+\cos(2x)}{1-\tan^2(x)\cos(2x)}\\[6pt]
&=\frac{\sin^2(x)+\cos^2(x)(\cos^2(x)-\sin^2(x))}{\cos^2(x)-\sin^2(x)(\cos^2(x)-\sin^2(x))}\\[6pt]
&=\frac{\sin^2(x)+\cos^4(x)-\sin^2(x)\cos^2(x)}{\cos^2(x)+\sin^4(x)-\sin^2(x)\cos^2(x)}\\[6pt]
&=\frac{\sin^4(x)+\cos^4(x)}{\cos^4(x)+\sin^4(x)}\\[12pt]
&=1
\end{align}
$$
That is,
$$
\arctan\left(\tan^2(x)\right)+\arctan\left(\cos(2x)\right)=\frac\pi4
$$
| {
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"url": "https://math.stackexchange.com/questions/1328806",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Determinants and monic polynomials I wish to show that
$$ \det \begin{pmatrix}
x & a & a & a\\
a & x & a & a\\
a & a & x & a\\
a & a & a & x
\end{pmatrix}=(x-a)^3(x+3a).$$
Obviously, I could expand it out and try to notice some factors as I go, but that seems rather tedious. Is there a particularly fast way of demonstrating the above identity?
| $$ \det \begin{pmatrix}
x & a & a & a\\
a & x & a & a\\
a & a & x & a\\
a & a & a & x
\end{pmatrix}~\quad R_1\to R_1-R_2~\&~ R_2\to R_2-R_3 ~\&~ R_3\to R_3-R_4\\$$
$$==- \det \begin{pmatrix}
x-a & a-x & 0 & 0\\
0 & x-a & a-x & 0\\
0 & 0 & x-a & a-x\\
a & a & a & x
\end{pmatrix}$$
$$=-(x-a)^3\det \begin{pmatrix}
1 & -1 & 0 & 0\\
0 & 1 & -1 & 0\\
0 & 0 & 1 & -1\\
a & a & a & x
\end{pmatrix}$$
Now if you expand this then you'll get answer
| {
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What is the exact value of $\eta(6i)$? Let $\eta(\tau)$ be the Dedekind eta function. In his Lost Notebook, Ramanujan played around with a related function and came up with some of the nice evaluations,
$$\begin{aligned}
\eta(i) &= \frac{1}{2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(2i) &= \frac{1}{2^{11/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(3i) &= \frac{1}{2\cdot 3^{3/8}} \frac{1}{(2+\sqrt{3})^{1/12}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(4i) &= \frac{1}{2^{29/16}} \frac{1}{(1+\sqrt{2})^{1/4}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(5i) &= \frac{1}{2\sqrt{5}}\left(\tfrac{1+\sqrt{5}}{2}\right)^{-1/2}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(6i) &=\; \color{red}{??}\\
\eta(7i) &= \frac{1}{2\sqrt{7}}\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)^{{1/4}}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(8i) &= \frac{1}{2^{73/32}} \frac{(-1+\sqrt[4]{2})^{1/2}}{(1+\sqrt{2})^{1/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(16i) &= \frac{1}{2^{177/64}} \frac{(-1+\sqrt[4]{2})^{1/4}}{(1+\sqrt{2})^{1/16}} \left(-2^{5/8}+\sqrt{1+\sqrt{2}}\right)^{1/2}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\end{aligned}$$
with the higher ones $>4$ added by this OP. (Note the powers of $2$.)
Questions:
*
*Similar to the others, what is the exact value of $\eta(6i)$?
*Is it true that the function,
$$F(\sqrt{-N}) = \frac{\pi^{3/4}}{\Gamma\big(\tfrac{1}{4}\big)}\,\eta(\sqrt{-N}) $$
is an algebraic number only if $N$ is a square?
P.S. It seems strange there is a function that yields an algebraic number for square input $N$ and a transcendental number for non-square $N$. (Are there well-known functions like that?) For an example of non-square $N$, we have,
$$\eta(\sqrt{-3}) = \frac{3^{1/8}}{2^{4/3}} \frac{\Gamma\big(\tfrac{1}{3}\big)^{3/2}}{\pi} = 0.63542\dots$$
and $F(\sqrt{-3})$ seems to be transcendental.
| Since we know the value of $\eta(3i)$, the point is just to compute the value of the product:
$$ \prod_{n\geq 0}(1+e^{-6\pi n})=\exp\sum_{n\geq 0}\log\left(1+e^{-6\pi n}\right)=\exp\sum_{n\geq 0}\int_{n}^{n+1}\frac{6\pi n}{1+e^{6\pi s}}\,ds$$
where:
$$\sum_{n\geq 0}\int_{n}^{n+1}\frac{6\pi n}{1+e^{6\pi s}}\,ds = \int_{0}^{+\infty}\frac{6\pi s\,}{1+e^{6\pi s}}-\int_{0}^{+\infty}\frac{6\pi\{s\}}{1+e^{6\pi s}}$$
and the first integral in the RHS equals $\frac{\pi}{72}$ by the residue theorem, while expanding the fractional part as its Fourier series, $\{s\}=\frac{1}{2}-\sum_{n\geq 1}\frac{\sin(2\pi n s)}{\pi n}$, we get:
$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{6\pi\{s\}\,ds}{1+e^{6\pi s}}&=&\frac{\log 2}{2}-\sum_{n\geq 1}\int_{0}^{+\infty}\frac{6 \sin(2\pi n s)}{n(e^{6\pi s}+1)}\,ds\\&=&\frac{\log 2}{2}-\sum_{n\geq 1}\frac{6}{n}\sum_{m\geq 0}(-1)^m\int_{0}^{+\infty}\sin(2\pi n s)\,e^{-6\pi m s}\,ds\\&=&\frac{\log 2}{2}-\frac{3}{\pi}\sum_{n\geq 1}\sum_{m\geq 0}\frac{(-1)^m}{9 m^2+n^2}\\&=&\frac{\log 2}{2}-\frac{\pi}{2}-\frac{3}{\pi}\sum_{m,n\geq 1}\frac{(-1)^m}{9m^2+n^2}\end{eqnarray*}$$
and the last series just depends on the number of ways to represent a positive integer $\not\equiv 2\pmod{3}$ through the binary quadratic form $n^2+9m^2$: it is, with minor manipulations, just a Dirichlet convolution. I have just applied the same techniques of this answer, just in reverse.
This shows a clear connection between the evaluation of the Dedekind eta function at quadratic irrationals and the class number problem: $\eta(\sqrt{-N})$ depends on $\sum_{n\geq 1}(-1)^n\frac{r(n)}{n}$, where $r(n)$ counts the number of ways to represent $n$ as $a^2+Nb^2$. If $N$ is a square or $a^2+Nb^2$ is the only reduced quadratic form of discriminant $-4N$ (class number one) we may explicitly compute such series, and it turns out that $F(\sqrt{-N})$ is an algebraic number. Otherwise, $\sum_{n\geq 1}(-1)^n\frac{r(n)}{n}$ is not even a convolution of Dirichlet series, hence your conjecture is very likely to hold.
Ultimately, the computation of $\eta(6i)$ can be carried on by recalling that:
$$j(\tau)=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^8+2^8\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^{3} $$
and by computing the Klein $j$-invariant $j(3i)$. The Wikipedia page gives:
$$ j(3i) = \frac{1}{27}(2+\sqrt{3})^2(21+20\sqrt{3})^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 5,
"answer_id": 3
} |
The number of ways to write $10$ as the sum of five natural numbers not equal to $3$ How many answers are there for the equation
$$x_1+x_2+x_3+x_4+x_5=10$$
given that $x_1,x_2\dots x_5\in\Bbb{Z^{0+}}\setminus\{3\}$.
| What's easy to compute is we know $j$ of them is equal to $3$, say the result is $S_j$
$$S_j=\binom{5}{j}\binom{10-3j+(5-j-1)}{(5-j-1)}=\binom{5}{j}\binom{14-4j}{4-j},$$
notice that $0\le j\le 3.$ And that $(10-3j)$, $(5-j-1)$ are what called Stars, Bars respectively. By exactly-IEP we have
$$\begin{align}E_0&=\sum_{j=0}^{3}(-1)^j\binom{j}{0}S_j\\
&=\sum_{j=0}^3(-1)^j\binom{5}{j}\binom{14-4j}{4-j}\\
&=+\binom{14}{4}-\binom{5}{1}\binom{10}{3}+\binom{5}{2}\binom{6}{2}-\binom{5}{3}\binom{2}{1}\\
&=531.\quad\quad\square
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Summation by Parts to Evaluate $\sum_{k=1}^{\infty}(2k+1)x^{2k}$ I need to evaluate $\sum_{k=1}^{\infty}(2k+1)x^{2k}$ using the Summation by Parts (SBP) method. It is given that $0 < |x| < 1$.
The notation our class uses for SBP is as follows:
$$ \sum_{i} u_i\cdot\triangle v_i = u_i\cdot v_i - \sum_{i} \triangle u_i \cdot v_{i+1}$$
where $u_i$ and $v_i$ are both sequences.
I have already evaluated the sum using differentiation, but when I attempt SBP I do not get the same result.
Here are my steps for differentiation, with the correct result:
$$ \frac{d}{dx} \left(x^{2k+1}\right) = (2k+1)x^{2k} $$
$$ \sum_{k=1}^{\infty} x^{2k+1} = \boldsymbol{x} + x^3 + x^5 + ... - \boldsymbol{x} = x(1 + x^2 + (x^2)^2 + ...) - x $$
$$ = x\left(\frac{1}{1-x^2}\right) - x = \frac{x^3}{1-x^2}$$
$$ \sum_{k=1}^{\infty} (2k+1)x^{2k} = \frac{d}{dx}\left( \sum_{k=1}^{\infty} x^{2k+1}\right) = \frac{d}{dx}\left( \frac{x^3}{1-x^2}\right) = \boxed{\frac{3x^2-x^4}{(1-x^2)^2}}$$
And here are my steps for SBP, with a different result:
$$\sum_{k=1}^{\infty}(2k+1)x^{2k} $$
$$ u_k = 2k+1, \triangle u_k = 2(k+1) + 1 - (2k + 1) = 2 $$
$$ \triangle v_k = v_{k+1} - v_k = \frac{x^{2(k+1)}}{x^2-1} - \frac{x^{2k}}{x^2-1} = \frac{x^{2k}(x^2-1)}{x^2-1} = x^{2k} $$
$$\sum_{k=1}^{\infty}(2k+1)x^{2k} = (2n+1)\left( \frac{x^{2n}}{x^2-1} \right)\bigg|_{n\rightarrow\infty} - \sum_{k=1}^{\infty} 2\left(\frac{x^{2k+2}}{x^2-1}\right)$$
$$ = 0 - \sum_{k=1}^{\infty} 2\left(\frac{x^{2k+2}}{x^2-1}\right) \textit{ since } |x| < 1$$
$$ = -\frac{2x^2}{x^2-1}\sum_{k=1}^{\infty} x^{2k} = -\frac{2x^2}{x^2-1}\left( \frac{x^2}{1-x^2} \right) = \frac{2x^2}{1-x^2}\left( \frac{x^2}{1-x^2} \right) = \boxed{\frac{2x^4}{(1-x^2)^2}}$$
Can anyone point out my mistake(s)?
| By setting $v_k = \frac{x^{2k}}{x^2-1}$ we have $v_{k+1}-v_k = x^{2k}$, hence:
$$\begin{eqnarray*} \sum_{k=1}^{n} (2k+1)(v_{k+1}-v_{k}) &=& (2n+1)(v_{n+1}-v_1)-\sum_{k=1}^{n\color{red}{-1}}2(v_{k+1}-v_1)\\&=&\color{red}{-v_1}+(2n+1)v_{n+1}-2\sum_{k=1}^{n-1}v_{k+1}\end{eqnarray*} $$
and by letting $n\to +\infty$:
$$ \sum_{k=1}^{+\infty}(2k+1)x^{2k} = \frac{x^2}{1-x^2}+\frac{2}{1-x^2}\sum_{k\geq 1}x^{2k}=\frac{x^2(1-x^2)+2x^2}{(1-x^2)^2}.$$
By the way, the fastest method (IMHO) to compute such a series is neither differentiation nor summation by parts. If we denote with $\Delta$ the backward difference operator, $\Delta a_{n} = a_{n}-a_{n-1} $, we have:
$$ (1-x)\sum_{n\geq 0} a_n x^n = \sum_{n\geq 0} \Delta a_n x^n. $$
Assuming $a_n=2n+1$, that is a polynomial of degree $1$ in $k$, $\Delta^2 a_n=0$ for any $n\geq 2$.
It follows that:
$$ (1-x^2)^2 \sum_{k\geq 1}(2k+1)x^{2k} = c_0 + c_1 x^2 + c_2 x^4$$
where $c_0,c_1,c_2$ are easy to find.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is the maximum of $\sum\limits_{\mathrm{cyc}} \frac{1}{3a + 5b + 7c}$ equal to $\frac{\sqrt{3}}{5}$?
Let $a,b,c>0$ such $ab+bc+ac=1$. Show that
$$\dfrac{1}{3a+5b+7c}+\dfrac{1}{3b+5c+7a}+\dfrac{1}{3c+5a+7b}\le\dfrac{\sqrt{3}}{5}.$$
(Note: $\mathrm{LHS} = \mathrm{RHS}$ when $a = b = c = 1/\sqrt 3$.)
since dear Mac sir,he solve with inequality $\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}\le\frac{\sqrt{3}}{4}$
use same methods,I think
$$(3a+5b+7c)^2\ge 75=75(ab+bc+ac)?$$
I would appreciate any help. Thanks in advance.
| Fact 1: If $x, y, z > 0$ with $-71(x^2 + y^2 + z^2) + 83(xy + yz + zx) = 2700$, then
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \le \frac{\sqrt 3}{5}.$$
(The proof is given at the end.)
Now, let
\begin{align*}
x &= 3a + 5b + 7c, \\
y &= 3b + 5c + 7a, \\
z &= 3c + 5a + 7b.
\end{align*}
We have $x, y, z > 0$ and
$$-71(x^2 + y^2 + z^2) + 83(xy + yz + zx)
= 2700(ab + bc + ca) = 2700.$$
By Fact 1, we have
$$\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}
\le\frac{\sqrt{3}}{5}.$$
We are done.
$\phantom{2}$
Proof of Fact 1:
Letting $x = 5u\sqrt 3, y = 5v\sqrt 3, z = 5w \sqrt 3$, it suffices to prove the following:
Fact 2: If $u, v, w > 0$ with $-71(u^2 + v^2 + w^2) + 83(uv + vw + wu) = 36$, then
$$3uvw \ge uv + vw + wu.$$
Let $p = u + v + w, q = uv + vw + wu, r = uvw$.
The condition $-71(u^2 + v^2 + w^2) + 83(uv + vw + wu) = 36$
is written as $-71(p^2 - 2q) + 83q = 36$ or
$$q = \frac{36 + 71p^2}{225}. \tag{1}$$
Using $p^2 \ge 3q$ and (1), we have
$$p \ge 3.$$
It suffices to prove that
$$3r \ge q.$$
Using degree three Schur, we have
$$r \ge \frac{4pq - p^3}{9}.$$
It suffices to prove that
$$3 \cdot \frac{4pq - p^3}{9} \ge q$$
or (using (1))
$$\frac{1}{225}(p - 3)(59p^2 - 36p + 36) \ge 0$$
which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $ $$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $$
How can I solve this one, I mean I get something like this:
$-3+\left(-1+2\cos ^2\left(x\right)\right)^22+2\left(-3\cos \left(x\right)+4\cos ^3\left(x\right)\right)^2+2\cos ^2\left(x\right)=0$
This equation seems rather hard to solve from here, any tips or other ways to come to an solution?
| Let $\mu = \cos 2x$ for shortness.
Squaring
$$
\cos (2x\pm x) = \cos 2x \cos x \mp \sin 2x \sin x
$$
we obtain
$$
\cos^2 (2x\pm x) = \cos^2 2x \cos^2 x + \sin^2 2x \sin^2 x \mp \text{cancelling terms} = \\
= \mu^2 \frac{1+\mu}{2}+(1-\mu^2)\frac{1-\mu}{2} \mp \text{cancelling terms} = \\ =
\mu^3 - \frac{\mu}{2} + \frac{1}{2} \mp \text{cancelling terms}
$$
So the original equation becomes
$$
2\mu^3 + \mu^2 - \mu + 1 = \frac{3}{2}\\
2\mu^3 + \mu^2 - \mu - \frac{1}{2} = 0\\
(2\mu+1)(2\mu^2 - 1) = 0
$$
with solutions
$$
\mu_1 = -\frac{1}{2},\quad \mu_{2,3} = \pm \frac{1}{\sqrt{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
What is the value of $x*y$?
Given that $$\left(\frac{x}{y}\right)^{-2} + \left(\frac{y}{x}\right)^{-2} = \frac{10}{3}$$
find the value of $x*y$.
My question is, can we calculate the value of $x*y$ or not? If yes, then how? If not, then why?
| As @Meelo said, if (x,y) is a solution so is (cx,cy) when c is not equal to zero.
$$\frac {y^2}{x^2} + \frac {x^2}{y^2} = \frac {y^4+x^4}{x^2y^2}$$
$$\frac{(cy)^2}{(cx)^2}+\frac{(cx)^2}{(cy)^2} = \frac{c^4(y^4+x^4)}{c^4x^2y^2} = \frac{x^4+y^4}{x^2y^2}$$
$$cx\cdot cy = c^2xy$$
Thus we can not find any single value for $xy$ because there are infinite solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$?
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$ ?
$7 \equiv 3 \pmod 4$
$7^2 \equiv 9 \pmod 4\equiv 1 \pmod 4$
$(7^2)^{16} \equiv 1^{16} \pmod 4$
i.e $7^{32} \equiv 1 \pmod 4$
Similarly $9 \equiv 1 \pmod 4$ implies $9^{45} \equiv 1 \pmod 4$.
But the problem arise with the coefficients and addition sign.
what to do?
| $\phi(4) = 2$ and since $(4,7), (4,9)$ are relatively prime we have $[7^2]_4 = [1]_4$, $[9^2]_4 = [1]_4$ and so
$[6 \cdot 7^{32}+7 \cdot 9^{45}]_4 = [6]_4+[7 \cdot 9]_4=[69]_4 = [1]_4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Least Common Denominator: $ \frac{\sqrt{x}}{x}+\frac{\ln\ x}{2\sqrt{x}} $ $$ \frac{\sqrt{x}}{x}+\frac{\ln \ x}{2\sqrt{x}} $$
I have tried combining these two fractions; however, I keep getting stuck.
$$\frac{2\sqrt{x}}{2\sqrt{x}}\cdot\frac{\sqrt{x}}{x}+\frac{\ln\ x}{2\sqrt{x}}\cdot \frac{x}{x}$$
So that we get
$$\frac {2x}{2\sqrt{x}\cdot(x)}+\frac {x\ln\ x}{2\sqrt{x}\cdot(x)}$$
What am I doing wrong? The answer is $$ \frac{2+\ln\ x}{2\sqrt{x}}$$
| Hint:
$\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
Sometimes one representation is more convenient for what you're trying to do than the other is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{ 5^{125}-1}{ 5^{25}-1}$ is a composite number Prove that $\dfrac {\left( 5^{125}-1\right)}{\left( 5^{25}-1\right)}$ is composite number using number theory. Do not use calculator or Wolfram alpha or anything like that.
| Put $x = 525$. Then $\frac{5^{125}-1}{5^{25}-1} = x^4+x^3+x^2+x+1 = (x^2+3^x+1)^2 - 5x(x+1)^2$ which has factor $(x^2+3x+1-513x-513)$ which is obviously $> 1$ and $< x^4+x^3+x^2+x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
} |
Logarithmic Integral I Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{ \ln^{2}x}{(x^{2} - x + 1)^{2}} \, dx.
\end{align}
It is speculated that the value is
\begin{align}
I = \frac{10 \, \pi^{3}}{3^{5} \, \sqrt{3}} - \zeta(2) + \frac{1}{18} \, \left( \psi^{(1)}\left(\frac{1}{3}\right) + \psi^{(1)}\left(\frac{1}{6}\right) \right).
\end{align}
Is it possible to obtain a demonstration of this result?
| Write $\dfrac{1}{(x^2-x+1)^2}$ as a Taylor series $\sum_{j=0}^\infty a_j x^j$. The coefficients can be written as
$$ a_j = \dfrac{j+1}{3} b_j + \dfrac{2}{3} c_j $$
where $b_j$ repeats $1,2,1,-1,-2,-1,\ldots$ for $j = 0,1,2,\ldots$
while $c_j$ repeats $1,1,0,-1,-1,0,\ldots$, both with period $6$.
Now $$\int_0^1 \log^2(x)\; x^k\; dx = \dfrac{2}{(k+1)^3}$$
so your integral becomes
$$ \dfrac{2}{3} \sum_{j=0}^\infty \dfrac{b_j}{(j+1)^2} + \dfrac{4}{3} \sum_{j=0}^\infty \dfrac{c_j}{(j+1)^3} $$
For $j \equiv 0 \mod 6$ we have $b_j = 1$, $c_j = 1$ and the contribution is $$ \eqalign{\dfrac{2}{3} \sum_{i=0}^\infty \dfrac{1}{(6i+1)^2} + \dfrac{4}{3} \sum_{i=0}^\infty \dfrac{1}{(6i+1)^3} &= \dfrac{1}{54} \sum_{i=0}^\infty \dfrac{1}{(i+1/6)^2} + \dfrac{1}{162} \sum_{i=0}^\infty \dfrac{1}{(i+1/6)^3}\cr & = \dfrac{\psi^{(1)}(1/6)}{54} + \dfrac{\psi^{(2)}(1/6)}{324}} $$
Similarly we can compute the contributions for $j = 1$ to $5 \mod 6$.
I get (with Maple's help) a total of
$$ \dfrac{\pi^2}{162} +
{\frac {\psi^{(1)} \left(1/6 \right) }{54}}-{\frac {\psi^{(2)} \left(1/6
\right) }{324}}+\dfrac{\psi^{(1)} \left( 1/3 \right)}{27} -{\frac {\psi^{(2)} \left(
1/3 \right) }{324}}-{\frac {\psi^{(1)} \left(
2/3 \right) }{54}}+{\frac {\psi^{(2)} \left( 2/3 \right) }{324}}-\dfrac{
\psi^{(1)} \left(5/6 \right)}{27} +{\frac {\psi^{(2)} \left( 5/6 \right) }{324}}
$$
Using the reflection identities for $\psi$ this becomes
$$
\dfrac{\psi^{(1)}(1/6)}{18} + \dfrac{\psi^{(1)}(1/3)}{18}
- \dfrac{\pi^2}{6} +{\frac {20\,{\pi }^{3}\sqrt {3}}{729}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I prove $\pi=e^{3/2}\prod_{n=2}^{\infty}e\left(1-\frac{1}{n^2}\right)^{n^2}$? I am interested about some infinite product representations of $\pi$ and $e$ like this.
Last week I found this formula on internet
$$\pi=e^{3/2}\prod_{n=2}^{\infty}e\left(1-\frac{1}{n^2}\right)^{n^2}$$
which
looks like unbelievable.
(I forgot the link but I am sure that this is the formula.)
How can I start to prove this formula?
Thank You.
| You may write, for $N \geq 2$,
$$
\begin{align}
e^{3/2}\prod_{n=2}^{N}e\left(1-\dfrac{1}{n^2}\right)^{n^2}&=e^{3/2}\times\prod_{n=2}^{N}e\times\prod_{n=2}^{N}\left(1-\dfrac{1}{n^2}\right)^{n^2}\\\\
&=e^{3/2}\times e^{N-1}\times\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{n^2}}\dfrac{(n+1)^{n^2}}{n^{n^2}}\\\\
&=e^{3/2}\times e^{N-1}\times\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{(n+1)^2}}\times\dfrac{(n+1)^{n^2}}{n^{(n-1)^2}}\times n^2\\\\
&=e^{N+1/2}\times\color{blue}{\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{(n+1)^2}}}\times\color{#C00000}{\prod_{n=2}^{N}\dfrac{(n+1)^{n^2}}{n^{(n-1)^2}}}\times \color{green}{\prod_{n=2}^{N}n^2}\\\\
&=e^{N+1/2}\times\color{blue}{\dfrac{1}{N^{(N+1)^2}}}\times\color{#C00000}{\dfrac{(N+1)^{N^2}}{2}}\times \color{green}{ (N!)^2}\\\\
&=\frac12\times e^{N+1/2}\times \left(1+\frac1N\right)^{N^2}\times\dfrac{ (N!)^2}{N^{2N+1}}. \tag1
\end{align}
$$ Then one may observe that, as $N \to +\infty$,
$$
N^2 \ln \left(1+\frac1N \right)=N-\frac{1}{2}+O\left(\frac1N\right)
$$ gives
$$
e^{N+1/2}\times\left(1+\frac1N\right)^{N^2}=e^{2N}\left(1+O\left(\frac1N\right)\right) \tag2
$$ and from the Stirling formula, we get
$$
\begin{align}
(N!)^2&=2\pi \;N^{2N+1}e^{-2N}\left(1+O\left(\frac1N\right)\right)
\end{align}
$$
$$
\begin{align}
\frac{(N!)^2}{N^{2N+1}}&=2\pi \;e^{-2N}\left(1+O\left(\frac1N\right)\right).\tag3
\end{align}
$$ By combining $(1)$, $(2)$ and $(3)$ we obtain, as $N \to +\infty$,
$$
e^{3/2}\prod_{n=2}^{N}e\left(1-\dfrac{1}{n^2}\right)^{n^2} = {\large \pi} \left(1+O\left(\frac1N\right)\right)
$$
leading to the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\int _0^\infty \frac{\ln x}{(x^2+1)^2}dx$ Calculate $$\int _0^\infty \dfrac{\ln x}{(x^2+1)^2}dx.$$
I am having trouble using Jordan's lemma for this kind of integral. Moreover, can I multiply it by half and evaluate $\frac{1}{2}\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$?
| Here is a 'real-analysis route'.
Step 1. We have $$ \int_0^{+\infty}\frac{\ln x}{x^2+1} dx=0 \tag1$$ as may be seen by writing
$$
\begin{align}
\int_0^{+\infty}\frac{\ln x}{x^2+1} dx&=\int_0^1\frac{\ln x}{x^2+1} dx+\int_1^{+\infty}\frac{\ln x}{x^2+1} dx\\\\
&=\int_0^1\frac{\ln x}{x^2+1} dx-\int_0^1\frac{\ln x}{\frac{1}{x^2}+1} \frac{dx}{x^2}\\\\
&=0.
\end{align}
$$
Step 2. Assume $a>0$. We have $$ \int_0^{+\infty}\frac{\ln x}{x^2+a^2} dx=\frac{\pi}{2}\frac{\ln a}{a} \tag2$$
as may be seen by writing
$$
\begin{align}
\int_0^{+\infty}\frac{\ln x}{x^2+a^2} dx&= \int_0^{+\infty}\frac{\ln (a \:u)}{(a \:u)^2+a^2} (a \:du)\\\\
&=\frac1a\int_0^{+\infty}\frac{\ln a +\ln u}{u^2+1} du\\\\
&=\frac{\ln a}{a}\int_0^{+\infty}\frac{1}{u^2+1} du+\frac1{a}\int_0^{+\infty}\frac{\ln u}{u^2+1} du\\\\
&=\frac{\ln a}{a}\times \frac{\pi}2+\frac1{a}\times 0\\\\
&=\frac{\pi}{2}\frac{\ln a}{a}.
\end{align}
$$
Step 3. Assume $a>0$. We have $$ \int_0^{+\infty}\frac{\ln x}{(x^2+a^2)^2} dx=\frac{\pi}{4}\frac{\ln a-1}{a^3} \tag3$$
since sufficient conditions are fulfilled to differentiate both sides of $(2)$.
Putting $a:=1$ in $(3)$ gives
$$ \int_0^{+\infty}\frac{\ln x}{(x^2+1)^2} dx=\color{blue}{-\frac{\pi}{4}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
The inequality $\frac{MA}{BC}+\frac{MB}{CA}+\frac{MC}{AB}\geq \sqrt{3}$ Given a triangle $ABC$, and $M$ is an interior point. Prove that:
$\dfrac{MA}{BC}+\dfrac{MB}{CA}+\dfrac{MC}{AB}\geq \sqrt{3}$.
When does equality hold?
| Another solution. Let $G$ be the centroid of the triangle $ABC$. It's well-known that $|GA|^2 = \dfrac{2(b^2+c^2)-a^2}{9}$, where $AB =c, BC =a, CA = b$. Thus,
$$\sum_{a,b,c}\frac{MA}{a} = \sum_{a,b,c}\frac{\sqrt 3|MA||GA|}{\sqrt 3a|GA|}\geq\sum_{a,b,c}\dfrac{2\sqrt 3|MA||GA|}{a^2+3|GA|^2} = \sum_{a,b,c}\frac{3\sqrt 3|MA||GA|}{a^2+b^2+c^2} \\ \geq
3\sqrt 3\sum_{a,b,c}\frac{\vec {MA}\cdot\vec {GA}}{a^2+b^2+c^2} = 3\sqrt 3\sum_{a,b,c}\frac{|GA|^2+|GB|^2+|GC|^2}{a^2+b^2+c^2} = \sqrt 3$$, where in the last chain we have used the fact that $\vec {GA}+\vec{GB}+\vec{GC} = \vec 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$ If $$x+y+z=7$$ and $$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{10}$$ Find the value of $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$$
I tried but I got nothing
| \begin{align*}
\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} &= \frac{7-(y+z)}{y+z} + \frac{7-(x+z)}{x+z} + \frac{7-(x+y)}{x+y} \\
&=\left( \frac{7}{y+z} - 1 \right)+ \left(\frac{7}{x+z} - 1 \right) + \left(\frac{7}{x+y} - 1 \right) \\
&=7 \left( \frac{1}{y+z} + \frac{1}{x+z} +\frac{1}{x+y} \right) - 3\\
&=7 \left(\frac{7}{10} \right) - 3\\
&= \boxed{\frac{19}{10}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Find the Laplace inverse of the following. $$
\frac{2s+5}{s^2+6s+34}
$$
I am stuck on this part:
Wolfram has the step by step showing that you simply split up the original fraction into
$$
\frac{2s}{s^2+6s+34} + \frac{5}{s^2+6s+34}
$$
and then it solves it. But that doesn't help. Could someone please help me understand how to do this problem?
| Consider the following.
\begin{align}
\frac{2s+5}{s^2+6s+34} = \frac{2 s + 5}{ (s+3)^{2} + 5^{2}} = 2 \, \frac{(s+3)}{(s+3)^{2} + 5^{2}} - \frac{1}{5} \, \frac{5}{(s+3)^{2} + 5^{2}}
\end{align}
Now, for
\begin{align}
f(s) \doteqdot \int_{0}^{\infty} e^{-st } \, f(t) \, dt,
\end{align}
then
\begin{align}
\frac{2s+5}{s^2+6s+34} \doteqdot 2 e^{-3t} \, \cos(5 t) - \frac{1}{5} \, e^{-3t} \, \sin(5t).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Definite integration. $$\int _0^1 \arctan(x^2-x+1)\,dx$$
ATTEMPT:
$\int \arctan(x^2-x+1)\,dx$
Let
$\arctan(x^2-x+1)=u$ and $dx=dv$:
$$du=\frac{2x−1}{(x^2−x+1)^2+1}dx= \frac{2x−1}{(x^2+1)(x^2-2x+2)}$$ and $v=x$.
Now from Integration by parts $I=uv-$$\int vdu.$
$$I=\arctan(x^2-x+1)x-\int \frac{x(2x−1)}{(x^2+1)(x^2-2x+2)}.$$
Now using partial fractions
$$
\int\frac{x(2x−1)}{(x^2+1)(x^2-2x+2)}=\int \frac{x}{(x^2+1)}-\frac{x}{x^2-2x+2} $$
Integrating term by term and finally arranging the result:
$$I=x\arctan(x^2−x+1)+\frac{1}{2}\ln(∣x^2+1∣)−\frac{1}{2}\ln(∣x^2−2x+2∣)−\arctan(x−1).$$
Now substituting the limit i got $I=\ln2.$
Is there any other way to approach this problem like by using the properties of definite integration as this method is long and tedious but though it works!!
| This response is similar in spirit to @nospoon's answer, but I think it is perhaps more elementary because it doesn't require the anti-derivative of arctangent.
$$\begin{align}
I
&=\int_{0}^{1}\arctan{\left(x^2-x+1\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}--\arctan{\left(\frac{1}{x^2-x+1}\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}-\arctan{\left(x\right)}-\arctan{\left(1-x\right)}\right]\,\mathrm{d}x\\
&=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}x-\int_{0}^{1}\arctan{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\arctan{\left(1-x\right)}\,\mathrm{d}x\\
&=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}x-2\int_{0}^{1}\arctan{\left(x\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}-2\arctan{\left(x\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\operatorname{arccot}{\left(x\right)}-\arctan{\left(x\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{2x}{1+x^2}\,\mathrm{d}x\\
&=\ln{\left(1+x^2\right)}\bigg{|}_{0}^{1}\\
&=\ln{(2)}.\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $ \lim_{x\to 0} ({\frac{1}{1-\cos(x)}} - {\frac{2}{x^2}})$ without L'hopital's rule? $$ \lim_{x\to 0} ({\frac{1}{1-\cos(x)}} - {\frac{2}{x^2}})$$
I know I can so some L'hopital's rule and get the answer, but is there another way, more clever? Like with Taylor expansion? I tried it and got $\infty*0$
| An alternative view is the following.
\begin{align}
\frac{1}{1- \cos x} - \frac{2}{x^{2}} &= \frac{1}{1 - \left(1 - \frac{x^{2}}{2!} - \frac{x^{4}}{4!} + \cdots \right)} - \frac{2}{x^{2}} \\
&= \frac{2}{x^{2}} \, \left[ \frac{1}{1 - \frac{2 \, x^{2}}{4!} + \frac{2 \, x^{4}}{6!} - \cdots } - 1 \right].
\end{align}
Now, by "long division" it is seen that
\begin{align}
\frac{1}{1 - \frac{2 \, x^{2}}{4!} + \frac{2 \, x^{4}}{6!} - \cdots } = 1 + \frac{2 \, x^{2}}{4!} + \frac{3 \, x^{4}}{6!} + \mathcal{O}(x^{6})
\end{align}
for which
\begin{align}
\frac{1}{1 - \cos x} - \frac{2}{x^{2}} &= \frac{2}{x^{2}} \, \left[ \left( 1 + \frac{2 \, x^{2}}{4!} + \frac{3 \, x^{4}}{6!} + \mathcal{O}(x^{6}) \right) - 1 \right] \\
&= \frac{1}{3!} \, \left[ 1 + \frac{x^{2}}{20} + \mathcal{O}(x^{4}) \right].
\end{align}
By taking the limit as $x \to 0$ leads to
\begin{align}
\lim_{x \to 0} \left\{ \frac{1}{1- \cos x} - \frac{2}{x^{2}} \right\} = \frac{1}{3!}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How do i evaluate this $\int^\frac{\pi}{2}_0 \frac{\sec^2x}{(\sec x+\tan x)^{n}}\,dx = ? ?$ Is there someone who can give me a fast way to evaluate this :$$\int^\frac{\pi}{2}_0 \frac{\sec^2x}{(\sec x+\tan x)^{n}}\,dx = ? $$
Note :what I think the result is : $\frac{n}{n²-1} $ but how if it were true ?
Thank you for any help.
| Maybe this answer from AOPS is helpful to you:
$I =\int_0 ^\frac{\pi}{2}\frac{\sec^{2}x}{(\sec x+\tan x)^{n}}\ dx = \int_0 ^\frac{\pi}{2}\frac{1}{(\sec x+\tan x)^{n}} \cdot \frac{1}{\cos^{2}x}\ dx$
$I = \int_0 ^\frac{\pi}{2}\frac{1}{\cos x}\cdot \frac{1}{(\sec x+\tan x)^{n}} \cdot \frac{1}{\frac{\sin x +1}{\cos x}}\cdot \frac{\sin x +1}{\cos^{2}x}\ dx$.
Integration by parts: $u =\frac{1}{\cos x}$ and $dv = \frac{1}{(\sec x+\tan x)^{n+1}} \cdot \frac{\sin x +1}{\cos^{2}x}\ dx$,
substitution: $\sec x+\tan x=t$, then $\frac{\sin x +1}{\cos^{2}x}dx =dt$,
so $v = \int t^{-n-1}\ dt$ and $v = \frac{t^{-n}}{-n}$.
$I = \left[\frac{1}{\cos x}\cdot \frac{-1}{n(\sec x+\tan x)^{n}}\right]_0 ^\frac{\pi}{2}+\frac{1}{n}\int_0 ^\frac{\pi}{2}\frac{1}{(\sec x+\tan x)^{n}}\cdot \frac{\sin x}{\cos^{2}x}\ dx$.
$I = \frac{1}{n}+\frac{1}{n}\int_0 ^\frac{\pi}{2}\frac{\sin x + 1 - 1}{(\sec x+\tan x)^{n}\cos^{2}x}\ dx$.
$I = \frac{1}{n}-\frac{1}{n}I +\frac{1}{n}\int_0 ^\frac{\pi}{2}\frac{\sin x + 1}{(\sec x+\tan x)^{n}\cos^{2}x}\ dx$.
The same substitution:
$I = \frac{1}{n}-\frac{1}{n}I +\frac{1}{n}\left[\frac{1}{(1-n)(\sec x+\tan x)^{n-1}}\right]_0 ^\frac{\pi}{2}$.
$n \cdot I = 1-I +\left[\frac{1}{(1-n)(\sec x+\tan x)^{n-1}}\right]_0 ^\frac{\pi}{2}$.
$(n+1)\cdot I = 1 -\frac{1}{n-1}\left[\frac{1}{(\sec x+\tan x)^{n-1}}\right]_0 ^\frac{\pi}{2}$.
$(n+1)\cdot I = 1 +\frac{1}{n-1} = \frac{n}{n-1}$.
$I = \frac{n}{n^{2}-1} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\tan \theta = 3\frac{15}{16}$, then find $\sin \theta$ If $\tan \theta = 3\cfrac{15}{16}$, then find $\sin \theta$.
| Let us rewrite $3 \frac{15}{16}$ as an improper fraction. So we get $3 \dfrac{15}{16} = \dfrac{16\cdot3+15}{16} = \dfrac{63}{16}$
We know that $\tan(\theta) = \dfrac{o}{a}$, where $o$ and $a$ are the opposite and adjacent sides of $\theta$. We also know through the Pythagorean theorem that the hypotenuse is $\sqrt{o^2+a^2} = \sqrt{63^2+16^2} = 65$. Recall that:
$$\sin(\theta) = \frac{o}{h}$$ where $h$ is the hypotenuse and $o$ is the opposite side. Subbing in the values we obtained and get:
$$\sin(\theta) = \frac{63}{65}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Divergent series whose terms converge to zero I've just begun to teach my class series. We usually have workshops where they work in groups on a tougher problem, and I was thinking of asking them to come up with a divergent series whose terms converge to $0$. While $\sum 1/n$ is the prototypical example, we haven't done the integral test yet. Do you know of any simpler examples?
| The harmonic series is pretty simple and you do not need the integral test to show it diverges. Here is a short proof by contradiction:
Suppose $\sum^{\infty } _{n=1}\frac{1}{n}$ converges to a finite number $L$. Then,
$L=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots +\cdots$ and this is greater than
$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}+\frac{1}{8}+\frac{1}{8}+\cdots$. Now group the terms:
$1+\frac{1}{2}+\left ( \frac{1}{4}+\frac{1}{4} \right )+\left ( \frac{1}{6}+\frac{1}{6} \right )+\left ( \frac{1}{8}+\frac{1}{8} \right )+\cdots $ and this latter item is
$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots =L+\frac{1}{2}$ so what we've shown is
$L\geq L+\frac{1}{2}$ which is absurd. Hence, the series does not converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0 \\
\implies x<3\\
x\in \emptyset $
Case $2$ ,for $\boxed{x< 0}\\
\dfrac{x^2+x-12}{x-3}\geq 2x\\
\implies \dfrac{(x-4)(x-3)}{x-3}\leq 0 \\
\implies x\leq 4\\
\implies x< 0\\
$
But the answer given is option $d.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.
| The equality is not true if $x=3$ so cases a),b) and c) can immediately be excluded.
Edit:
The following statements are equivalent:
*
*$\frac{x^{2}-\left|x\right|-12}{x-3}\geq2x$
*$\frac{x^{2}-6x+\left|x\right|+12}{x-3}\leq0$
*$\left[\frac{x^{2}-5x+12}{x-3}\leq0\wedge x\geq0\right]\vee\left[\frac{x^{2}-7x+12}{x-3}\leq0\wedge x<0\right]$
*$\left[\frac{x^{2}-5x+12}{x-3}\leq0\wedge x\geq0\right]\vee\left[\frac{\left(x-3\right)\left(x-4\right)}{x-3}\leq0\wedge x<0\right]$
We find a negative discriminant for $x^{2}-5x+12$ and conclude that
this expression is positive. Then we proceed with the following equivalent
statements:
*
*$\left[x-3<0\wedge x\geq0\right]\vee\left[x-4\leq0\wedge x\neq3\wedge x<0\right]$
*$\left[0\leq x<3\right]\vee\left[x<0\right]$
*$x<3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solving for trigonometric values Find $a+2b$ if $\tan a = 1/7$ and $\sin b=\frac{1}{\sqrt{10}}$. I had tried to solve it by trigonometric ratios but i could not. Please solve it by a method of class10 standards.
| Hint:
$\tan a = \frac{1}{7}$
$\tan b = \frac{1}{3}$
$\tan(a+2b) = \frac{\tan a + \tan 2b}{1-\tan a \tan 2b}$
$\tan 2b = \frac{2tanb}{1-tan^2 b}$
You get $\tan 2b = \frac{6}{8}$
$\tan (a + 2b) = 1$
$a+2b = \arctan{1} = \frac{\pi}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Points on Surface, Distance Optimized How do I find the points on the surface:
$$x^3+y^3+z^3=1$$
such that the distance to the origin is minimized?
My Thoughts:
Perhaps we can minimize the distance squared? Not sure.
| If we take $x=\rho\sin\theta,y=\rho\cos\theta\sin\phi,z=\rho\cos\theta\cos\phi$, then we have to find the minimum of $\rho$ under the constraint:
$$ \rho^3(\sin^3\theta+(\sin^3\phi+\cos^3\phi)\cos^3\theta)=1 $$
or, by switching to the dual problem, the maximum of:
$$ f(\theta,\rho) = \sin^3\theta+(\sin^3\phi+\cos^3\phi)\cos^3\theta.$$
The stationary points of $f(\theta,\rho)$ are given by:
$$ \left\{\begin{array}{rcl}\frac{\partial}{\partial \theta}\,f(\theta,\phi)=-3\cos\theta\sin\theta\left(-\sin\theta+\cos^3\phi\cos\theta+\cos\phi\cos^3\theta\right)&=&0\\\frac{\partial}{\partial \phi}\,f(\theta,\phi)=-3\cos\phi\sin\phi\left(-\sin\phi+\cos^3\phi\cos\theta+\cos\phi\cos^3\theta\right)&=&0\end{array}\right.$$
so, at least in principle, we can locate them. Anyway, since:
$$ 1-\left(\cos^3 x+\sin^3 x\right) = 4\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\sin^2\left(\frac{x}{2}\right)\left(2+\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)\geq 0$$
$$ 1+\left(\cos^3 x+\sin^3 x\right) = 4\sin^2\left(\frac{\pi}{4}+\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)\left(2-\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)\geq 0$$
the range of $g(x)=\sin^3 x+\cos^3 x$ is $[-1,1]$, so the maximum of $f(\theta,\phi)$ is one and the minimum of $\rho$ is $\color{red}{1}$, too. Obviously, it is attained at $(1,0,0),(0,1,0),(0,0,1)$.
For non-believers, here it is a picture of the unit sphere with our surface being a nice hat:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why $\frac{1}{9}$ becomes $\frac{1}{3}$ in $\frac{1}{3} \int \frac{1}{s^2+1} \, ds$? I'm trying to compute:
$$\int\frac{1}{x^2+4x+13} \; dx$$
And I'm following the steps given by W|A here:
I don't understand why the $\cfrac{1}{9}$ becomes $\cfrac{1}{3}$ when the new substitution is made. If I switch the $s$ back to what it was, I'll get $u^2/9$ and I still need to multiply it by 9 to get $u^2+9$. I don't get why I need to change it to $1/3$.
EDIT:
$$u=x+2$$
$$s=\frac{u}{3}=\frac{1}{3}\cdot u$$
$$ds=\overbrace{\left[\frac{1}{3}\right]'u+\frac{1}{3}[u]'}^{\text{Product rule}}$$
$u=x+2$ and $u'=1$, I guess the trick is here: Instead of writing $u'$ in the product rule above as $1$, you write it as $du$.
$$ds=\left[\frac{1}{3}\right]'u+\frac{1}{3}[u]'=0\cdot u+\frac{1}{3}\cdot du=\frac{1}{3} \cdot du$$
$$ds=\frac{1}{3}du$$
"Solving" for $du$, we have:
$$3 ds=du$$
And then we switch the $ds$ with $yds$, where $y$ is whatever we have as a coefficient of $ds$.
| You have
$$
\frac{1}{9}\int \frac{1}{\frac{u^2}{9} + 1} \; du = \frac{1}{9}\int \frac{1}{\left(\frac{u}{3}\right)^2 + 1} \; du.
$$
With the substitution $s = \frac{u}{3}$ you get $ds = \frac{1}{3}du$ or $\color{green}{3 ds} = \color{red}{du}$. So
$$\begin{align}
\frac{1}{9}\int \frac{1}{\left(\frac{u}{3}\right)^2 + 1} \; \color{red}{du}
&=\frac{1}{9}\int \frac{1}{s^2 + 1} \; \color{green}{3ds}\\
&= \frac{\color{green}{3}}{9}\int \frac{1}{s^2 + 1} \; \color{green}{ds}\\
&= \frac{1}{3}\int \frac{1}{s^2 + 1} \; ds
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Solving following quartic equation Solve in $\mathbb{R}$ :
$$(x^2+2)^2+8x^2=6x (x^2+2) $$
My try: I tried to make the graph by calculating values for $x=1, 2, 3, 4$ and I found that the function is positive at $x=0$ but negative at $x=1$, so the graph must have crossed the $x$ axis, thus there will be a root between $0$ and $1$ and similarly this was the case between $3$ and $4$, but I was unable to solve it precisely. I was also unable to find about the other two roots.
What is the way to solve it just by algebra or rough plotting with the help of pen and paper and not using any computational tools?
| $$(x^2+2)^2+8x^2=6x(x^2+2)\Longleftrightarrow$$
$$x^4+4x^2+4+8x^2=6x(x^2+2)\Longleftrightarrow$$
$$x^4+12x^2+4=6x(x^2+2)\Longleftrightarrow$$
$$x^4+12x^2+4=6x^3+12x\Longleftrightarrow$$
$$x^4+12x^2-6x^3-12x+4=0\Longleftrightarrow$$
$$(x^2-4x+2)(x^2-2x+2)=0\Longleftrightarrow$$
$$(x^2-4x+2)=0\vee (x^2-2x+2)=0\Longleftrightarrow$$
$$x^2-4x=-2\vee x^2-2x=-2\Longleftrightarrow$$
$$x^2-4x+4=-2+4\vee x^2-2x+1=-2+1\Longleftrightarrow$$
$$(x-2)^2=2\vee (x-1)^2=-1\Longleftrightarrow$$
$$x-2=\sqrt{2}\vee x-2=-\sqrt{2}\vee -\Longleftrightarrow$$
$$x=2+\sqrt{2}\vee x=2-\sqrt{2}\vee -\Longleftrightarrow$$
"$-$" $\rightarrow$ $(x-1)^2=-1$ has no real solutions
So:
$$(x^2+2)^2+8x^2=6x(x^2+2)\Longleftrightarrow$$
$$x=2+\sqrt{2} \vee x=2-\sqrt{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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How can $p^{q+1}+q^{p+1}$ be a perfect square? How can one find all primes $(p,q)$ such that $p^{q+1}+q^{p+1}$ is a perfect square
I considered it $\mod 2$ and found a trival solution .
Im curious about an eventual answer Diophantine equations are extremely hard.
This seems harder than IMO Q2 of this year .
edit1:
I think one should consider it $\mod 4$ .
| Claim: $(p,q)=(2,2)$ is the only pair of prime solutions.
Proof.
Consider $$ p^{q+1} + q^{p+1} = n^2.$$
With $(p,q)=(2,3),(3,2)$ the LHS equals $43$, a prime number itself.
With $(p,q)=(3,3)$ it equals $2\cdot3^4$, which obviously is not a square as the exponent of $2$ is odd.
With one equal to $2$ and the other $\ge29$, say respectively $p$ and $q$, it is not a square because $$\left(2^{\frac{q+1}{2}}\right)^2<2^{q+1}+q^3<2^{q+1}+2^{\frac{q+3}{2}}+1=\left(2^{\frac{q+1}{2}}+1\right)^2;$$
one can rule out $5\le q < 29$ by trial and error.
Suppose there exist solutions with $p,q\ge5$. They are respectively equal to $6a\pm1$ and $6b\pm1$, for some positive integers $a,b$, so substituting for each of the four cases we have $$\begin{align}(6a+1)^{6b+2}+(6b+1)^{6a+2}&=n^2 \\ \left((6a+1)^{3b+1}\right)^{2}+\left((6b+1)^{3a+1}\right)^{2}&=n^2 \tag{1}\end{align}$$ or $$ \begin{align}(6a-1)^{6b+2}+(6b+1)^{6a}&=n^2 \\ \left((6a-1)^{3b+1}\right)^{2}+\left((6b+1)^{3a}\right)^{2}&=n^2 \tag{2}\end{align}$$ or $$\begin{align} (6a+1)^{6b}+(6b-1)^{6a+2}&=n^2 \\ \left((6a+1)^{3b}\right)^{2}+\left((6b-1)^{3a+1}\right)^{2}&=n^2 \tag{3}\end{align}$$ or $$\begin{align}(6a-1)^{6b}+(6b-1)^{6a}&=n^2 \\ \left((6a-1)^{3b}\right)^{2}+\left((6b-1)^{3a}\right)^{2}&=n^2. \tag{4}\end{align}$$
All of them are pithagorean identities, and as such, one and only one of the numbers squared in their LHS is odd. But each one of them is a power of an odd number, hence an odd number itself. Thus we have a contradiction, and the claim is proved. $ \ \ \ \ \ \ \ \ \ \ \ \text{QED}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Approximation of integration I want to estimate the integral $\int_0^{1/2}\ln(1+ \frac{x^2}{4})$
with error at most $10^{-4}$.
Any help will be appreciated.
I have calculated the power series of $\ln(1+ \frac{x^2}{4})$ which is
$$\sum_{n=0}^{n= \infty} \frac{(-1)^n}{(2n+2)\cdot 2^{2n+1}}x^{2n+2},$$
and radius of convergence $= 1/2$. Now calculate
$$\int_0^{1/2}\sum_{n=0}^{n= \infty} \frac{(-1)^n}{(2n+1)\cdot 2^{2n+1}}x^{2n+1}\,dx = \sum_{n=0}^{n= \infty} \frac{(-1)^n}{(2n+3)(2n+2)\cdot 2^{2n+1}}(\frac{1}{2})^{2n+3}.$$
| The taylor series you put in the comment is wrong. You should get
$$
\ln\left(1+\frac{x^2}{4}\right)= \frac{x^2}{4}-\frac{x^4}{32}+\frac{x^6}{192}-\frac{x^8}{1024}+O(x^{10})
$$
Integrating up to the 2nd power you get
$$
\int_0^{1/2} \frac{x^2}{4}dx = \frac{1}{96} \simeq 0.010417
$$
Integrating up to the 4th power you get
$$
\int_0^{1/2} \frac{x^2}{4}-\frac{x^4}{32} dx = \frac{ 157}{15360 } =0.01022135416
$$
Meanwhile the real result is
$$
\int_0^{1/2} \ln\left(1+\frac{x^2}{4}\right) dx = \frac12 \left(-2+\ln(17/16)+8 \tan^{-1}(1/4)\right) \simeq 0.010227
$$
Since when we integrate up to the 4rth power we left out the terms with order $O(x^{2n})$ for $n> 2$, we get a result with error of order less than $10^{-4}$, indeed
$$
0.010227-0.01022135416= 0.00000564584< 0.0001= 10^{-4}
$$
| {
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Epsilon-delta proof of $\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2}$ I have been doing $\varepsilon$-$\delta$ proofs for fun and I challenged myself to prove $$\displaystyle\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2},\quad a,b\in\mathbb{R}$$
The definition says: We say that $\displaystyle\lim_{x\to\infty}f(x)=l$ if for any positive number $\varepsilon$ we can find a positive number $N$ (depending on $\varepsilon$ in general) such that $|f(x)-l|<\varepsilon$ whenever $x>N$.
So I started with: $\left|\sqrt{(x+a)(x+b)}-x-\dfrac{a+b}{2}\right|<\varepsilon$ whenever $x>N$.
Manipulating the first inequatlity
\begin{gather*}
-\varepsilon<\sqrt{(x+a)(x+b)}-x-\dfrac{a+b}{2}<\varepsilon\\
-\varepsilon+\dfrac{a+b}{2}<\sqrt{(x+a)(x+b)}-x<\varepsilon+\frac{a+b}{2}
\end{gather*}
At this point I thought about adding $x$, squaring the expressions and then expanding them. I did it and I got: $$\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}+ax+bx+x^2-a\varepsilon-b\varepsilon-2x\varepsilon+\varepsilon^2<x^2+ax+bx+ab<\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}+ax+bx+x^2+a\varepsilon+b\varepsilon+2x\varepsilon+\varepsilon^2$$
And here I'm not sure how to proceed. Am I on the right track?
Thanks for any help / hints.
Note: There might be other ways to prove this, but I'd like to do it using just algebra if possible, even if it's not the best method.
| This is not an epsilon delta proof, but here it goes:
$$\left(\sqrt{(x+a)(x+b)}-x\right)= x(\sqrt{(1+\frac{a}{x})(1+\frac{b}{x})}-1) = x(f(x) -1)$$
expanding $g(z) := f(\frac{1}{x})$ around $z=0$: $g(z) = 1 + \frac{a+b}{2}z +\mathcal{O}(z^2)$ therfore
$$ x(f(x) -1) = x(1 + \frac{a+b}{2}\frac{1}{x} + \mathcal{O}(\frac{1}{x^2}) -1) = \frac{a+b}{2} + \mathcal{O}(\frac{1}{x}) \overset{x\rightarrow \infty}{\longrightarrow} \frac{a+b}{2} $$
Ok just now I read your note. Please don't down vote because I'm unable to read posts till the end :)
| {
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Problem using trigonometric substitution: Domain of $\theta$ I'm studying calculus from Rogawski's Calculus. In trigonometric substitution $x=a\sec \theta$ he made a note:
In the substitution $x = a \sec θ$ , we choose $0\le θ \le π$
2 if $x \ge a$ and $π \le θ < \frac{3π}2$ if
$x \le −a$. With these choices, $a \tan \theta$ is the
positive square root $\sqrt{x^2 − a^2}$.
When I work on the integral:
$$\int \frac {\mathrm{d}x}{x\sqrt{x^2-9}}$$
Using the substitution of $x=3\sec\theta$ with the domain of $\theta$ shown above, the integration will be :
$$\int \frac {dx}{x\sqrt{x^2-9}}= \int \frac {3\sec\theta\tan\theta d\theta}{(3\sec\theta)\sqrt{9\sec^2-9}}=\int \frac {\tan\theta d\theta}{3\sqrt{\tan^2 \theta}}$$
$$= \int \frac{d\theta}{3}= \frac\theta 3+ \mathrm{C}$$
$$\int \frac {dx}{x\sqrt{x^2-9}}= \frac 13 \sec^{-1}\left(\frac x3\right)+ \mathrm{C}$$
Which is very wrong in the negative part of the domain of $x$ as shown in this graph:
Notice that the slope of the blue function in the negative domain should be positive not negative!.
The problem of this substitution is $2$ things:
$1)$ The domain of $\theta$ is chosen so that inverse cannot be done, since $\theta =\sec^{-1}x\notin (\pi,\frac{3\pi}2) $ which is the domain chosen for $\theta $.
$2)$ Depending on first problem, we should choose $\theta \in (0,\pi)-\{\frac\pi 2\}$ which makes $\sqrt{\tan^2 \theta}=|\tan \theta|$. The integral then should be re-written as a piecewise function:
$$\int \frac {dx}{x\sqrt{x^2-9}}= \int \frac {3\sec\theta\tan\theta d\theta}{(3\sec\theta)\sqrt{9\sec^2-9}}$$
$$\int \frac {\tan\theta d\theta}{3\sqrt{\tan^2 \theta}} = \begin{cases} = \int \frac {d\theta}{3} = \frac 13 \sec^{-1}(\frac x3)+ C & \text{if $\theta \in (0,\frac \pi 2)$} \equiv x>3 \\= \int \frac {-d\theta}{3} = \frac {-1}3 \sec^{-1}(\frac x3)+ C & \text{if $\theta \in (\frac \pi 2,\pi)$}\equiv x<-3 \end{cases}$$
My questions are:
Is this thinking right ? there is mistake to choose the domain of $\theta$ as mentioned in the book ?
Mathematica gives me the answer of $-\dfrac {1}{3} \tan^{-1} \left(\dfrac{3}{\sqrt{x^2-9}}\right) +\mathrm{C}$ which is right when graphed. But how to construct this integral ?
Thanks for help.
| I am not going to answer your problem on the domain part. Instead, I am suggesting an alternate way to integrating the very last integral.
Edited
Claim-1) $P = \int { \frac {dx}{\sqrt {1 - x^2}}} = ... = \sin ^{-1} x$.
Let $x = \sin u$. Then, $dx = du$.
∴ $P = \int {\frac {(\cos u)du}{\sqrt {1 - \sin ^2 u}}} = u = \sin ^{-1} x$
[The whole Claim-2 can be ignored. I just don't want to delete it.]
Claim-2) $Q = \int {\frac {dx}{\sqrt {a^2 – x^2}}} = … = \sin ^{-1} \frac {x}{a}$.
Let $x = av$. Then, $dx = (a)dv$.
∴ $Q = \int {\frac {(a)dv}{\sqrt {a^2 – (av)^2}}}$
$ = \int {\frac {dv}{\sqrt {1 - v^2}}}$, which is essentially $P$.
$ = \sin ^{-1} v$
$ = \sin ^{-1} (\frac {x}{a})$
Claim-3) $R = \int {\frac {dx}{x \sqrt {x^2 – a^2}}} = … = \frac {-1}{a} \sin ^{-1} \frac {a}{x}$.
Further edited
Let $w = \frac {a}{x}$. Then, $dx = \frac {(-a) dw}{w^2}$
∴ $R = \int {\frac {-a (dw)/(w^2)}{(a/w) \sqrt {(a/w)^2 - a^2}}}$
$= - \int {\frac {dw}{w \sqrt{(a/w)^2 -a^2}}}$
$= - \int {\frac {dw}{aw \sqrt {(1/w)^2 - 1^2}}}$
$= \frac {-1}{a} \int {\frac {dw}{\sqrt {1^2 - w^2}}}$
$= \frac {-1}{a} P$
$= \frac {-1}{a} \sin ^{-1} w$
$= \frac {-1}{a} \sin ^{-1} \frac {a}{x}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove using $ \varepsilon-\delta $ that $ \lim_{(x,y)\to(1,1)} \frac {x^2+2xy-3y^2}{x^2-y^2} = 2 $
Prove limit using $ \varepsilon-\delta $ definition that: $$ \lim_{(x,y)\to(1,1)} \frac
{x^2+2xy-3y^2}{x^2-y^2} = 2 $$
I've been reading quite a lot about how to prove limits; so I want to show what I've done so far in this one, so you can tell me any suggestion (tricks) and even point out any mistake.
What I've tried:
I want to find $ \delta $ for every $ \varepsilon $ that verifies: $$ 0 < \left \| (x,y) - (1,1) \right \| < \delta \Rightarrow \left | \frac {x^2+2xy-3y^2}{x^2-y^2} - 2 \right | < \varepsilon $$
So here is my attempt:
$$ \begin{align*}
\left | \frac {x^2+2xy-3y^2}{x^2-y^2} - 2 \right | &\overset{(1)}{=} \left | \frac {y-x}{y+x} \right | \\
&\overset{(2)}{\leq} \frac {|x|+|y|}{|x+y|} \\
&\overset{(3)}{\leq} \frac {2\left \| (x,y)-(1,1) \right \|}{|x+y|} \\
&\overset{(4)}{\leq} 4\left \| (x,y)-(1,1) \right \| \\
& < \ 4 \delta
\end{align*} $$
So I can take $ \delta = \varepsilon / 4 $. Is this right?
Justifications:
(1) Basic operations.
(2) Triangle inequality.
(3) I used $ |x| \leq \left \| (x,y)-(1,1) \right \| $.
(4) I supposed $ |x| < 1/2 $ and also $ |y| < 1/2 $ then $ |x+y| < 1/2 $. (I don't understand why this step holds though).
(5) The metric I'm using is: $ \left \| (x,y) \right \| = \sqrt{x^2 + y^2} $
| I always like to let
variables go to zero.
So,
if we let
$x = u+1$
and
$y = v+1$,
$\begin{array}\\
\frac {x^2+2xy-3y^2}{x^2-y^2}
&=\frac {(u+1)^2+2(u+1)(v+1)-3(v+1)^2}{(u+1)^2-(v+1)^2}\\
&=\frac {u^2+2u+1+2(uv+u+v+1)-3(v^2+2v+1)}{(u^2+2u+1)-(v^2+2v+1)}\\
&=\frac {u^2+4u+2uv-3v^2-4v)}{u^2+2u-v^2-2v}\\
&=\frac {u^2+2uv-3v^2+4u-4v}{u^2-v^2+2u-2v}\\
&=\frac {(u-v)(u+3v)+4(u-v)}{(u-v)(u+v)+2(u-v)}\\
&=\frac {(u+3v)+4}{(u+v)+2}
\quad\text{(for }u \ne v)\\
\end{array}
$
The limit is,
therefore,
$\lim_{u \to 0, v \to 0} \frac {(u+3v)+4}{(u+v)+2}
=\frac{4}{2}
=2
$.
In this case,
it was harder.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then.. If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then
(A) $a=c$
(B) either $a=c$ or $a+b+c+d=0$
(C) $a+b+c+d=0$
(D) $a=c$ and $b=d$
I solved $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ and got $a(a+b+d)=c(c+b+d)$ and so I thought that (A) is the correct option. But the correct answer is (B).
I'm how $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ if $a+b+c+d=0$. Please help.
| Using this, $$\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}=\dfrac{1(a+b)+1(c+d)}{1(b+c)+1(d+a)}$$
$\implies(a+b+c+d)\{a+b-(b+c)\}=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Creating a PDF for a discrete random variable with a countably infinite set of values? I am unsure how to transition from discrete random variables with a finite set of values to ones with a countably infinite set of values.
The question that spawned this problem:
A bucket has two white and one black marble. You will continuously draw marbles from the bucket until you get a white, but if you draw a black you put it and an extra black back into the bucket. If we let X = # of draws, find P(X=k) and show the function is a PDF.
X seems to me to be a discrete random variable as the set of values it can take on is {1, 2, 3 ... }, the set of positive integers. However, the set of values is countably infinite, which is preventing me from using the simple steps to show the probability distribution when the set finite.
| The simple thing to do is to calculate small cases, detect a pattern, and generalize.
For instance, the probability that you would only need to draw once is $$\Pr[X = 1] = \frac{2}{3},$$ because the initial state of the bucket is $2$ white and $1$ black. Thus the chance you get a white on the first draw is $2/3$, and the drawing stops upon the first draw of a white marble.
Next, the probability that you would need to draw exactly twice is $$\Pr[X = 2] = \frac{1}{3} \cdot \frac{2}{4} = \frac{1}{6},$$ because the only way to stop at 2 draws is to first fail to draw a white with probability $1/3$, and then to draw a white on the second try, which occurs with probability $2/4$ (as there are now two black marbles as a result of the first failed draw).
And now we can also easily calculate $$\Pr[X = 3] = \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{2}{5} = \frac{1}{15},$$ and $$\Pr[X = 4] = \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{2}{6}.$$ And now the pattern is evident:
$$\Pr[X = n] = \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \ldots \cdot \frac{n-1}{n+1} \cdot \frac{2}{n+2}.$$ Can we simplify this product? Sure: $$\Pr[X = n] = \frac{4(n-1)!}{(n+2)!} = \frac{4}{n(n+1)(n+2)}.$$
And this makes sense, because the probability that exactly $n$ draws are required to obtain the first white marble is to note that we must have $n-1$ successive draws of a black marble, followed by drawing a white. The $k^{\rm th}$ draw of a black marble, conditioned on the previous $k-1$ failures, occurs with probability $\frac{1}{3} \frac{2}{4} \frac{3}{6} \cdots \frac{k}{k+2}$.
The only thing remaining to check is that $$\sum_{n=1}^\infty \Pr[X = n] = 1;$$ in other words, that this probability mass function is in fact a valid one. This is not terribly difficult to do--here is a hint: $$\frac{4}{n(n+1)(n+2)} = 2 \left( \frac{1}{n} - \frac{1}{n+1} - \frac{1}{n+1} + \frac{1}{n+2} \right).$$
| {
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Find the side of an equilateral triangle inscribed in a circle.
Excuse the poor drawing.
$\triangle CDE$ is an equilateral triangle inscribed inside a circle, with side length $16$. Let $F$ be the midpoint of $DE$. Points $G$ and $H$ are on the circle so that $\triangle FGH$ is an equilateral. Find the side length of $\triangle FGH$ and express it as $a\sqrt{b}-c$ where $a, b, c$ are positive integers.
My attempt
$$\theta \text{ in the Large } \triangle = \theta \text{ in the Small } \triangle = 60^{\circ}$$
$$(\text{side of the Large } \triangle) \cdot \sin 60^{\circ} + (\text{side of the Small } \triangle )\cdot \sin 60^{\circ} = 2\cdot(\text{radius of circle})$$
Solving for the small side, I got $\dfrac{64}{3}-16$, which doesn't match the required form.
Thanks for any help.
| This answer may not be the simplest, but it is straightforward.
I particularized the problem by making one side of the large equilateral triangle the segment between the points $(-8,0)$ and $(8,0)$. Then simple geometry tells us the third vertex is at $(8\sqrt{3},0)$, the circumcenter of the triangle is at $A(0,\dfrac{8\sqrt{3}}3)$, and the radius of the circumcircle is $\dfrac{16\sqrt{3}}3$.
The equation of the circumcircle is then
$$x^2+\left(y-\frac{8\sqrt{3}}3\right)^2=\left(\frac{16\sqrt{3}}3\right)^2$$
$$x^2+\left(y-\frac{8\sqrt{3}}3\right)^2=\frac{256}3$$
The one side of the small equilateral triangle, $\overline{FH}$, is on the line $y=-\sqrt{3}x$. That gives us two equations in two unknowns for the coordinates of point $H$ which is on both the circle and the line. Substitute the expression for $y$ in the linear equation into the quadratic equation and we get a quadratic equation for $x$:
$$x^2+\left(-\sqrt 3x-\frac{8\sqrt 3}3\right)^2=\frac{256}3$$
$$4x^2+16x-64=0$$
$$x^2+4x-16=0$$
Solving this gives us this positive value for $x$:
$$x=2\sqrt 5-2$$
The side of the small equilateral triangle is twice the $x$-coordinate of point $H$, so the side of the triangle is
$$4\sqrt 5-4$$
The final answer to your problem, given the side is $a\sqrt b-c$, is
$$a=4, \quad b=5, \quad c=4$$
I checked this answer numerically with Geogebra, the source of my diagram above, and my answer checks.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this double infinite sum (Catalan number) Let $C_n = \dfrac{1}{n+1}\binom{2n}{n}$.
Is it possible to find the exact value of this infinite sum ?
$$\sum_{n=1}^\infty \sum_{k=n}^\infty \frac{\left(C_{n+1}-2C_n\right)\left(C_{k+1}-C_k\right)}{4^{n+k}}$$
In general, is it possible to evaluate
$$\sum_{n=1}^\infty \sum_{k=n}^\infty \frac{\left(C_{n+1}-2C_n\right)\left(C_{k+1}-C_k\right)}{x^{n+k}}$$
in term of $x$?
Thanks in advances.
| We have:
$$ C_{k+1}-C_k = \frac{3k\, 4^k\, \Gamma\left(k+\frac{1}{2}\right)}{\sqrt{\pi}\,\Gamma(k+3)}\tag{1} $$
hence:
$$ \sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} =\frac{(6n+2)\,\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}\,\Gamma(n+2)} \tag{2}$$
and:
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{C_{n+1}-2C_n}{4^n}\sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} &=& \frac{4}{\pi}\sum_{n\geq 1}\frac{(n-1)\,\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+3)}\cdot \frac{(3n+1)\,\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+2)}\\&=&\frac{4}{\pi}\sum_{n\geq 1}\frac{(n-1)(3n+1)\,\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+3)}\cdot B\left(n+\frac{1}{2},n+\frac{1}{2}\right)\\&=&\frac{4}{\pi}\int_{0}^{1}f\left(\frac{u(1-u)}{1-u+u^2}\right)\,\frac{du}{\sqrt{u(1-u)}}\tag{3}\end{eqnarray*}$$
where:
$$ f(x)=\sum_{n\geq 1}\frac{(n-1)(3n+1)\,\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+3)}\,x^n=\frac{5-5 \sqrt{1-4 x}+2 x^2-2 x \left(4+\sqrt{1-4 x}+8\log 2\right)+16 x\,\log\left(1+\sqrt{1-4 x}\right)}{4 x^2}\tag{4}$$
gives, after some extra work,
$$ \sum_{n\geq 1}\frac{C_{n+1}-2C_n}{4^n}\sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} = \color{red}{\frac{592}{3\pi}-62}.\tag{5}$$
| {
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New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| Let $$ f(x) = \sum_{n = 1}^{\infty} nx^{n-1}, \quad |x| < 1. $$ Then $$f'(x) = \sum_{n = 2}^{\infty} n(n-1)x^{n-2} = 2 \left( \sum_{n=2}^{\infty} \frac{n(n-1)}2x^{n-2} \right ).$$ Note $\frac{n(n-1)}2 = \dbinom{n}2$ so $$f'(x) = 2 \sum_{n=2}^{\infty} \dbinom{n}2x^{n-2} .$$ Now consider the coefficient of $x^k$ in $\frac{1}{(1-x)^3} = (1+x+x^2+ \cdots )^3$.The coefficient of $x^k$ is the number of ways to solve the equation $a+b+c = k$ where $0 \le a,b,c \le k$. Imagine this as $k$ dots where we need to place $2$ bars.
This gives us $\dbinom{k+2}2$ ways. Thus, we have $$ \sum_{n=2}^{\infty} \dbinom{n}2x^{n-2} = \frac{1}{(1-x)^3} $$ so $$f'(x) = \frac{2}{(1-x)^3}$$ so $$f(x) = \int \frac{2}{(1-x)^3} dx = \frac{1}{(1-x)^2} +C. $$ Letting $x = 0$ gives $C = 0$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 8
} |
$\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
*given $$ \frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d} = 65, \frac {1} {a^2} + \frac {1} {b^2} + \frac {1} {c^2} + \frac {1} {d^2} = 209$$
find $$\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$$
(A) $2006$ (B) $2007$ (C) $2008$ (D) $2009$ (E) $2010$
I've use the direct way (make them become 1 fraction)
$$\frac {abc+abd+acd+bcd} {abcd} = 65$$
$$\frac {a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2} {a^2b^2c^2d^2} = 209$$
$$\frac {a^2b^2c^2d^2(ab+ac+ad+bc+bd+cd)} {a^3b^3c^3d^3}$$
$$\frac {(ab+ac+ad+bc+bd+cd)} {abcd}$$
but it doesn't make sense with these power $abcd$ thing
| $(1/a+1/b+1/c+1/d)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}+2\times \frac{1}{a}\times\frac{1}{b}+2\times\frac{1}{a}\times\frac{1}{c}+2\times\frac{1}{a}\times\frac{1}{d}+2\times\frac{1}{b}\times\frac{1}{c}+2\times\frac{1}{b}\times\frac{1}{d}+2\times\frac{1}{c}\times\frac{1}{d}$
$(65)^2=209+2(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})$
$4225-209=2(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})$
$(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})=2008$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Basis for the field extension $\mathbb{Q}(\zeta_{12})$ Consider the cyclotomic field $\mathbb{Q}(\zeta_{12})$ where $\zeta_{12}$ represents the $12$-th primitive root of unity. Since the minimal polynomial of $\zeta_{12}$ is given by $\Phi_{12}(x)$ which has degree $4$, we have that, $$ [ \mathbb{Q}(\zeta_{12}) : \mathbb{Q} ] =4$$ And so I would think that $$ a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3$$ would be a basis for $\mathbb{Q}(\zeta_{12})$.
Now consider the isomorphism that maps $5 \in (\mathbb{Z}_{12})^{\times}$ to $\rho_5 (\zeta_{12}) = \zeta_{12}^5$.
I compute the following, $$\rho_5 (a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3) = a + b \zeta_{12}^5 + \zeta_{12}^{10} + \zeta_{12}^{15}= a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3$$
I'm taking the exponents to be modulo $4$.
I don't think my calculation is correct since this isomorphism should not fix the entire field by Galois Theory. What am I doing wrong ?
| What makes you think that
$$a + b \zeta_{12}^5 + c\zeta_{12}^{10} + d\zeta_{12}^{15}= a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3$$
is true? The powers of $\zeta_{12}$ repeat modulo 12. In fact since the $12$th cyclotomic polynomial is $\Phi_{12}=x^4-x^2+1$, we have that
$$\begin{align*}
\zeta_{12}^5&=\zeta_{12}^5+0\\\\
&=\zeta_{12}^5+(-\zeta_{12})\underbrace{(\zeta_{12}^4-\zeta_{12}^2+1)}_{0}\\
&=\zeta_{12}^3-\zeta_{12}
\end{align*}$$
and by a similar computation, we can find that $\zeta_{12}^{10}=-\zeta_{12}^2+1$. Lastly, since the powers of $\zeta_{12}$ repeat modulo 12, we have $$\zeta_{12}^{15}=\zeta_{12}^{12}\cdot \zeta_{12}^3=1\cdot \zeta_{12}^3=\zeta_{12}^3$$
Thus the correct decomposition in the basis $\{1,\zeta_{12},\zeta_{12}^2,\zeta_{12}^3\}$ is
$$a + b \zeta_{12}^5 + c\zeta_{12}^{10} + d\zeta_{12}^{15}= (a+c) - b \zeta_{12} - c \zeta_{12}^2 + (b+d) \zeta_{12}^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $\int\frac{8x+9}{(2x+1)^3}\,dx$. Do I split $\displaystyle\int\frac{8x+9}{(2x+1)^3}\,dx$ into partial fractions? Or do I use $(2x+1)^{-3}$ by itself? Not sure what to do. Please advice.
The answer given is $\dfrac{-16x+13}{4(2x+1)^2} +C$
| Hint
$$\frac{8x+9}{(2x+1)^3} = \frac{4(2x+1)}{(2x+1)^3}+\frac5{(2x+1)^3} = \frac{4}{(2x+1)^2}+\frac5{(2x+1)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
find $\left( \frac{x}{x+y} \right)^{2007} + \left( \frac{y}{x+y} \right)^{2007}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
if $x, y$ are non-zero numbers satisfying $x^2 + xy + y^2 = 0$, find the value of $$\left(\frac{x}{x+y} \right)^{2007} + \left(\frac{y}{x+y} \right)^{2007}$$
(A). $2$ (B). $1$ (C). $0$ (D). $-1$ (E). $-2$
expanding it would give us $$ \frac { x^{2007} + y^{2007}} {(x+y)^{2007}}$$
how do I calculate this? Very appreciate for all of those who had helped me
| Set $x=ry$
$\implies y^2(r^2+r+1)=0\implies r^2+r+1=0\implies r^3-1=(r-1)(r^2+r+1)=0$
$\implies r^3=1\ \ \ \ (1)$
$\dfrac x{x+y}=\dfrac{ry}{y+ry}=\dfrac r{1+r}$
$\dfrac y{x+y}=\dfrac y{y+ry}=\dfrac 1{1+r}$
As $2007\equiv3\pmod6=6a+3$ where $a=334$( in fact $a$ can be any integer)
The required sum $=\dfrac{r^{6a+3}+1}{(1+r)^{6a+3}}=\dfrac{(r^3)^{2a+1}+1}{(-r^2)^{6a+3}}=\dfrac{(r^3)^{2a+1}+1}{-(r^3)^{2(2a+1)}}$
Use $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Calculus: Finding Arc Length--Squaring the Derivative Where did the -1/2 come from? Math Example about finding the arc length.
I have gotten the derivative of the equation.
Here is the equation.
$$f(x)=\frac{x^5}{5} + \frac{1}{12x^3}$$
Derivative of the equation is:
$$f'(x) = x^4 - \frac{1}{4x^4}$$
The next step is to square the derivative:
$$f'(x)^2 = x^8 -\frac12 + \frac {1}{16x^8}$$
My question is where did $\left(-\frac 12\right)$ come from after squaring the derivative?
Thank you.
| In case, we don't remember formula then it can calculated by simple multiplication as follows $$f'(x)=x^4-\frac{1}{4x^4}$$ $$\implies (f'(x))^2=\left(x^4-\frac{1}{4x^4}\right)^2$$ $$=\left(x^4-\frac{1}{4x^4}\right)\left(x^4-\frac{1}{4x^4}\right)$$ $$=(x^4)(x^4)+\left(-\frac{1}{4x^4}\right)(x^4)+(x^4)\left(-\frac{1}{4x^4}\right)+\left(-\frac{1}{4x^4}\right)\left(-\frac{1}{4x^4}\right)$$ $$=x^8-\frac{1}{4}-\frac{1}{4}+\frac{1}{16x^8}$$ $$=x^8\color{blue}{-\frac{1}{2}}+\frac{1}{16x^8}$$ **
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove an equality If $a+b+c=0$
prove that
$\frac {(a^4 +b^4 +c^4)}{2}=\frac {(a^2+b^2+c^2)}{2^2}^2$
I have expanded the right side and have got this far:
$a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)$
I need $a^2=b^2=c^2$ to prove the equality.
Any ideas?
| One has $$0 = (a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab+bc+ca).$$
Then, $$(\frac{a^2 +b^2 +c^2}{2})^2 = (ab+bc+ca)^2 = a^2b^2 + b^2c^2 +c^2a^2 + 2abc(a+b+c) = a^2b^2 + b^2c^2 +c^2a^2.$$
Finally, one has
$$(a^2+b^2+c^2)^2 = a^4+b^4+c^4 + 2(a^2b^2 + b^2c^2 +c^2a^2) = a^4+b^4+c^4 + 2(\frac{a^2 +b^2 +c^2}{2})^2$$
Thus, $$\frac{1}{2}(a^4+b^4+c^4) = (\frac{a^2 +b^2 +c^2}{2})^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
find x in $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$
let $a = 6+\sqrt x , b=6-\sqrt x$
cube each side
\begin{align}
(\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\
(\sqrt[3]{a^2} + 2\sqrt[3]{ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b) &= 3 \\
\sqrt[3]{a^3} + \sqrt[3]{3a^2b} + \sqrt[3]{3ab^2} + \sqrt[3]{b^3} &= 3 \\
a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3
\end{align}
There's still had cube root, how do I remove it?
| go to power of three both side
$$(\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3})^3 \\\xrightarrow[(a+b)^3=a^3+b^3+3ab(a+b)]{} 6+\sqrt{x} +6-\sqrt{x} +3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})(\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} \sqrt[3] {3})=3\\
$$ we can subsitute $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$ so we have $$ 6+\sqrt{x} +6-\sqrt{x} +3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})\sqrt[3] {3}=3\\$$simplifying
$$ 3(\sqrt[3]{6+\sqrt x})(\sqrt[3]{6-\sqrt x})\sqrt[3] {3}=3-12\\(\sqrt[3]{36- x})\sqrt[3] {3}=-3$$ to the power of three
$$ 3(36- x)=-27\\36-x=-9\\x=45$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Sine/cosine series $$\frac{\sin²(1°) + \sin²(2°) + \sin²(3°) + .. + \sin²(90°)}{\cos²(1°) + \cos²(2°) + \cos²(3°) + .. + \cos²(90°)} = ?$$
I tried to use multiple identities but I couldn't simplify the expression. Where should I start?
| $$\frac{( sin²(1) + sin²(2) + sin²(3) + .. + sin²(90) )}{ ( cos²(1) + cos²(2) +cos²(3) + .. + cos²(90) )} = \\ \frac{2}{2}*\frac{( sin²(1) + sin²(2) + sin²(3) + .. + sin²(90) )}{ ( cos²(1) + cos²(2) +cos²(3) + .. + cos²(90°) )}$$ so try by $sin^2x=\frac{1-cos2x}{2} \\cos^2x=\frac{1-cos2x}{2}$
so
$$\frac{1-cos2 +1-cos 4+1-cos 6+ ...+1-cos 180}{1+cos2 +1+cos 4+1+cos 6+ ...+1+cos 180}=\\ \frac{90-(cos 2+cos 4+cos 6+ ...+cos 180)}{90+(cos 2+cos 4+cos 6+ ...+cos 180)}$$
now see that $$(cos 2+cos 4+cos 6+ ...+cos 180)=\\(cos 2+cos 178 )+(cos 4 +cos 176) +...(cos 44 +cos 46) +cos 90 +cos 180 =0 +0 +0+0+...+cos 90 +cos 180 =0+0+(-1)$$ so $$\frac{90-(cos 2+cos 4+cos 6+ ...+cos 180)}{90+(cos 2+cos 4+cos 6+ ...+cos 180)}=\frac{90-(-1)}{90+(-1)}=\frac{91}{89}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Evaluating the indefinite integral $\int\sqrt{16-9x^2}\,dx$ I need to solve the integral below, but I just can't figure how.
$$\int \sqrt{16-9x^2}\,dx$$
I have tried to replace $9x^2$ with $16\sin^2\theta$. I get to a point where I have the function below. Please let me know whether I'm on the right track, and please explain to me how to finish it...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$
| Integrate by parts
$$
\begin{eqnarray}
\color{blue}{ \int \sqrt{ 16 - 9 x^2 } d x } &=& x \sqrt{ 16 - 9 x^2 }
+ \int \frac{9 x^2}{ \sqrt{ 16 - 9 x^2 } } dx\\
&=& x \sqrt{ 16 - 9 x^2 }
- \int \frac{16 - 9 x^2}{ \sqrt{ 16 - 9 x^2 } } dx
+ \int \frac{16}{ \sqrt{ 16 - 9 x^2 } } dx\\
&=& x \sqrt{ 16 - 9 x^2 }
- \underbrace{ \color{blue}{
\int \sqrt{ 16 - 9 x^2 } dx
} }_{\displaystyle \text{This is the same!}}
+ \int \frac{16}{ \sqrt{ 16 - 9 x^2 } } dx.
\end{eqnarray}
$$
Rearange
$$
\begin{eqnarray}
\color{blue}{ \int \sqrt{ 16 - 9 x^2 } d x }
&=& \frac{1}{2} x \sqrt{ 16 - 9 x^2 }
+ \int \frac{8}{ \sqrt{ 16 - 9 x^2 } } dx\\
&=& \frac{1}{2} x \sqrt{ 16 - 9 x^2 }
+ \frac{8}{3} \underbrace{ \color{green}{
\int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big)
} }_{\displaystyle \text{This should be familiar!}^{(1)}}\\
\\
&=& \bbox[16px,border:2px solid #800000] { \color{#800000}{
\frac{1}{2} x \sqrt{ 16 - 9 x^2 }
+ \frac{8}{3} {\sin^{-1}}\big( 3 x / 4\big) } }
\end{eqnarray}
$$
Footnotes
(1) In case
$$
\color{green}{
\int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big)
}
$$
is not familiar...
$$
\begin{eqnarray}
\color{green}{
\int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big)
} &\stackrel{\color{magenta}{3 x / 4 = \sin(\phi)}}=&
\int \frac{1}{\sqrt{1 - \sin^2(\phi)} } d\sin(\phi)\\
&=& \int \frac{\cos(\phi)}{\cos(\phi)} d\phi\\
&=& \phi\\
&\stackrel{\color{magenta}{\phi = {\sin^{-1}}\big( 3 x / 4\big)}}=&
{\sin^{-1}}\big( 3 x / 4\big).
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 4
} |
compute the integral $\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$ Compute this integral:
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$
my solution is I used derivative of Cauchy integral formula, which is
$$f^{(n)}(z_0) = \frac{n!}{2\pi i}\int \frac{f(z_0)}{(z-z_0)^{n+1}}$$
Then, I got
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$
$$ = \int_{|z|=1} \frac{(z-2)^3}{(2z-1)^3}dz$$
$$ = \frac{2 \pi i}{2!}\left(\frac{1}{2} - 2\right)^3 = \frac{27\pi i}{8}$$
That's what I get from the Cauchy integral formula. So, am I correct for this approach. If not, can someone please show me how to get the right one? Thank you
| Notice,
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz=\frac{1}{8}\oint\frac{(z-2)^3}{\left(z-\frac{1}{2}\right)^3}dz$$ We see that $z=2$ & $z=\frac{1}{2}$ are two points out of which point $z=\frac{1}{2}$ is a pole of third order as it is lying inside the unit circle $|z|=1$. Hence, $f(z)=(z-2)^3$
hence, using cauchy integral formula, we have $$\frac{1}{8}\oint\frac{(z-2)^3}{\left(z-\frac{1}{2}\right)^3}dz$$ $$=\frac{1}{8}\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2f(z)}{dz^2}\right]_{z=\frac{1}{2}}$$ $$=\frac{1}{8}\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2}{dz^2}(z-2)^3\right]_{z=\frac{1}{2}}$$ $$=\frac{\pi i}{8}\left[6\left(z-2\right)\right]_{z=\frac{1}{2}}$$ $$=\frac{3\pi i}{4}\left[\frac{1}{2}-2\right]$$ $$=\frac{3\pi i}{4}\left[\frac{-3}{2}\right]$$ $$=\color{blue}{-\frac{9\pi i}{8}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For Pythagorean triple $x^2+y^2=z^2$, if $x=13$ and $y+z=169$, then how can I determine all possible $y$ and $z$?
If I know that $x=13$, that $x^2+y^2=z^2$ and $y+z=169$, how can I determine all possible values for $y$ and $z$?
I know that one possibility (if not only one) is $84$ and $85$, but was curious as to how this would be found.
| Here's how I would do it.
With the first equation:
\begin{gather}
x = 13\\
x^2 = 169 \\
\end{gather}
Hence the second equation becomes:
\begin{gather}
169 + y^2 = z^2 \\
y^2 - z^2 = -169\\
(y + z)\cdot(y - z) = -169\\
169 (y - z ) = -169\\
y -z = -1\\
z = y+1\\
\end{gather}
And for the third equation, we obtain:
\begin{gather}
y + (y+1) = 169\\
2y = 168\\
y = 84\\
\end{gather}
Hence we selected only one solution, $(x=13,y=84,z=85)$, even over $\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Does the centroid of a triangle ever fall outside of its Morley's triangle? Let $T$ be a triangle, and $M$ its (first) Morley triangle:
(Image from Bruce Shawyer web page.)
Q1. Does the centroid $c$ of $T$ ever fall outside of $M$?
Let $m$ be the centroid of $M$.
Q2. Among all triangles $T$ of unit area, which achieve the maximum
separation $\| c-m \|$ between the centroid of $T$ and the centroid of $M$?
| An approach to Q2 ...
Divide each angle into sixths, with $A = 6\alpha$, $B = 6 \beta$, $C = 6 \gamma$, so that $\alpha + \beta + \gamma = 30^\circ$. Position $\triangle ABC$ on the coordinate plane with $A$ at the origin and the positive $x$-axis bisecting $\angle A$, as shown. Then, with $r$ the circumradius of $\triangle ABC$, we have
$$\begin{align}
A &= (0,0) \\
B &= c\,(\cos 3\alpha,-\sin 3\alpha) = 2 r \sin 6\gamma\,(\cos 3\alpha,-\sin 3\alpha) \\
C &= b\,( \cos 3 \alpha,\phantom{-}\sin 3\alpha ) = 2 r \sin 6\beta\,(\cos 3\alpha,\phantom{-}\sin 3\alpha)
\end{align}$$
The centroid, $K$, is given by
$$\begin{align}
K :&= \frac{1}{3}(A+B+C) \\[4pt]
&= \frac{2r}{3}\left(\; \cos 3\alpha \left( \sin 6\beta + \sin 6\gamma \right), \sin 3\alpha \left( \sin 6\beta - \sin 6\gamma \right) \;\right) \\[4pt]
&= \frac{4r}{3} \left(\;
\cos 3\alpha\,\sin 3 (\beta + \gamma)\,\cos 3 (\beta - \gamma)\;,
\;\sin 3\alpha\,\sin 3 (\beta - \gamma)\,\cos 3 (\beta + \gamma)
\;\right) \\[4pt]
&= \frac{4r}{3} \left(\;
\sin^2 3(\beta+\gamma)\,\cos 3 (\beta - \gamma)\;,
\;\cos^2 3(\beta+\gamma)\,\sin 3 (\beta - \gamma)
\;\right)\end{align}$$
Converting from the expressions for $BP$ and $BR$ in Cut-the-Knot's Proof 1 of Morley's Miracle, we have, for circumradius $r$,
$$E = |AE|\,(\cos\alpha,\phantom{-}\sin\alpha) = 8 r\,\sin 2\beta\,\sin 2 \gamma\,\sin(60^\circ + 2\beta)\;(\cos\alpha,\phantom{-}\sin\alpha)$$
$$F = |AF|\,(\cos\alpha,-\sin\alpha) = 8 r\,\sin 2\beta\,\sin 2 \gamma\,\sin(60^\circ + 2\gamma)\;(\cos\alpha,-\sin\alpha)$$
and we can deduce
$$D = 4 r \sin 2\beta \, \sin 2\gamma\;(\;
\cos(\beta-\gamma) ( 4 \cos^2(\beta+\gamma) - 1 )\;,
\;\sin(\beta-\gamma) ( 4 \sin^2(\beta+\gamma) - 1 )
\;)$$
so that the Morley center, $M$, is (after some simplification effort)
$$\begin{align}
M :&= \frac{1}{3}(D+E+F) \\[6pt]
&= \frac{16r \sqrt{3}}{3} \; \sin 2 \beta \sin 2 \gamma \;\left(
\cos\alpha\,\cos(\beta-\gamma)\,\cos(\beta+\gamma),
-\sin\alpha\,\sin(\beta-\gamma)\,\sin(\beta+\gamma)
\right)
\end{align}$$
From here, one can arrive at an expression for $|K-M|$, and set about maximizing the distance subject to the unit-area condition:
$$1 = |\triangle ABC| = \frac{1}{2} b c \sin A = 2 r^2 \sin 6\alpha\,\sin 6 \beta\,\sin 6 \gamma$$
Unfortunately, the trigonometric manipulations get pretty hairy at this point, and I haven't been able to derive anything useful from them yet. Perhaps this information will give someone else a nice head start on a full solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that type question of Trigonometric Identities
If $3\sin A + 5\cos A = 5$, then prove that:
$$5\sin A + 3\cos A = ±3.$$
| The question as stated seems to be incorrect. Most likely it is asking for $$\color{blue}{5 \sin A} \color{red}{-} \color{blue}{3 \cos A}.$$
If such is the case then from the first equation if we divide throughout by $\sqrt{34}$, we get
$$\frac{3}{\sqrt{34}} \sin A + \frac{5}{\sqrt{34}} \cos A=\frac{5}{\sqrt{34}}.$$
Let $\sin \alpha=\frac{3}{\sqrt{34}}$, then we get
$$\cos(A-\alpha)=\frac{5}{\sqrt{34}}.$$
But then we also get
$$\cos(A-\alpha)=\frac{5}{\sqrt{34}}=\cos \alpha.$$
This implies that
$$A-\alpha = 2n \pi \pm \alpha, \qquad \text{ for } n \in \mathbb{Z}.$$
Thus
$$A=2n \pi \qquad \text{ or } \qquad A=2n\pi+2\alpha.$$
Let $\color{blue}{5 \sin A} \color{red}{-} \color{blue}{3 \cos A}=x$.
*
*If $A=2n \pi$, then $x=-3$.
*If $A=2n\pi+2\alpha$, then $5 \sin A \color{red}{-} 3 \cos A=5 \sin 2\alpha \color{red}{-} 3 \cos 2\alpha$.
Using the values of $\sin \alpha$ and $\cos \alpha$, we get $x=3$.
Thus $x = \pm 3$.
NOTE: If the actual problem (with $+$ sign) stands as is then $x=-3$ or $x=\frac{99}{17}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral problem Find $$ \int e^{x \sin x+\cos x} \left(\frac{x^4\cos^3 x-x \sin x+\cos x}{x^2\cos^2 x}\right) \, dx$$
My attempt:I tried putting $x \sin x+\cos x=t$ and cannot express it in the form of $\int e^t(f(t)+f'(t)) \, dt$
| \begin{align}
& \int e^{x\sin x+\cos x}(\frac{x^4\cos^3x-x\sin^2x+\cos x}{x^2\cos^2x})dx\\
& \hspace{5mm} =\int e^{x\sin x+\cos x}(x^2\cos x-\frac{x\sin^2x-\cos x}{x^2\cos^2x})dx\\
& \hspace{5mm} =\int e^{x\sin x+\cos x}(x^2\cos x-\frac{x\tan^2x-\sec x}{x^2})dx
\end{align}
Realize that $\frac{x\tan^2x-\sec x}{x^2}=\frac{d}{dx}\frac{\sec x}{x}$ and that you can make $\frac{\sec x}{x}$ appear elsewhere by factoring $x^2\cos x-1$ into $(x-\frac{\sec x}{x})(x\cos x)$. So the above is equal to:
\begin{align}
& \int e^{x\sin x+\cos x} \left((x-\frac{\sec x}{x})(x\cos x)+1-\frac{x\tan^2x-\sec x}{x^2}\right) \, dx\\
&=\int \left[e^{x\sin x+\cos x}\left(x-\frac{\sec x}{x}\right)(x\cos x)+e^{x\sin x+\cos x}\left(1-\frac{x\tan^2x-\sec x}{x^2}\right)\right]dx
\end{align}
Now realize that $e^{x\sin x+\cos x}(x\cos x)=\frac{d}{dx}e^{x\sin x+\cos x}$. The above is equal to:
$$\int \left[\left(x-\frac{\sec x}{x}\right)\frac{d}{dx}(e^{x\sin x+\cos x})+e^{x\sin x+\cos x}\frac{d}{dx}\left(x-\frac{\sec x}{x}\right)\right]\,dx
$$
Now, this looks exactly looks like the product rule with $u=x-\frac{\sec x}{x}$ and $v=e^{x\sin x+\cos x}$. So the integral is equal to $$(x-\frac{\sec x}{x})e^{x\sin x+\cos x}+C$$
(To be honest, I did use WolframAlpha to evaluate the integral and work backward to take the derivative by hand, and then reverse each step, but I don't see any other way of evaluating such a difficult integral by hand...)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Trig substitution for integral of $z/(x^2+z^2)$? So I have an integral $\int_1^4\int_y^4\int_0^z\frac{z}{x^2+z^2}\,dx\,dz\,dy$ but I can't figure out what trig substitution to use on the first step. When I try $z=\cos$ and $x=\sin$, I end up with $\int\cos$ but the book comes up with $z\cdot\frac{1}{z}\cdot\arctan \frac{x}{z}$ so I know I screwed up.
Can someone show me how the book took this step?
Thanks!
| The innermost integral is with respect to $x$, treating $z$ as a constant.
Since this is the case, you might as well go with: $$ x = z\tan\theta \implies \text{d}x = z \sec^2\theta \text{ d}\theta \qquad (*)$$ $$ x = 0 \implies \theta = 0, \ x = z \implies \theta = \dfrac{\pi}{4} $$
$$ \int_{0}^{z} \dfrac{z}{x^2+z^2} \text{ d}x = \int_{0}^{\pi/4} \dfrac{z}{z^2 \sec^2\theta} \ z\sec^2\theta \text{ d}\theta = \dfrac{\pi}{4} $$
In the case of the indefinite integral, we have that:
$$ \begin{aligned} \int \dfrac{z}{x^2+z^2} \text{ d}x \ \overset{(*)}= \ \int \dfrac{z}{z^2 \sec^2\theta} \ z\sec^2\theta \text{ d}\theta \ = \int z \dfrac{1}{z} \text{ d}\theta \ & = \ z \dfrac{1}{z} \theta + \mathcal{C} \ = \ z \dfrac{1}{z} \arctan \left( \dfrac{x}{z} \right) + \mathcal{C} \end{aligned} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Condition for common roots of two Quadratic equations: $px^2+qx+r=0$ and $qx^2+rx+p=0$ The question is:
Show that the equation $px^2+qx+r=0$ and $qx^2+rx+p=0$ will have a common root if $p+q+r=0$ or $p=q=r$.
How should I approach the problem? Should I assume three roots $\alpha$, $\beta$ and $\gamma$ (where $\alpha$ is the common root)? Or should I try combining these two and try to get a value for the Discriminant? Or should I do something else altogether?
| If the two quadratic polynomials $f(x) = px^2 + qx + r$ and $g(x) = qx^2 + rx + p$ have a common root then this is also a root of the linear polynomial
$$h(x) = qf(x) - pg(x) = (q^2-pr)x - (p^2-qr)$$
which means that either $q^2-pr=0$ and $p^2-qr=0$ for which $p=q=r$ or that $x = \frac{p^2-qr}{q^2-pr}$ is the common root. Inserting this into either $f(x)$ of $g(x)$ gives us
$$p^3 + q^3 + r^3 = 3pqr$$
Now
$$p^3+q^3 + r^2 - 3pqr = (p+q+r)\left[\frac{3(p^2+q^2+r^2) - (p+q+r)^2}{2}\right]$$
so $p+q+r = 0$ or $(p+q+r)^2 = 3(p^2+q^2+r^2)$.
By Cauchy-Schwarz we have $(p+q+r)^2 \leq 3(p^2+q^2+r^2)$ with equality only when $p=q=r$ so the two polynomials have a common root if and only if $p+q+r=0$ or $p=q=r$.
${\bf Assumptions}$: This answer assumes $p,q\not=0$ and real polynomials; $p,q,r\in\mathbb{R}$. The first condition is assumed since the polynomials is said to be quadratic and otherwise the only-if statement is not true since $p=0\implies x = -\frac{r}{q}$ is a common root for all $q,r$ and if $q=0$ we need $p=r=0$ to have a common root. If the second condition is relaxed allowing for complex coefficients then there are other solutions as shown in the answer by Blue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Two infinite radicals question Hello I have stucked with theese two questions:
$\sqrt{a:\sqrt{a:\sqrt{a: \cdots}}} + \sqrt[3]{a\cdot\sqrt[3]{a\cdot\sqrt[3]{a\cdots}}} = 12$
$a=\text{ ?}$
$\sqrt{6+\sqrt{7+\sqrt{6-\sqrt{6-\sqrt{6- \cdots}}}}}=?$
| $\sqrt{a/\sqrt{a/\sqrt{a/ ...}}}=a^\frac{1}{2}/(a^{\frac{1}{4}}/(a^{\frac{1}{8}}...))=a^{\sum_{i=1}^{\inf}\frac{1}{2*(-2)^{i-1}}}=a^{\frac{1}{3}} $
$\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a ...}}}=a^{\sum_{i=1}^{inf}\frac{1}{3i}}=a^{\frac{1}{2}}
$
Then we have $a^{\frac{1}{3}}+a^{\frac{1}{2}}=12$
Function $f(a)=a^{\frac{1}{3}}+a^{\frac{1}{2}}$ is strictly increasing so equation can't have more than 1 solution. It's easy to see that solution is 64.
For second part let $t=\sqrt{6-\sqrt{6-..}}$
But then $ t^2=6-\sqrt{6-\sqrt{6-..}}=6-t $
$t^2+t-6=0$
Using t>0 we have t=2
then we get $\sqrt{6+\sqrt{7+t}}=\sqrt{6+\sqrt{9}}=\sqrt{6+3}=3 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction: $\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}$
Prove
$$\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}.$$
My problem with this is that it doesn't hold for the base case: $n=1$. This question is from the book "Abstract Algebra" by Charles Pinter (Page 212, exercise 7). When I go about proving it, I can't reach my goal which is $\frac{(n+1)!-1}{(n+1)!}=1-\frac{1}{(n+1)!}$, for $n=n+1$.
| $$\frac{1}{2!}+\frac{2}{3!}+...+\frac{n}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!} $$
Name as $p(n)$
1 step:$$n=1 \to p(1):\frac{1}{2!}=\frac{2!-1}{2!} $$
2nd step :assume $$n=k \to p(k):\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}=\frac{(k+1)!-1}{(k+1)!} $$
3rd step :prove for n=k+1 $$n=k+1 \to p(k+1)=\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}+\frac{k+1}{(k+2)!}=\frac{(k+2)!-1}{(k+2)!}$$ so substitute $\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}={\color{Red}{\frac{(k+1)!-1}{(k+1)!} }}$ into p(k+1)
$${\color{Red}{\frac{(k+1)!-1}{(k+1)!} }}+\frac{k+1}{(k+2)!}=\\\frac{(k+1)!-1}{(k+1)!}*\frac{k+2}{k+2}+\frac{k+1}{(k+2)!} =\\\frac{(k+1)!(k+2)-1(k+2)}{(k+2)!}+\frac{k+1}{(k+2)!}=\\ \frac{(k+1)!(k+2)-1(k+2)+(k+1)}{(k+2)!} =\\\frac{(k+1)!(k+2)-1}{(k+2)!} =\\\frac{(k+2)!-1}{(k+2)!} $$ the proof is complete now (by induction)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Approximating $\sqrt{1+\frac{1}{n}}$ by $1+\frac{1}{2n}$ I was wondering how to approximate $\sqrt{1+\frac{1}{n}}$ by $1+\frac{1}{2n}$ without using Laurent Series.
The reason why I ask was because using this approximation, we can show that the sequence $(\cos(\pi{\sqrt{n^{2}-n}})_{n=1}^{\infty}$ converges to $0$. This done using a mean-value theorem or Lipschitz (bounded derivative) argument where
$$
|\cos(\pi{\sqrt{n^{2}-n}})-\cos(\pi{n}+\pi/2)|=|\cos(\pi{\sqrt{n^{2}-n}})|
\leq \pi|\sqrt{n^{2}-n}-n-1/2| = \pi |\frac{-1/4}{n^{2}-n+n+1/2}|
$$
I looked up $\sqrt{1+\frac{1}{n}}$ and saw that this approximation can be obtained using Laurent series at $x=\infty$. I am not familiar with Laurent series since I have not had any complex analysis yet, but I was wondering if there was another naive way to see this?
| Start with
$(1+x/2)^2-(1+x)
=1+x+x^2/4-(1+x)
=x^2/4
$.
Then,
$(1+x/2)^2-(1+x)
\ge 0
$,
so
$1+x/2
\ge \sqrt{1+x}
$.
Going the other way,
which is harder,
$(1+x/2)^2-x^2/4
=(1+x)
$,
so,
for $x \ge 0$,
$\begin{array}\\
\sqrt{1+x}
&=\sqrt{(1+x/2)^2-x^2/4}\\
&=(1+x/2)\sqrt{1-(x/(2(1+x/2)))^2}\\
&\ge (1+x/2)(1-(x/2(1+x/2))^2)
\quad\text{since }\sqrt{1-a}>1-a\text{ for }0 < a < 1\\
&=1+x/2-\frac{(1+x/2)x^2}{4(1+x/2)^2}\\
&=1+x/2-\frac{x^2}{4(2+x)}\\
&>1+x/2-x^2/8\\
\end{array}
$
This isn't the best,
but it is within a constant factor
of the $x^2$ term
and it is gotten by going forward,
not working backwards from
a known result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
For what values of $ a, b$ does the equation have real roots?
For what values of $a,b$ does the equation
$${ x }^{ 2 }+2\left( 1+a \right) x+\left( 3{ a }^{ 2 }+4ab+4{ b }^{ 2 }+2 \right) = 0$$
have real roots?
For it to have real roots, the discriminant has to be $>0$, correct? (Or equal to, I suppose, since the question didn't specify distinct or not) So I tried using the values, which gave me ${ \left( 2+2a \right) }^{ 2 }-4\left( 3{ a }^{ 2 }+4ab+4{ b }^{ 2 } +2\right) $ but I'm not sure where to go after that.
| You have the right idea.
You just need to continue
expanding that expression.
From
$x^2 + 2(1+a)x + (3a^2 + 4ab + 4b^2 + 2) = 0
$,
the discriminant is
(ignoring the factor of 2
since we are concerned
only about the sign)
$\begin{array}\\
d
&=(1+a)^2-(3a^2 + 4ab + 4b^2 + 2)\\
&=a^2+2a+1-(3a^2 + 4ab + 4b^2 + 2)\\
&=-2a^2+2a-1-4ab-4b^2\\
&=-a^2+2a-1-a^2-4ab-4b^2
\quad\text{This is the key step}\\
&=-(a-1)^2-(2b+a)^2\\
\end{array}
$
So,
in the miraculous way of many homework problems,
this is the negative
of a sum of two squares.
So $d \le 0$
and,
for $d = 0$,
we must have
$a=1$ and
$2b+a=0$,
which means
$b = -1/2$.
For any other values of
$a$ and $b$,
the discriminant is negative,
and so there are no real roots.
For these values of $a$ and $b$,
there is a repeated root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that the trigonometric equation has no solution Prove that the trigonometric equation $$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$ has no solution. I tried applying $T2's$ lemma to contradict but could only do so for the first and third quadrant values if $x$. There must be some good proof without the restrictions in the values of $x$. Thanks.
| The minimum of $\frac{\sin^3}{1-\sin{x}}$ is $-\frac12$, so, except for $\sin{x}=-1$, $\frac{\sin^3}{1-\sin{x}}\gt-\frac12$.
The minimum of $\frac{\cos^3}{1-\cos{x}}$ is $-\frac12$, so $\frac{\cos^3}{1-\cos{x}}\ge-\frac12$.
So, except for $\sin{x}=-1$, $\frac{\sin^3}{1-\sin{x}}+\frac{\cos^3}{1-\cos{x}}\gt-1$.
If $\sin{x}=-1$, then $\cos{x}=0$, so $\frac{\sin^3}{1-\sin{x}}+\frac{\cos^3}{1-\cos{x}}=-\frac12+0$.
So, the equation has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$, prove that $ x^2+y^2=1$. If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$,
prove that $$ x^2+y^2=1$$
I tried taking sine of both the sides, I only come to this result: $$x^2 + y^2 -2x^2y^2 + 2xy\sqrt{(1-y^2)(1+x^2)}=1.$$
| Let, $\sin^{-1}x=\alpha\iff x=\sin \alpha$
& $\sin^{-1}x=\beta\iff y=\sin\beta$ then we have$$\alpha+\beta=\frac{\pi}{2}$$
Now, we have
$$x^2+y^2=(\sin\alpha)^2+(\sin\beta)^2$$
$$x^2+y^2=\sin^2\alpha+\sin^2\beta$$
Substituting $\beta=\frac{\pi}{2}-\alpha$, we get
$$x^2+y^2=\sin^2\alpha+\sin^2\left(\frac{\pi}{2}-\alpha\right)$$
$$=\sin^2\alpha+\cos^2\alpha=1$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x^2+y^2=1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Multiply inverse of matricies Let ${A}$ be a $2 \times 2$
matrix such that ${A}$ * $\begin{pmatrix} 3 \\ -8 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad {A} * \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$
Find
${A}^{-1} *\begin{pmatrix} -2 \\ -1 \end{pmatrix}.$
What is the easiest way to start this problem?
Thanks
| First, notice that $A$ is invertible since it transforms a basis of $\Bbb R^2$ to a basis of $\Bbb R^2$. Now write
$$A^{-1 }\cdot(-2,-1)^T=-2A^{-1}\cdot(1,0)^T-A^{-1}\cdot(0,1)^T$$
and the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Knowing that $x+y+z=3$ and $x,y,z\ge0$, how to prove the following inequality? Let $x,y,z\ge 0$, if $x+y+z=3$, show that
$$\dfrac{1}{\sqrt{x^2+xy+y^2}}+\dfrac{1}{\sqrt{y^2+yz+z^2}}+\dfrac{1}{\sqrt{z^2+zx+x^2}}\ge \dfrac{12+2\sqrt{3}}{9}$$
Using this inequality:
$$\left(\sum_{cyc}\dfrac{1}{\sqrt{x^2+xy+y^2}}\right)\cdot\left(\sum_{cyc}\sqrt{x^2+xy+y^2}\right)\ge 9$$
it is enough to check that
$$\dfrac{9}{\sum_{cyc}\sqrt{x^2+xy+y^2}}\ge \dfrac{12+2\sqrt{3}}{9}$$
And I can't go further.I can't observer something,only find this inequality $=$ iff $x=y,z=0$
| Let $g(u,v)=u^{2}+v^{2}+uv$.
Let $f(x,y,z)=g^{-\frac{1}{2}}(x,y)+g^{-\frac{1}{2}}(x,z)+g^{-\frac{1}{2}}(y,z)$ which simply describes the sums you are interested in.
The basic idea here is that we will minimize the function $f(x,y,z)$ constrained to $x+y+z=3$ and show that the absolute minimum is precisely $\frac{12+2\sqrt{3}}{9}$.
So simply compute a partial,
$\quad \quad \quad \quad \quad \quad \frac{\partial f}{\partial x} = -(\frac{x+\frac{y}{2}}{g^{\frac{3}{2}}(x,y)}+\frac{x+\frac{z}{2}}{g^{\frac{3}{2}}(x,z)})$
Now since we are confined to $x,y\geq0$, we know that the leftmost term and rightmost term are positive semi-definite, so the whole expression (because of the minus in front) is negative semi definite. Thus for this partial derivative to be equal to 0, it must be that $x=y=z=0$. But this can't happen because this would imply that $x+y+z=0\neq3$. A local extremum must have all partials equal to zero, so clearly that can't happen if $\frac{\partial f}{\partial x}\neq 0$ (and in fact, by the symmetry of the function, you'd get the same contradiction for the other partials). Therefore, the only critical points within this region must lie on the boundary of the region defined by $x+y+z=3$. These are the three lines defined by
\begin{align*}
z=0, y=3-x, x\in[0,3]
\end{align*}
\begin{align*}
y=0, z=3-x, x\in[0,3]
\end{align*}
\begin{align*}
x=0, z=3-y, y\in[0,3]
\end{align*}
Again, by the symmetry, I'll just look at one of these lines, specifically $z=0, y=3-x$. Thus we are now looking at one dimensional function $h(x)=f(x, 3-x, 0)$, which through some simplication leads to $h(x)=\frac{1}{x}+\frac{1}{3-x}+\frac{1}{\sqrt{x^{2}-3x+9}}$. From here, we attempt to find extrema for $h(x)$, and I'll skip the computational details of teh derivatives. If you take a derivative, you can see easily that $\frac{3}{2}$ is a solution to make $h'(x)=0$. Then when you compute $h''(x)$, you will see that it is positive definite, meaning $h'(x)$ is strictly increasing, so $x=\frac{3}{2}$ is the ONLY solution to $h'(x)=0$. Now since $h''(x)$ is strictly positive, you now know that the ABSOLUTE minimum on this line must be at $\frac{3}{2}$, and note that $h(\frac{3}{2}$)=$\frac{12+2\sqrt{3}}{9}$. Thus by the previous argument, since it's an absolute min on the boundary, it's an absolute min over the entire surface. You can check for yourself that the other two parts of the boundary give you the exact same result. Thus by definition of absolute minimum, the result you wanted has been proven.
| {
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"url": "https://math.stackexchange.com/questions/1390518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots Prove that $x^4 + x^3 + x^2 + x + 1$ has no roots.
I can see two ways of proving it.
The first one is to notice, that $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$. It has the only root 1. And it is not the root of $x^4 + x^3 + x^2 + x + 1$. So, $x^4 + x^3 + x^2 + x + 1$ does not have roots.
Another way is to solve it as a palindromic polynomial. It does not have roots.
But is there any way to directly manipulate the expression to show that it is always greater than zero?
| $$f(x)=x^4+x^3+x^2+x+1$$
if $x=1$ then $f(1)=5$ hense $f(1)>0$
if $x\neq 1$ then $f(x)=\frac{x^5-1}{x-1}$
$(x^5-1)$ and $(x-1)$ have the same sign any $x$. Hence $f(x)>0$
| {
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Simplifying some square numbers expressions I need help to fix theese out. Thank you.
$ \frac { \frac {1} {\sqrt {3} } - \sqrt {12} } { \sqrt {3} } $
$ \frac { \sqrt {x} } {\sqrt[3] {3} } + \frac {\sqrt[4] {x}} {\sqrt {x} } $
$ \sqrt {5} - \sqrt {3} = n \Rightarrow \sqrt {5} + \sqrt {3} = ? $
| If $$\left(\sqrt{5}-\sqrt{3}\right) = n\;,$$ Then $$\left(\sqrt{5}-\sqrt{3}\right)\cdot \left(\sqrt{5}+\sqrt{3}\right)= n\cdot \left(\sqrt{5}+\sqrt{3}\right)$$
So we get $$\displaystyle \left(\sqrt{5}+\sqrt{3}\right) = \frac{2}{n}$$
| {
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How to integrate $\int (x-1)\sqrt{x} \, \text{d}x$ How do I find this integral:
$$\int (x-1)\sqrt{x} \, \text{d} x$$
I thought to use use substitution, but am not sure what I should use as $u$.
| $$\int (x-1)\sqrt{x} \, \text{d} x=\int (x\sqrt{x}-\sqrt{x}) \, \text{d} x=\int (x\sqrt{x})\, \text{d} x-\int(\sqrt{x}) \, \text{d} x=$$
$$\int x\sqrt{x}\, \text{d} x-\int\sqrt{x} \, \text{d} x=\int x^{\frac{3}{2}}\, \text{d} x-\int x^{\frac{1}{2}} \, \text{d} x=\frac{1}{\frac{5}{2}}x^{\frac{3}{2}+1}-\int x^{\frac{1}{2}} \, \text{d} x=$$
$$\frac{2}{5}x^{\frac{5}{2}}-\int x^{\frac{1}{2}} \, \text{d} x=\frac{2x^{\frac{5}{2}}}{5}-\int x^{\frac{1}{2}} \, \text{d} x=$$
$$\frac{2x^{\frac{5}{2}}}{5}-\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}=\frac{2x^{\frac{5}{2}}}{5}-\frac{2}{3}x^{\frac{3}{2}}=\frac{2x^{\frac{5}{2}}}{5}-\frac{2x^{\frac{3}{2}}}{3}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area under a curve subintervals For the graph $y=-x^2+10x+24$ for $(-1\le x \le 3)$
How many subintervals would I need to estimate the area to within 0.1 unit²
I ended up drawing a crazy graph and came up with 800, which is wrong...
| Using $n$ intervals of equal length $\frac{4}{n}$, an estimate of the area by upper rectangles is given by
$$A(n) = \sum_{k=1}^n f\left(-1+k \cdot \frac{4}{n}\right)\cdot \frac{4}{n}.$$
Since $f(x) = -x^2+10x+24$, we get
$$A(n) = \frac{4}{n} \sum_{k=1}^n \left[-\left(-1+k \cdot \frac{4}{n}\right)^2 + 10 \left(-1+k \cdot \frac{4}{n}\right) +24\right]$$
or simplified
$$A(n) = \frac{4}{n} \sum_{k=1}^n \left[- k^2 \cdot \frac{16}{n^2} + k \cdot \frac{48}{n}+13\right].$$
We want
$$\left|\int_{-1}^3 f(x) \, dx - A(n)\right| \leq 0.1.$$
Let us simplify the LHS before we solve for $n$. Using
$$\sum_{k=1}^n k = \frac{n(n+1)}{2}, \, \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
we can find
$$A(n) = \frac{4}{n} \left( -\frac{8}{n} \frac{(n+1)(2n+1)}{3} + 24(n+1) + 13n\right) = \frac{380n^2+192n-32}{3n^2}$$
and therefore $\int_{-1}^3 f(x) \, dx = \lim_{n \to \infty} A(n) = \frac{380}{3}$. Hence the inequality becomes
$$\left|\frac{380}{3} - \frac{380n^2+192n-32}{3n^2} \right|\leq 0.1$$
which eventually yields
$$n \geq 639.833,$$
so $n = 640$ should work.
You can check that $\frac{380}{3} = 126.666\dots$ and for $n = 640$ we get $A(n) \approx 126.76664$ whereas for $n = 639$ we have $A(n) \approx 126.76680$.
| {
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Can someone show me HOW to do this, I don't just want the answer $4n$ to the power of $3$ over $2 = 8$ to the power of negative $1$ over $3$
Written Differently for Clarity:
$$(4n)^\frac{3}{2} = (8)^{-\frac{1}{3}}$$
EDIT
Actually, the problem should be solving $4n^{\frac{3}{2}} = 8^{-\frac{1}{3}}$. Another user edited this question for clarity, but they edited it incorrectly to add parentheses around the right hand side, as can be seen above.
| $$4n^{\frac32} = 8^{-\frac{1}{3}}= \frac{1}{8^{\frac{1}{3}}}$$
$$4\sqrt{n^3}= \frac{1}{\sqrt[3]{8}}= \frac{1}{2}$$
$$\sqrt{n^3}= \frac{1}{8}$$
$$\left(\sqrt{n^3}\right)^2= \left(\frac{1}{8}\right)^2$$
$$\left|n^3\right|= \frac{1}{8^2}=\frac{1}{64}$$
Note that $n^{\frac32}$ or $\sqrt{n^3}$ implies that $n^3\geq 0$ and we can drop the absolute value bars. So now we have
$$n^3=\frac{1}{64}$$
$$\sqrt[3]{n^3}=\sqrt[3]{\frac{1}{64}}$$
$$n=\frac{1}{\sqrt[3]{64}}=\frac14$$
| {
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If $x\cos(\theta)-\sin(\theta)=1$ then what is the value of $x^2+(1+x^2)\sin(\theta)=1$ The question given is,
If $x\cos(\theta)-\sin(\theta)=1$ then find the value of $x^2+(1+x^2)\sin(\theta)$.
There are four options given $1$, $-1$, $0$ and $2$. I tried using $\sin^2+\cos^2=1$. I also tried to isolate $x$ and put its value in the second equation but things didn't get simplified.
Please explain how should I solve this question.
| The original claim $x^2+(1+x^2)\sin\theta=1$ is false. When $\sin(\theta)=-1$, and $\cos(\theta)=0$ (say at $\theta=-\pi/2$), then for any $x$, we have $x\cos(\theta)-\sin\theta=1$ but
$$
x^2+(1+x^2)\sin\theta=x^2-(1+x^2)=-1\neq 1.
$$
Edit for the updated question: The case $\cos\theta=0$ gives us $\sin(\theta)=-1$ and so
$$
x^2+(1+x^2)\sin\theta=-1.
$$
When $\cos\theta\neq 0$, we have $x\cos\theta-\sin\theta=1\implies x=\frac{1+\sin\theta}{\cos\theta}$ and so
$$
x^2=\frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta},\quad 1+x^2=\frac{2+2\sin\theta}{\cos^2\theta}
$$
which implies
$$
x^2+(1+x^2)\sin\theta=\frac{1+2\sin\theta+\sin^2\theta+(2\sin\theta+2\sin^2\theta)}{\cos^2\theta}=\frac{2(1+2\sin\theta+\sin^2\theta)+\sin^2\theta-1}{\cos^2\theta}=-1+2x^2.
$$
This means your expression is in general not identically equal to any constant.
| {
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Define the function $f:(0,1)\to (0,1)$
Define the function $f:(0,1)\to (0,1)$ by $$ \displaystyle f(x) = \left\{ \begin{array}{lr} x+\frac 12 & \text{if}\ \ x < \frac 12\\ x^2 & \text{if}\ \ x \ge \frac 12 \end{array} \right.$$ Let $a$ and $b$ be two real numbers such that $0 < a < b < 1$. We define the sequences $a_n$ and $b_n$ by $a_0 = a, b_0 = b$, and $a_n = f( a_{n -1})$, $b_n = f (b_{n -1} )$ for $n > 0$. Show that there exists a positive integer $n$ such that $$(a_n - a_{n-1})(b_n-b_{n-1})<0.$$
it's just for sharing a new ideas, thanks:)
| My solution:
Suppose, on the contrary, that $(a_n-a_{n-1})(b_n-b_{n-1}) \geq 0$ for all $n\in \mathbb{N}$.
Then, $a_n \geq a_{n-1}$ and $b_n \geq b_{n-1}$ or $a_n \leq a_{n-1}$ and $b_n \leq b_{n-1}$.
It is easy to prove that $a_{n-1}<\frac{1}{2}$ and $b_{n-1}<\frac{1}{2}$ or $a_{n-1}\geq\frac{1}{2}$ and $b_{n-1}\geq\frac{1}{2}$.
Lemma $1$: $a_n \geq \frac{1}{2}$ for infinitely many $n$.
Proof: Just note that if $a_n < \frac{1}{2}$, then $a_{n+1} \geq \frac{1}{2}$.
Lemma $2$4: $a_n \geq \frac{3}{4}$ and $b_n \geq \frac{3}{4}$ are both true for infinitely many $n$.
Proof: Use lemma $1$ and our initial assumption.
Lemma $3$: Let $g(n)=b_n-a_n$. Then, $g(n)>\frac{1}{2}$ for some $n$.
Proof: Note that either $g(n) \geq g(n-1)$ or $g(n) \geq \frac{3}{2} g(n-1)$, depending on which side of $\frac{1}{2}$ $a_n$ and $b_n$ are on. Use lemma 2 to show that there exists a $k$ such that $g(k) \geq (\frac{3}{2})^{\lceil log_\frac{3}{2} \frac{1}{2g(1)} \rceil} g(1) > \frac{1}{2}$.
We get a contradiction to our initial assumption since $a_n$ and $b_n$ are on the same side of $\frac{1}{2}$ and are between $0$ and $1.$ Thus, the problem is proved.
And I hope to see more solutions, thanks
| {
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Remainder of division of ${6}^{7^n}$ to $43$
What is the Remainder of the division of ${6}^{7^n}$ to $43$?
I've tried with Fermat's little theorem, but it haven't work.
Update : lab bhattacharjee gave a nice proof. But I want to know if this proof is correct.
${6^7}^n = ({6^7})^{7^{n-1}}$
$6^7 = ({6^3})^2$ * 6
$6^3 = 216 = 1(mod 43) => ({6^3})^2 = 1^2 = 1 (mod 43) => 6^7 = 6 (mod 43) => ({6^7})^{7^{n-1}} = 6^{7^{n-1}} (mod 43) <=> {6^7}^n = {6^7}^{n-1} (mod 43)$
The same way we obtain $6^{7^{n-1}} = 6^{7^{n-2}} (mod 43)$
$6^{7^{n-2}} = 6^{7^{n-3}} (mod 43)$
...
${6^7}^2 = {6^7}^1 (mod 43)$
${6^7}^1 = {6^7}^0 = 6 (mod 43)$
So ${6^7}^n = 6 (mod 43)$
| As $\phi(43)=42$
$\implies6^{(7^n)}\equiv6^{7^n\pmod{42}}\pmod{43}$
let use find $7^n\pmod{42}$
Now as $(7^n,42)=7$ for integer $n\ge1$
we shall find $7^{n-1}\pmod6$
$7\equiv1\pmod6\implies7^{n-1}\equiv1^{n-1}\equiv1$
$\implies7^n\equiv7\cdot1\pmod{6\cdot7}\equiv7$
$\implies6^{(7^n)}\equiv6^7\pmod{43}$
Now use $6^3\equiv1\pmod{43}$ and $6^7=(6^3)^2\cdot6\equiv?\pmod{43}$
| {
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What is the derivative of $|x^3|$? Let $f(x)=|x^3|$.
I found two ways to differentiate this function. Apparently method 2 is wrong, but I cannot figure out why. So the question is, is method two wrong and why?
Method 1 (according to wolfram mathematica)
$$f(x)=|x^3|=|x|^3$$
$$f'(x)=3|x|^2 \text{Sgn}(x)=3|x^2|\frac{x}{|x|}=3|x|x$$
Method 2 (using chain-rule)
$$f'(x)= \text{Sgn}(x^3)\cdot3x^2$$
With method 2 it seems $f'(0)$ is not defined, since $\text{Sgn}(0)$ is not defined. However, this is not true according to WolframAlpha.
| I learned that, ...
... whenever you want to compute the derivative of a function involving an absolute value, first convert the absolute value into its definition
$$
|x| = \sqrt{x^2} = (x^2)^{\frac{1}{2}}.
$$
So, $f$ can be re-written as
$$
f(x) = |x^3| = ((x^3)^2)^{\frac{1}{2}}.
$$
If we now apply the chain rule multiple times to compute its derivative, we obtain
$$
\begin{eqnarray}
\dfrac{df(x)}{dx}
&=& \left( \dfrac{df(x)}{d((x^3)^2)} \right) \cdot \left( \dfrac{d((x^3)^2)}{dx} \right) \\
&=& \left( \dfrac{df(x)}{d((x^3)^2)} \right) \cdot \left( \dfrac{d((x^3)^2)}{d(x^3)} \right) \cdot \left( \dfrac{d(x^3)}{dx} \right) \\
&=& \left( \dfrac{1}{2} \dfrac{1}{((x^3)^2)^{\frac{1}{2}}} \right) \cdot \left( 2x^3 \right) \cdot \left( 3x^2 \right) \\
&=& \dfrac{3x^5}{((x^3)^2)^{\frac{1}{2}}}.
\end{eqnarray}
$$
Now there are two options of how to proceed:
*
*Combine $(x^3)^2 = x^6$ to yield $\dfrac{df(x)}{dx} = \dfrac{3x^5}{\sqrt{x^6}}$.
*Or use again the definition of the absolute value to yield $\dfrac{df(x)}{dx} = \dfrac{3x^5}{|x^3|}$.
The computer algebra system Maple seems to prefer the first option; I would prefer the second one, because we started with an absolute value in the original function and end with an absolute value in its derivative. Nonetheless, both are equivalent.
| {
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Minimum sum of factors of a natural number Let's say I have a natural number $N$. $a$ and $b$ are two factors of $N$. How can I find $a$ and $b$ such that $a + b$ is minimum.
Examples:
*
*$N = 12$, $a = 3$, $b = 4$
*$N = 13$, $a = 1$, $b = 13$
| consider the factors of 36 and their sums
1 + 36 = 37
2 + 18 = 20
3 + 12 = 15
4 + 9 = 13
6 + 6 = 12
So $a = b = 6 = \sqrt{36}$
Consider the factors of 12 and their sums
1 + 12 = 13
2 + 6 = 8
3 + 4 = 7
So $a = 3 \lt \sqrt{12} \lt b = 4$
It should be fairly clear that the two factors that are "closest to" $sqrt N$ are the two factors that you are looking for.
To prove this, you would have to show that
$1 \le a_1 \lt a_2 \le \sqrt N$ implies that
$\sqrt N \le \dfrac{N}{a_2} \lt \dfrac{N}{a_1} \le N$
| {
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Determinant of $ n \times n$ matrix and its characteristic polynomial. Suppose, $M_4, M_5,..M_n$ is as follows then determinant and characteristic polynomial of $M_n$. $M_4=\left(
\begin{array}{cccc}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
\end{array}
\right),M_5=\left(
\begin{array}{ccccc}
0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
1 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 \\
\end{array}
\right),M_6=\left(
\begin{array}{cccccc}
0 & 0 & 1 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 & 1 \\
1 & 1 & 0 & 0 & 0 & 1 \\
1 & 1 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 & 0 \\
\end{array}
\right), \quad M_8=\left(
\begin{array}{cccccccc}
0 & 0 & 1 & 1 & 1 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\
1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\
\end{array}
\right)$
| The matrix $M_n$ is a circulant matrix where $c_0=c_1=c_{n-1}=0$ and other $c_k=1$. Eigenvalues and eigenvectors of circulant matrices can be readily calculated.
| {
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$\displaystyle\int_{0}^{\pi} \frac{\sin^2 x}{a^2 - 2ab\cos x +b^2} \,\mathrm dx$ An integration:
$$\int_{0}^{\pi} \frac{\sin^2 x}{a^2 - 2ab\cos x +b^2} \,\mathrm dx$$
I am stuck with this definite integral. Will putting $\,\cos x = z\,$ help out here?
| The antiderivative calculated by Wolfram is
$$\frac{x(a^2+b^2)-2(a^2-b^2)\tan^{-1}\bigg(\frac{(a+b)\tan\left(\frac{x}{2}\right)}{a-b}\bigg)+2ab\sin(x)}{4a^2b^2}$$
For $a=b$, the intgegral becomes
$$\int_0^{\pi} \frac{\sin^2(x)}{2a^2(1-\cos(x)}\mathrm dx=\frac{\pi}{2a^2}$$
For $a>b$ the $\tan^{-1}$-term tends to $\frac{\pi}{2}$ because the argument tends to infinite for $x=\pi$. For $x=0$, the antiderivate is $0$.
The final result is $\dfrac{\pi}{2a^2}$
If $a<b$, either change the roles of $a$ and $b$ or use $-\dfrac{\pi}{2}$ for the problematical term in the antiderivate. The result is now $\dfrac{\pi}{2b^2}$
So, the integral is $\dfrac{\pi}{2\max(a,b)^2}$.
| {
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Finding an expression for the diameter of a circle. The figure shows two circles of radius $1$ that touch at $P$. 1]1 $T$ is a common tangent line; $C_1$ is the circle that touches $C$, $D$, and $T$; $C_2$ is the circle that touches $C$, $D$, and $C_1$; $C_3$ is the circle that touches $C$, $D$, and $C_2$. This procedure can be continued indefinitely and produces an infinite sequence of circles $\{C_n\}$. Find an expression for the diameter of $C_n$.
| Notice, the radius say $r_1$of circle $C_1$ is given by the formula (see derivation) $$\color{red}{r_1=\frac{ab}{(\sqrt a+\sqrt b)^2}}$$ setting radii $a=b=1$ of larger identical circles, we get $$r_1=\frac{1\cdot1}{(1+1)^2}=\frac{1}{4}$$
Now, the radius $r_2$ of circle $C_2$ externally touching the circles $C$, $D$ & $C_1$ is given by the standard formula (see derivation)
$$\color{red}{r_2=\frac{abc}{2\sqrt{abc(a+b+c)}+(ab+bc+ca)}}$$ Setting the values of $a=1, b=1, c=r_1=\frac{1}{4}$, we get radius of circle $C_2$ as follows $$r_2=\frac{1\cdot1\cdot\frac{1}{4}}{2\sqrt{1\cdot1\cdot\frac{1}{4}\left(1+1+\frac{1}{4}\right)}+1\cdot1+1\cdot \frac{1}{4}+\frac{1}{4}\cdot 1}=\frac{1}{12}$$
Similarly,
the radius $r_3$ of circle $C_3$ externally touching the circles $C$, $D$ & $C_2$ is calculated by setting the values of $a=1, b=1, c=r_2=\frac{1}{12}$ in the above standard formula as follows $$r_3=\frac{1\cdot1\cdot\frac{1}{12}}{2\sqrt{1\cdot1\cdot\frac{1}{12}\left(1+1+\frac{1}{12}\right)}+1\cdot1+1\cdot \frac{1}{12}+\frac{1}{12}\cdot 1}=\frac{1}{24}$$ Thus, continuing the same procedure, we can calculate radius of any circle $C_n$ using above standard formula.
| {
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Why does this "miracle method" for matrix inversion work? Recently, I answered this question about matrix invertibility using a solution technique I called a "miracle method." The question and answer are reproduced below:
Problem: Let $A$ be a matrix satisfying $A^3 = 2I$. Show that $B = A^2 - 2A + 2I$ is invertible.
Solution: Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, by power series expansion, we would simply be looking for
$$ \frac{1}{B} = \frac{1}{A^2 - 2A + 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$
where the coefficient of $A^n$ is
$$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$
But we know that $A^3 = 2$, so
$$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$
and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that
$$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$
Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, compute the product $BB^{-1}$, and you'll find that, miraculously, this answer works!
I discovered this solution technique some time ago while exploring a similar problem in Wolfram Mathematica. However, I have no idea why any of these manipulations should produce a meaningful answer when scalar and matrix inversion are such different operations. Why does this method work? Is there something deeper going on here than a serendipitous coincidence in series expansion coefficients?
| If $A$ is any element in a $\mathbb{Q}$-algebra $\mathfrak{A}$ satisfying the equation $x^3-2\cdot1_{\mathfrak{A}}=0$, then $\mathbb{Q}[A]\cong\mathbb{Q}[\sqrt[3]{2}]$. Now you can use a standard technique for finding the inverse of an element in the field $\mathbb{Q}[\sqrt[3]{2}]$ in order to get the inverse of $B$ (it is clear that $gcd(x^3-2,x^2-2x+2)=1$, so that there exist $a(x),b(x)$ such that $1=a(x)(x^3-2)+b(x)(x^2-2x+2)$. Then $1=b(A)(A^2-2I+2)$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "71",
"answer_count": 8,
"answer_id": 7
} |
Find Closed-form expression of a integer sequence. We have a integer sequence $u_{n+1}=2p.u_{n}-u_{n-1}$ (p is a positive integer, $n\geq 3$ )
*
*When $u_{1}=1; u_{2}=p$ then with a regular way I can find Closed-form expression of the sequence.
$u_{n+1}=((p+\sqrt{p^{2}-1})^{n}+(p-\sqrt{p^{2}-1})^{n}).\frac{1}{2}$
*
*But when $u_{1}=1; u_{2}=2p-1$, how much I try, I can not find out it's Closed-form expression.
So can someone show me how to find it or explain to me why we can not find it
| One way to find a general solution is to use generating functions. Define:
$$
U(z) = \sum_{n \ge 0} u_n z^n
$$
Rewrite the recurrence shifting indices:
$$
u_{n + 2} = 2 p u_{n + 1} - u_n
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$, recognize some sums:
$$
\frac{U(z) - u_0 - u_1 z}{z^2}
= 2 p \frac{U(z) - u_0}{z} - U(z)
$$
Solve for $U(z)$, split into partial fractions:
$\begin{align}
U(z)
&= \frac{u_0 + (u_1 - 2 p u_0) z}{1 - 2 p z + z^2} \\
&= \frac{u_0 (\sqrt{p^2 - 1} - p) + u_1}
{2 \sqrt{p^2 - 1} (1 - (\sqrt{p^2 - 1} + p) z)}
+ \frac{u_0 (\sqrt{p^2 - 1} - p) - u_1}
{2 \sqrt{p^2 - 1} (1 + (\sqrt{p^2 - 1} - p) z)}
\end{align}$
This is just two geometric series:
$$
u_n
= \frac{u_0 (\sqrt{p^2 - 1} - p) + u_1}{2 \sqrt{p^2 - 1}}
\cdot \left( \sqrt{p^2 - 1} - p \right)^n
+ \frac{u_0 (\sqrt{p^2 - 1} - p) - u_1}{2 \sqrt{p^2 - 1}}
\cdot \left( \sqrt{p^2 - 1} + p \right)^n
$$
Edit: The above is valid for $p \ne 1$ only. In case $p = 1$:
$\begin{align}
U(z)
&= \frac{u_0 + (u_1 - 2 u_0) z}{1 - 2 z + z^2} \\
&= \frac{u_0 + (u_1 - 2 u_0) z}{(1 - z)^2}
\end{align}$
Here we could got the partial fraction route too, but we have another option:
$\begin{align}
U(z)
&= (u_0 + (u_1 - 2 u_0) z) \sum_{n \ge 0} (-1)^n \binom{-2}{n} z^n \\
&= (u_0 + (u_1 - 2 u_0) z) \sum_{n \ge 0} \binom{n + 2 - 1}{2 - 1} z^n \\
&= (u_0 + (u_1 - 2 u_0) z) \sum_{n \ge 0} (n + 1) z^n \\
&= \sum_{n \ge 0} u_0 (n + 1) z
+ \sum_{n \ge 0} (u_1 - 2 u_0) (n + 1) z^{n + 1}
\end{align}$
The coefficient of $z^n$ is now:
$$
u_n = u_0 + (n_1 - u_0) n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a power series representation for $\left(\frac{x}{2-x}\right)^3$
Find a power series representation for $\displaystyle\left(\frac{x}{2-x}\right)^3$
My approach is in finding something similar to $\displaystyle\left(\frac{x}{2-x}\right)^3$ to which I can easily find the power series representation of.
I use $\displaystyle\frac{1}{2-x}$, noting that $\displaystyle\left(\frac{1}{2-x}\right)'=\frac{1}{(2-x)^2} \text { and } \left(\frac{1}{2-x}\right)''=\frac{2}{(2-x)^3}$.
So
$$\displaystyle \frac{1}{2-x}=\int\frac{1}{(2-x)^2}dx \iff \frac{1}{2}\sum^{\infty}_{n=0}\left(\frac{x}{2}\right)^n=\int\frac{1}{(2-x)^2}dx$$
and differentiating both sides, I get the power series representation of the first derivative
$$\frac{1}{2}\sum^{\infty}_{n=1}n\left(\frac{x}{2}\right)^{n-1}=\frac{1}{(2-x)^2}$$
for the second derivative,
$$\frac{1}{(2-x)^2}=\int\frac{2}{(2-x)^3} \iff \frac{1}{2}\sum^{\infty}_{n=1}n\left(\frac{x}{2}\right)^{n-1}=\frac{1}{2}\sum^{\infty}_{n=0}(n+1)\left(\frac{x}{2}\right)^n=\int\frac{2}{(2-x)^3}dx$$
differentiating both sides, I get the power series representation of the second derivative
$$\displaystyle \frac{1}{2}\sum^{\infty}_{n=1}(n+1)n\left(\frac{x}{2}\right)^{n-1}=\frac{2}{(2-x)^3}$$
Is this so far correct? If it is, in the end I would multiply the power series representation of $\displaystyle\frac{2}{(2-x)^3}$ by $\displaystyle\frac{x^3}{2}$ to cancel out the $2$ and get the power series for $\displaystyle\left(\frac{x}{2-x}\right)^3$.
| You're approach is fine (besides the missing integration constant according to @Winther's comment).
Hint: Note, that you could also obtain a power series representation by applying the binomial series to your expression, since
\begin{align*}
\left(\frac{x}{2-x}\right)^3
=\left(\frac{x}{2}\right)^3\left(1-\frac{x}{2}\right)^{-3}
=\ldots
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Completing the square help The textbook gives this equation:
${12x^2 + 24x -8x = 0}$ with an answer of ${x = 0}$ or ${x = -{4\over3}}$
But I suspect it should be ${12x^2 + 24x -8 = 0}$
So in order to solve this, I would first isolate the x terms on one side of the equation by adding 8 to both sides:
${12x^2 + 24x = 8}$
I would then divide both sides by the coefficient of the ${x^2}$ or 12 in this case which gives:
${x^2 + 2x = {8\over12}}$
I then divide the coefficient of x by 2 and square the result and add it to both sides
${x^2 + 2x + 1 = {8\over12}}$
=> ${(x + 1)^2 = {8\over12}}$
=> ${x + 1 = \pm \sqrt{8\over12}}$
=> ${x + 1 = \pm \sqrt{2\over3}}$
=> ${x = - 1\pm \sqrt{2\over3}}$
I've taken a wrong turn somewhere, I'm not sure how to get to ${x = 0}$ or ${x = -{4\over3}}$.
| $12x^2+24x-8x=0\Rightarrow 12x^2+16x=0 \Rightarrow 3x^2+4x=0 \Rightarrow x(3x+4)=0$
So, $x=0$ or $x=- \frac 43$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$ Prove that $$\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx=\frac{\pi\sqrt{2}}{2}\log\left(1+\frac{\sqrt{2}}{2}\right).$$
I managed to prove this result with some rather roundabout complex analysis (writing the log term as an infinite sum involving nested logs), but I am hoping for a more direct solution via complex or real methods. The log term seems to require a rather complicated branch cut, so I am unsure as to how to solve the problem with a different technique.
| We can transform the integral into a form that has only one simple branch point that needs to be considered. First, let $x=u^2$, and the integral is
$$\int_0^\infty dx \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) = 2 \int_0^{\infty} du \frac{u^2}{1+u^4} \log{\left ( \frac{1+u^2}{2 u} \right )} $$
which may be rewritten as
$$-2 \log{2} \int_0^{\infty} du \frac{u^2}{1+u^4} + 2 \int_0^{\infty} \frac{du}{u^2+\frac1{u^2}} \log{\left ( u+\frac1{u} \right )} \qquad (*)$$
Let's worry about the latter integral. By subbing $v=u+1/u$, we get (exercise for the reader):
$$2 \int_0^{\infty} \frac{du}{u^2+\frac1{u^2}} \log{\left ( u+\frac1{u} \right )} = 2 \int_2^{\infty} dv \frac{v}{\sqrt{v^2-4}} \frac{\log{v}}{v^2-2} $$
Finally, sub $v^2=y^2+4$, so we get
$$2 \int_0^{\infty} \frac{du}{u^2+\frac1{u^2}} \log{\left ( u+\frac1{u} \right )}= \frac12 \int_{-\infty}^{\infty} dy \frac{\log{(y^2+4)}}{y^2+2} $$
Now we have an integral ripe for the residue theorem. Consider
$$\oint_C dz \frac{\log{(z^2+4)}}{z^2+2} $$
where $C$ is a semicircle of radius $R$ in the upper half plane, with a detour about the branch point at $z=i 2$ and up and down the imaginary axis. (For a picture, see this.) By the residue theorem, the contour integral is equal to
$$\underbrace{\int_{-\infty}^{\infty} dx \frac{\log{(x^2+4)}}{x^2+2}}_{\text{Integral over the real axis}} \underbrace{-i 2 \pi (i) \int_2^{\infty} \frac{dy}{2-y^2}}_{\text{integral over imaginary axis about branch point } z=i 2} = i 2 \pi \underbrace{\frac{\log{2}}{i 2 \sqrt{2}}}_{\text{residue at pole }z=i \sqrt{2}} $$
Thus, the second integral in (*) is equal to
$$\pi \int_2^{\infty} \frac{dy}{y^2-2} + \frac{\pi}{2 \sqrt{2}} \log{2} $$
and
$$\int_2^{\infty} \frac{dy}{y^2-2} = \frac{1}{2 \sqrt{2}} \log{(3+2 \sqrt{2})} $$
For the first integral in (*), we may use the residue theorem again, this time over a simple quarter-circle $Q$ in the upper right half plane. Thus, the integral we seek is, by the residue theorem,
$$\oint_Q dz \frac{z^2}{1+z^4} = (1+i) \int_0^{\infty} du \frac{u^2}{1+u^4} = i 2 \pi \frac{e^{i 2 \pi/4}}{4 e^{i 3 \pi/4}} $$
Thus,
$$\int_0^{\infty} du \frac{u^2}{1+u^4} = \frac{\pi}{2 \sqrt{2}} $$
Putting this all together, we finally get that the original integral is equal to
$$\int_0^\infty dx \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) =-\frac{\pi}{\sqrt{2}} \log{2} + \frac{\pi}{2 \sqrt{2}} \log{(3+2 \sqrt{2})} + \frac{\pi}{2 \sqrt{2}} \log{2} $$
or
$$\int_0^\infty dx \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) = \frac{\pi}{\sqrt{2}} \log{\left ( 1+\frac1{\sqrt{2}} \right )} $$
as was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$ Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$.
I tried to solve it.But since $n$ is given to be $\leq$ 5,my calculations went lengthy.
Applying integration by parts repeatedly,we get
\begin{align}
\int e^x(x-1)^n \, dx &= \left[(x-1)^ne^x-n(x-1)^{n-1}e^x+n(n-1)(x-1)^{n-2}e^x \right. \\
& \hspace{5mm} \left.-n(n-1)(n-2)(x-1)^{n-3}e^x+n(n-1)(n-2)(n-3)(x-1)^{n-4}e^x \right. \\
& \hspace{5mm} \left.-n(n-1)(n-2)(n-3)(n-4)e^x\right]
\end{align}
\begin{align}
\int_{0}^{1}e^x(x-1)^n \, dx &= -n(n-1)(n-2)(n-3)(n-4)e-(-1)^n+n(-1)^{n-1}-n(n-1)(-1)^{n-2} \\
& \hspace{5mm} +n(n-1)(n-2)(-1)^{n-3}-n(n-1)(n-2)(n-3)(-1)^{n-4} \\
& \hspace{5mm} +n(n-1)(n-2)(n-3)(n-4) \\
&=16-6e
\end{align}
Now solving this is very difficult,is there another simple and elegant method to find $n=3.$
| I would define
$$a_n:=\int_0^1 e^x (x-1)^n dx.$$
Then we simply get $a_0=e-1$ and for $n>0$ using integrations by parts we get
$$a_n = [e^x(x-1)^n]_0^1 - n\int_0^1 e^x (x-1)^{n-1}dx = (-1)^{n+1}-n a_{n-1}.$$
With that recursive formula you compute $a_1$, $a_2$ and $a_3$ very quickly and get your result $a_3=16-6e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
} |
How to Prove : $\frac{2}{(n+2)!}\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^{n+2}=\frac{n(3n+1)}{12}$ While I calculate an integral
$$
\int\limits_{[0,1]^n}\cdots\int(x_1+\cdots+x_n)^2\mathrm dx_1\cdots\mathrm dx_n
$$
I used two different methods and got two answers. I am sure it's equivalent, but how can I prove it?
$$\displaystyle\dfrac{2}{(n+2)!}\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^{n+2}=\dfrac{n(3n+1)}{12}$$
Sincerely thanks!
| Suppose we seek to evaluate
$$\sum_{k=0}^n (-1)^k {n\choose k} (n-k)^{n+2}.$$
We introduce
$$(n-k)^{n+2} =
\frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}} \exp((n-k)z) \; dz.$$
This yields for the sum
$$\frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}} \exp(nz)
\sum_{k=0}^n (-1)^k {n\choose k} \exp(-kz) \; dz
\\ = \frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}} \exp(nz)
(1-\exp(-z))^n \; dz
\\ = \frac{(n+2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}}
(\exp(z)-1)^n \; dz.$$
This is
$$(n+2)! [z^{n+2}] (\exp(z)-1)^n.$$
Observe that $\exp(z)-1$ starts with $z + \frac{1}{2} z^2 + \frac{1}{6} z^3 +\cdots.$
There are two possibilities here depending on which term from the series expansion is included in the product of the $n$ terms, the first is a single term with $z^3$ and the rest being $z$ which yields
$${n\choose 1} 1^{n-1} \frac{1}{6} = \frac{1}{6} n.$$
The other possibility is two terms in $z^2$ with the rest being $z$
which yields
$${n\choose 2} 1^{n-2} \frac{1}{2^2} = \frac{1}{8} n (n-1).$$
Adding these we obtain
$$\frac{n(3n+1)}{24}$$
for a final answer of
$$(n+2)! \frac{n(3n+1)}{24}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
} |
Integrate $ \int \sqrt{\frac{x^3-3}{x^{11}}}\ dx$ $$ \int \sqrt{\frac{x^3-3}{x^{11}}}\ dx $$
It seems like substitution could not do it. Is there another way?
| $x^3-3=y^2 \Rightarrow x=(y^2+3)^{\frac 1 3} \Rightarrow dx= \dfrac {2y} 3 \cdot {(y^2+3)^{\frac {-2} 3}}\cdot dy$
$$\int \sqrt{\dfrac {x^3-3} {x^{11}}}dx=\dfrac 2 3 \int\dfrac {y^2} {{(y^2+3)}^{\frac 5 2}}dy=\dfrac 23 \{\int \dfrac 1 {(y^2+3)^{\frac 3 2}}dy-3\int \dfrac 1 {(y^2+3)^{\frac 5 2}}dy \}= $$
$$=\dfrac 2 3 \{\dfrac {y} {3(y^2+3)^{\frac 1 2}}-\dfrac {y(2y^2+9)} {9(y^2+3)^{\frac 3 2}} \}+c=\dfrac {2y^3} {27(y^2+3)^{\frac 3 2}}+c=$$
$$=\dfrac {2(x^3-3)^{\frac 3 2}} {27x^{\frac 9 2}}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How can $f(x,y)= x^4+x^3y+x^2y^2+xy^3+y^4$ be factorized into a product of two polynomials? Let $x,y$ be 2 coprime integers. I assume the following polynomial:$$f(x,y)= x^4+x^3y+x^2y^2+xy^3+y^4$$ is not irreducible. So there must be at least 2 other polynomials of degree $\leq 4$ such that: $$f(x,y)=g(x,y)h(x,y)$$
How can one find $g(x,y)$ and $h(x,y)$?
| The important thing to notice
is that
$f(x, y)$
is homogeneous;
every term
of the form
$x^ay^b$
has the same value of
$a+b$.
Therefore,
dividing by
$y^n$,
where $n$ is this sum,
$\begin{array}\\
f(x,y)
&= x^4+x^3y+x^2y^2+xy^3+y^4\\
&= y^4 \left((x/y)^4+(x/y)^3+(x/y)^2+(x/y)+1 \right)\\
&= y^4 \left(r^4+r^3+r^2+r+1 \right)
\qquad\text{where } r=x/y\\
\end{array}
$
Now the problem is reduced
to factoring
$r^4+r^3+r^2+r+1
$.
| {
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"url": "https://math.stackexchange.com/questions/1409863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $((a+b)/2)^n\leq (a^n+b^n)/2$ Struggling with this proof.
Prove that $$\left(\frac{a+b}{2}\right)^n≤\frac{a^n+b^n}{2},$$ where $a$ and $b$ are real numbers such that $a+b≥0$ and $n$ is a positive integer.
What technique would you use to prove this (e.g. induction, direct, counter example). How would you go about proving it?
Thanks in advance.
| A much easier (than my original) answer
Let $c>0$ and let $$p_c(x)=(c+x)^n +(c-x)^n=2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}c^{n-2k}x^{2k}$$ Note that $p_c(x)$ has only positive coefficients and every term has even power. So $p_c(0)\leq p_c(x)$ for all $x$.
Then let $c=\frac{a+b}{2}$ and $x=\frac{a-b}{2}$. Then $c+x=a$ and $c-x=b$, and we get:
$$a^n + b^n =p_c(x)\geq p_c(0) = c^n + c^n=2\left(\frac{a+b}{2}\right)^n$$
My original answer
Preliminary
Let $$f(x,y)=\sum_{i=0}^{n-1} x^{i}y^{n-1-i}.$$ Then we can easily show that $$f(x,y)=f(y,x)\tag{1}$$
$$(x-y)f(x,y)=x^n-y^n\tag{2}$$
$$f(x,y_1)\leq f(x,y_2)\text{ when } x\geq 0\text{ and } |y_1|\leq y_2\tag{3}$$
Proof
Let $a\leq b$ and let $c=\frac{a+b}{2}$. Note $b-c=c-a\geq0.$
Now, since $a\leq b$ and $-a< b$, we have $|a|\leq b$ so we have, by $(1)$ and $(3)$:
$$f(c,a)\leq f(c,b)=f(b,c)$$
Multiplying this inequality by $c-a=b-c\geq 0$, we get, by $(2)$:
$$c^n-a^n\leq b^n-c^n.$$
Hence $$\frac{a^n+b^n}{2} \geq c^n,$$
which is the inequality we want.
Sadly, this doesn't work for non-integers $n\geq 1$.
| {
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"url": "https://math.stackexchange.com/questions/1410604",
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"source": "stackexchange",
"question_score": "3",
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Evaluation of $\int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$ Evaluation of $\displaystyle \int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx\;,$$ Now Put $x=t^2\;,$ Then $dx = 2tdt$
Then Integral $$\displaystyle I = \int\frac{(2+t)\cdot 2t}{(t^2+t+1)^2}dt = 2\int\frac{(2t+t^2)}{(t^2+t+1)^2}dt$$
Now We can write Integral $$\displaystyle I = 2\int\frac{(2t^{-3}+t^{-2})}{(1+t^{-1}+t^{-2})^2}dt\;,$$
Now Put $(1+t^{-1}+t^{-2})=u\;,$ Then $(-t^{-2}-2t^{-3})dt=du\Rightarrow (t^{-2}+2t^{-3})dt=-du$
So Integral $$\displaystyle I = -2\int \frac{1}{u^2}dt = \frac{2}{u}+\mathcal{C} = \frac{2t^2}{t^2+t+1}+\mathcal{C} = \frac{2\sqrt{x}}{x+\sqrt{x}+1}+\mathcal{C}$$
My question is why my answer is different from wolframalpha.com
http://www.wolframalpha.com/input/?i=integration+of+%282%2Bsqrt%28x%29%29%2F%28x%2Bsqrt%28x%29%2B1%29%5E2
Thanks
| You have a typo at the very last step.
Note that
$$\frac{2t^2}{t^2+t+1}=\frac{2\color{red}{x}}{x+\sqrt x+1}$$
and that
$$\frac{2x}{x+\sqrt x+1}-2=\frac{2x-2(x+\sqrt x+1)}{x+\sqrt x+1}=\frac{-2(\sqrt x+1)}{x+\sqrt x+1}$$
| {
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"url": "https://math.stackexchange.com/questions/1410667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Give the equations that are a tangent to the parabola $y = x^2 + 5x + 6$ and pass through $(1,1)$ I have been given the question:
Give the equations that are a tangent to the parabola: $y = x^2 + 5x + 6$ and pass through the point $(1,1)$
I have tried two different methods for solving this. The first I don't know why am I incorrect (I would appreciate if someone would explain to me where I went wrong).
First Attempt:
$y = x^2 + 5x + 6$
Use $y - y_1 = m(x - x_1)$ to create lines that go through $(1,1)$
$y - 1 = m(x - 1)$
$y = mx + 1 - m$
Find when these lines intersect with the parabola:
$x^2 + 5x + 6 = mx + 1 - m$
$x^2 + x(5 - m) + (5 + m) = 0$
Find for what values of M these lines only intersect once using $b^2 - 4ac = 0$
$(5-m)^2 - 4(5+m) = 0 = m^2 - 14m + 5$
$(m - 7)^2 - 49= -5$
$m = \pm \sqrt{44} + 7$
Giving the equations:
$$
y = (\sqrt{44} + 7)x - \sqrt{44} + 1
$$
$$
y = (-\sqrt{44} + 7)x + \sqrt{44} + 1
$$
With the second attempt I am lost as to what that last equation is representing so I just got confused.
second:
$ y = x^2 + 5x + 6 $
$ y' = 2x + 5 $
All the lines that have the same gradient as a the parabola and pass through $(1,1)$
$ y - 1 = (2x + 5)(x - 1) $
$y = 2x^2 + 3x - 4$
Find when these intersect the parabola:
$x^2 + 5x + 6 = 2x^2 + 3x - 4$
which gave:
$x = \pm3 + 1$
From here I got lost and didn't know what was going on.
I would appreciate it if someone could explain why my first method is wrong, tell me if either of the methods are even on track to solving the issue and give an answer to the problem.
Thanks
-Kingpulse
Solution for method one
The first method was actually correct apart from at the very end.
The mistake was the incorrect substitution into $-m$, I was substituting $-\sqrt{44}$ into $-m$ not $-(\sqrt{44}+7)$.
Proof that it is a tangent:
Parabola : $y = x^2 + 5x + 6$
Find intersections of lines
$x^2 + 5x + 6 = (\sqrt{44} + 7)x - (\sqrt{44} + 7) + 1$
$x^2 + x(-2-\sqrt{44}) + (12 + \sqrt{44}) = 0$
$b^2 - 4ac = 0$ if it is a tangent (only one root)
$(-2-\sqrt{44})^2 - 4(12+\sqrt{44})$
$4+4\sqrt{44} + 44 - 48 -4\sqrt{44} = 0$
It is a tangent.
The equations are:
$$ y = (\sqrt{44} + 7)x - (\sqrt{44} + 7) + 1$$
and
$$ y = (-\sqrt{44} + 7)x - (-\sqrt{44} + 7) + 1$$
| HINT:
For the second use parametric form $(t,t^2+5t+6)$
$$\dfrac{y-(t^2+5t+6)}{x-t}=(2x+5)_{(\text{ at }x=t)}$$
Now set $x=1,y=1$ to find the two values of $t$ to b $$1\pm\sqrt{11}$$
$$m=(2x+5)_{(\text{ at }x=t)}=2t+5=?$$
| {
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"url": "https://math.stackexchange.com/questions/1412441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $(a, b) \mid ((a + b), (a - b))$ I tried this:
Suppose $(a, b) = d$. Then $ax + by = d$.
Let $((a + b), (a – b)) = e$.
Then $$\begin{align}e& = (a + b)u + (a – b)v\\
&= au + bu + av – bv\\
&= a(u + v) + b(u – v)\end{align}$$
Let $u + v = x$ and $u – v = y$, then $d = e$. So, $d \mid e$.
What are the possible errors?
| If $(a, b) = d$, then we have
$$a+b = d\cdot\left(\frac{a}{d} + \frac{b}{d}\right)\text{ and }a-b = d\cdot\left(\frac{a}{d} - \frac{b}{d}\right)$$
and obviously then
$$((a+b), (a-b)) = \left(d\cdot\left(\frac{a}{d} + \frac{b}{d}\right), d\cdot\left(\frac{a}{d} - \frac{b}{d}\right)\right) = d\cdot\left(\left(\frac{a}{d} + \frac{b}{d}\right), \left(\frac{a}{d} - \frac{b}{d}\right)\right),$$
which even more obviously is dividable by $d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Diophantine equation $(x+y)(x+y+1) - kxy = 0$ The following came up in my solution to this question, but buried in the comments, so maybe it's worth a question of its own. Consider the Diophantine equation
$$ (x+y)(x+y+1) - kxy = 0$$
For $k=5$ and $k=6$ the positive integer solutions are given by successive terms of OEIS sequences A032908 and A101265 respectively.
Are there positive integer solutions for any other positive integers $k$?
I know there are solutions if you remove the positivity requirement, but I've been unable to find any with positive integers.
If my programming is correct, I've been able to rule out all $k \le 500$.
| What I got so far:
First, solve for $x$:
$$
x=\frac{k y-2 y-1\pm\sqrt{k^2 y^2-4 k y^2-2 k y+1}}{2}
$$
$x$ is only an integer, if the discriminant is a perfect square, so:
$$
k^2 y^2-4 k y^2-2 k y+1=z^2
$$
Solving this for $y$:
$$
y=\frac{k\pm\sqrt{k^2 z^2-4 k z^2+4 k}}{k^2-4 k}
$$
This can only happen if:
$$
k^2 z^2-4 k z^2+4 k=w^2\Rightarrow w^2-(k-4)kz^2=4k
$$
Which is a Pell equation.
If we have $w$ and $z$, then (if I made no mistakes):
$$
x=\frac{1}{4} \left(\frac{4-w}{k-4}-\frac{w}{k}+2 z\right)\qquad y=\frac{k-w}{(k-4) k}
$$
$$
x=\frac{1}{4} \left(\frac{w}{k}+\frac{w+4}{k-4}-2 z\right)\qquad y=\frac{k+w}{(k-4) k}
$$
Will give us the solutions
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 3
} |
Solve L'Hopitals problem $$\lim_{x\rightarrow \frac{\pi}{2}} \frac{\sec x}{{\sec^2 3x}} $$
I used LH: $$\lim_{x\rightarrow \frac{\pi}{2}} \frac{\sec x \tan x}{6\sec 3x \sec 3x \tan 3x}$$
then: $$\lim_{x\rightarrow \frac{\pi}{2}} \frac{\sec x\tan x}{ 6 \sec^2 3x \tan 3x}$$
Now I'm stuck there.
What should I do next?
| Solution With out Using $\bf{D-Lhopital \; Rule}$
Let $$\displaystyle y = \lim_{x\rightarrow \frac{\pi}{2}}\frac{\sec x}{(\sec 3x)^2} = \lim_{x\rightarrow \frac{\pi}{2}}\frac{(\cos 3x)^2}{\cos x}.$$
Now Using $$\bullet\; \cos 3x = 4\cos^3 x-3\cos x$$
We get $$\displaystyle y = \lim_{x\rightarrow \frac{\pi}{2}}\frac{(4\cos^3 x-3\cos x)^2}{\cos x} = \lim_{x\rightarrow \frac{\pi}{2}}\frac{\cos^2 x\cdot(4\cos^2 x-3)}{\cos x} = 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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is it true every left inverse of a matrix is also right inverse of it? I am wondering that, consider there are $m$ linear equations with $n$ unknowns. We can represent it as $AX=B$. Let $L$ is the left inverse of $A$ therefore $LA=I$. Again from $AX=B$, we get $LAX=LB$ implies $X=LB$. Till this I have no problem but from $X=LB$, multiplying it by $A$ we get $AX=ALB$ implies $B=ALB$. So does it imply also $AL=I$ ?
| It's not true for non-square matrices. Consider
$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 0 & 0 \\ 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
and
$$\begin{pmatrix} 1 & -1 \\ 0 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find a formula for the inverse of the function. $f(x) = \frac{4x − 1}{2x + 3}$ Please tell me know if my answer is right and whether the steps are correct? Thanks.
$f(x) = \frac{4x − 1}{2x + 3}$
Step 1: Write $y=f(x)$
$y=\frac{4x-1}{2x+3}$
Step 2: Solve this equation for $x$ in terms of $y$ (if possible)
2(a) Multiply both sides by $2x+3$
$(2x+3)\cdot(y)\ =\frac{4x-1}{2x+3}\cdot(2x+3)$
2(b) Distribute y term
$2xy+3y = 4x-1$
2(c) Isolate $x$ and $y$ terms
$2xy + 3y = 4x - 1 $
$-2xy+1 = -2xy$
$3y+1 = 4x-2xy$
$3y+1 = x (4-2y)$
2(d) Divide both sides by $(4-2y)$
$x=\frac{3y+1}{4-2y}$
$f^{-1}=\frac{3x+1}{4-2x} $
| $y=\dfrac{4x-1}{2x+3}$
swap $ x,y $
$x=\dfrac{4y-1}{2y+3}$
Solve back $y$ in terms of $x$
$y=\dfrac{3x+1}{-2x+4}, $ done.
EDIT1:
It is an interesting bi-linear or fractional linear function. Notice that coefficients in the left diagonal got swapped and signs of right diagonal elements changed,
$$ \dfrac{a x + b }{c x + d} \rightarrow \dfrac{d x - b }{-c x + a} $$
leaving $ (a d - b c) $ unaltered.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\langle 5, x^2+x +1 \rangle$ is maximal ideal in $\mathbb{Z}[x]$. Here is my try, of which I'm rather skeptical.
Let $I$ an ideal such that $\langle 5, x^2+x +1 \rangle \subset I \subset \mathbb{Z}[x]$. Because of the containment, there must be some $\alpha \in I$ such that:
$$\alpha \mid (5 + x^2+x+1) \implies (\alpha \mid 5) \wedge (\alpha \mid x^2+x+1).$$
Since both $5$ and $x^2 +x+1$ are irreducible, then $\alpha=1$, thus $I=\mathbb{Z}[x]$.
Thanks!
| Assume by contradiction
$$ \langle 5, x^2+x +1 \rangle \subsetneq I \subsetneq \mathbb{Z}[x] $$
Pick some $P(x) \in I$ which is not in $\langle 5, x^2+x +1 \rangle$.
By long division you can write $P(x) =(X^2+X+1)Q(X)+aX+b$.
As $X^2+X+1 \in \langle 5, x^2+x +1 \rangle \subsetneq I$ you get $aX+b \in I$ and $aX+b \notin \langle 5, x^2+x +1 \rangle$.
Case 1: If $5|a$ then $aX \in \langle 5, x^2+x +1 \rangle$ from which you get $b \in I$ and $b \notin \langle 5, x^2+x +1 \rangle$.
The former implies that $5$ doesn't divide $b$, and hence $1$ can be written as a linear combination of $5,b$, thus $1 \in I$ (contradiction).
Case 2: If $5 \nmid a$, then there exists a $k$ such that $ak=5n+1$. Then, as $5 \in I$,
$$akx+bk \in I \implies X+bk \in I$$
By dividing $bk$ by $5$ you get that $X+c \in I$ for some $c \in \{ 0, \pm 1 , \pm 2 \}$.
Then by long division of $X^2+X+1$ to $X+C$ the remainder is an integer which must be relatively prime to $5$. Thus again you get $1 \in I$, contradiction.
Alternately for the long division at the end, observe that:
If $X \in I$ then
$$1 =X^2+X+1-X(X+1) \in I$$
If $X + 1 \in I$ then
$$1=X^2+X+1-(X+1)X \in I$$
If $X - 1 \in I$ then
$$1=2(X^2+X+1)-2(X-1)(X+2)-5\in I$$
If $X +2 \in I$ then
$$1=2(X^2+X+1)-2(X-1)(X+2) -5\in I$$
If $X - 2 \in I$ then
$$1=3(X^2+X+1)-3(X+3)(X-2)-20 \in I$$
| {
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