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A circle tangent to two circles touching internally and line Find the radius of a circle touching two circle $x^2+y^2+3\sqrt{2}(x+y)=0$ and $x^2+y^2+5\sqrt{2}(x+y)=0$ and also touching the common diameter of the two given circles.
The two circles touch internally and the common diameter is $x-y=0$.
Let centre of required circle be $(h,k)$
$$\sqrt{\left(h-\frac{3}{\sqrt{2}}\right)^2+\left(k-\frac{3}{\sqrt{2}}\right)^2}=r+3$$
$$\sqrt{\left(h-\frac{5}{\sqrt{2}}\right)^2+\left(k-\frac{5}{\sqrt{2}}\right)^2}=5-r$$
and
$$r=\frac{h-k}{\sqrt{2}}$$
assuming $h>k$
Squaring,
$$h^2+k^2+3\sqrt{2}h+3\sqrt{2}k=r^2+6r$$
$$h^2+k^2+5\sqrt{2}h+5\sqrt{2}k=r^2-10r$$
substituting the value of $r$ and subtracting the two equations,
$$k=\frac{5h}{3}$$
But $h>k$. Where am I making the mistake?
I think there are two such circles which are mirror immages about $y=x$. One of them has $h>k$.
| I am posting a diagram based on my understanding of the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove (or check) the expression is positive given constraints on variables? The following proof problem have taken me a few days. Perhaps it is too hard for me to overcome it. Can you help me?
The expression is by the following:
\begin{equation}
\begin{split}
&2\,x{c}^{x-1}\ln \left( c \right) -{2}^{x}\ln \left( 2 \right) +{c}^
{x}\ln \left( c \right) +{c}^{x}{2}^{x}\ln \left( 2 \right) -x{2}^{x
}\ln \left( 2 \right) +2\,{x}^{2}{c}^{x-1}\ln \left( c \right) -{c}^
{x}\ln \left( c \right) {x}^{2}\\
&-{c}^{x}\ln \left( c \right) {2}^{x}+
2\,{c}^{x-1}+{2}^{x}-2\,{c}^{x}+{c}^{x}\ln \left( c \right) x{2}^{x}-
2\,x{c}^{x-1}\ln \left( c \right) {2}^{x}-{c}^{x}x{2}^{x}\ln \left(
2 \right) \\
&+2\,x{c}^{x-1}{2}^{x}\ln \left( 2 \right) +{c}^{x}{2}^{x}-2
\,{c}^{x-1}{2}^{x},
\end{split}
\end{equation}
where $x\in[2,+\infty)$, and $1<c<2$.
Our goal is to prove that the aforementioned expression is positive .
To facilitate subsequent view, I give each terms in the expression a unique sequence number by the following:
*
*$\qquad$$2\,x{c}^{x-1}\ln \left( c \right)$
*$\qquad$$ -{2}^{x}\ln \left( 2 \right)$
*$\qquad$${c}^{x}\ln \left( c \right)$
*$\qquad$${c}^{x}{2}^{x}\ln \left( 2 \right)$
*$\qquad$$-x{2}^{x}\ln \left( 2 \right)$
*$\qquad$$2\,{x}^{2}{c}^{x-1}\ln \left( c \right)$
*$\qquad$$-{c}^{x}\ln \left( c \right) {x}^{2}$
*$\qquad$$-{c}^{x}\ln \left( c \right) {2}^{x}$
*$\qquad$$2\,{c}^{x-1}$
*$\qquad$${2}^{x}$
*$\qquad$$-2\,{c}^{x}$
*$\qquad$${c}^{x}\ln \left( c \right) x{2}^{x}$
*$\qquad$$-2\,x{c}^{x-1}\ln \left( c \right) {2}^{x}$
*$\qquad$$-{c}^{x}x{2}^{x}\ln \left(2 \right)$
*$\qquad$$2\,x{c}^{x-1}{2}^{x}\ln \left( 2 \right)$
*$\qquad$${c}^{x}{2}^{x}$
*$\qquad$$-2\,{c}^{x-1}{2}^{x}$
Maybe the right way is $\cdots\quad$Try showing that each term is $>0$. If some are $< 0$, try combining two or more. This will get you closer to the desired proof.
But HOW?
| The following is a false proof. However, the method may be right. So it still has certain reference significance.
Combine term $9$, $10$ and $11$, we get
$$
2\,{c}^{x-1}+{2}^{x}-2\,{c}^{x}=\left( \frac{2}{c}-1 \right) {c}^{x}+\left({2}^{x}-{c}^{x}\right)>0
$$;
Combine term $4$ and $8$, we get
$$
{c}^{x}{2}^{x}\ln \left( 2 \right) -{c}^{x}\ln \left( c \right) {2}^{x}=c^x2^x\left(\ln(2)-\ln(c)\right)>0
$$;
Combine term $(12,14) + (13,15) $, we get
\begin{equation}
\begin{split}
\,&\{{c}^{x}\ln \left( c \right) x{2}^{x}-{c}^{x}x{2}^{x}\ln \left(2 \right)\}+\{-2\,x{c}^{x-1}\ln \left( c \right) {2}^{x}+2\,x{c}^{x-1}{2}^{x}\ln \left( 2 \right)\}\\
&=-c^xx2^x\left(\ln2-\ln c\right)+2xc^{x-1}2^x\left(\ln2-\ln c\right)\\
&=\left(\frac{2}{c}-1\right)2^xxc^x\left(\ln2-\ln c\right)>0
\end{split}
\end{equation}
;
Combine term $1,3,6,7$, we get
\begin{equation}
\begin{split}
&2\,x{c}^{x-1}\ln \left( c \right)+{c}^{x}\ln \left( c \right)+2\,{x}^{2}{c}^{x-1}\ln \left( c \right)-{c}^{x}\ln \left( c \right) {x}^{2}\\
&=\left(\left(\frac{2}{c}-1\right)x^2+\frac{2}{c}x+1\right)\cdot c^x\ln c
\end{split}.
\end{equation}
As the equation
$$\left(\frac{2}{c}-1\right)x^2+\frac{2}{c}x+1=0$$ has two roots $-1$ and $-\frac{1}{\frac{2}{c}-1}$, which has the following inequality relation
$$-\frac{1}{\frac{2}{c}-1}<-1,$$
So we have $$\left(\frac{2}{c}-1\right)x^2+\frac{2}{c}x+1>0$$
for $1<c<2$ and $x>2$
.
At this point, four terms have not been considered, namely $ 2,5,16$ and $17$. One can easily deduce that 2+5 and 16+17 are both negative quantities.
\begin{equation}
\begin{split}
&\{-{2}^{x}\ln \left( 2 \right) -x{2}^{x}\ln \left( 2 \right)\} +\{{c}^{x}{
2}^{x}-2\,{c}^{x-1}{2}^{x}\}\\
&=(-1-x)2^x\ln 2 +(1-\frac{2}{c})c^x2^x<0
\end{split}
\end{equation}
So something went wrong.
| {
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"url": "https://math.stackexchange.com/questions/1787212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one parameterize $x^2 + xy + y^2 = \frac{1}{2}$?
Parameterize the curve $C$ that intersects the surface
$x^2+y^2+z^2=1$ and the plane $x+y+z=0$.
I have this replacing equations:
$$ x^2+y^2+(-x-y)^2=1$$
and clearing have the following:
$$ x^2+xy+y^2=1/2$$
which it is the equation of an ellipse but I find it difficult parameterization values
Any advice will be of much help, thanks in advance
| Since $2 x^2 + 2 x y + 2 y^2$ is a quadratic form, we write
$$\begin{bmatrix} x\\ y\end{bmatrix}^T \begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} = 1$$
We now compute the eigendecomposition of the matrix above
$$\begin{array}{rl} \begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix} &= \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{bmatrix} \begin{bmatrix} 3 & 0\\ 0 & 1\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{bmatrix}\\\\ &= 3 \begin{bmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{bmatrix}^T + \begin{bmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{bmatrix}^T\end{array}$$
Thus, $2 x^2 + 2 x y + 2 y^2 = 1$ can be written as
$$3 \left(\frac{x+y}{\sqrt{2}}\right)^2 + \left(\frac{x-y}{\sqrt{2}}\right)^2 = 1$$
It should now be easy to parameterize the ellipse.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Integrate $ \int \frac{1}{1 + x^3}dx $ $$ \int \frac{1}{1 + x^3}dx $$
Attempt:
I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.
| You can factor the denominator using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$, and in our case $a=x$ and $b=1$
$$\int \dfrac{1}{(x+1)(x^2-x+1)} dx$$
Assume $\dfrac{1}{(x+1)(x^2-x+1)} = \dfrac {A}{x+1} + \dfrac {Bx+C}{x^2-x+1} $
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.
| As you observed $$\sum_{n=1}^{\infty} \frac{n(n+1)}{n!}$$ further reduces to,
$$\begin{align}
&\sum_{n=1}^{\infty} \frac{n(n+1)}{n(n-1)!} \\
=&\sum_{n=1}^{\infty} \frac{n+1}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{n}{(n-1)!}+\frac{1}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{(n-1)}{(n-1)!} + \frac{2}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{1}{(n-2)!} +\sum_{n=1}^{\infty} \frac{2}{(n-1)!},
\end{align}$$
Both summations equal $e$(euler's number) whose expansion is given by,
$$e=\sum_{n=1}^{\infty} \frac{1}{(n-1)!}.
$$
So $$\sum_{n=1}^{\infty} \frac{n(n+1)}{n!} = 3e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Evaluation of $\lim\limits_{x\rightarrow 0}\frac1x\left((1+2x+3x^2)^{1/x}-(1+2x-3x^2)^{1/x}\right) $
Evaluation of $$\lim_{x\rightarrow 0}\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} $$
$\bf{My\; Try::}$ Let $$l=\lim_{x\rightarrow 0}\frac{e^{\frac{\ln(1+2x+3x^2)}{x}}-e^{\frac{\ln(1+2x-3x^2)}{x}}}{x}$$
Using $$\bullet \; \frac{\ln(1+x)}{x}=x-\frac{x^2}{2}+\frac{x^3}{3}-.....\infty$$
But I am not Getting answer.
Now How can I solve after that, Help me
Thanks
| Consider $$A=\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} =\frac{B^{\frac{1}{x}}-C^{\frac{1}{x}}}x$$ using $$B=(1+2x+3x^2) \qquad , \qquad C=(1+2x-3x^2)$$ So $$\log(B^{\frac{1}{x}})=\frac{1}{x}\log(B)$$ and now, using Taylor series $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ replace $y$ by $(2x+3x^2)$ to get $$\log(B)=2 x+x^2-\frac{10 x^3}{3}+O\left(x^4\right)$$ which gives $$\log(B^{\frac{1}{x}})=2+x-\frac{10 x^2}{3}+O\left(x^3\right)$$ Now, using Taylor again, $$B^{\frac{1}{x}}=e^{\log(B^{\frac{1}{x}})}=e^2+e^2 x-\frac{17 e^2 x^2}{6}+O\left(x^3\right)$$ Doing the same with the second term, you should arrive to $$C^{\frac{1}{x}}=e^2-5 e^2 x+\frac{127 e^2 x^2}{6}+O\left(x^3\right)$$ All of that makes $$A=6 e^2-24 e^2 x+O\left(x^2\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Show $\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$? Prove that:
$$\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$$
The LHS is irrational number and RHS is rational number. May be this is almost a $\frac{1}{2}$.
According to the calculator it is a $\frac{1}{2}$
I tried to cube both sides but it looked too messy.
| Here's the proof I would use... note that I skip a few steps, specifically on simplifying the far right term on the LHS. Likewise, I apologize for any typos... my computer crashed thrice while writing this
$$\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$$
$$= 2^{2/3} - \frac{1}{2(\sqrt[3]2-1)} + \left(\frac{9}{4}\left(2^{1/3} - 1\right)\right)^{1/3}$$
$$= 2^{2/3} - \frac{1}{2(\sqrt[3]2-1)} + 1+\frac{1}{2^{2/3}}-\frac{1}{2^{1/3}}$$
$$= \frac{1}{2}\left(2^{2/3}(2-1) - \frac{1}{\sqrt[3]2-1} + 2+2^{1/3}\right)$$
$$= \frac{1}{2}\left(2^{2/3} - \frac{1}{2^{1/3}-1} + 2+2^{1/3}\right)$$
$$= \frac{1}{2(\sqrt[3]2-1)}\left(2-2^{2/3} - 1 + 2^{4/3} - 2+2^{2/3} - 2^{1/3}\right)$$
$$= \frac{1}{2(2^{1/3}-1)}\left( 2^{1/3} - 1\right)$$
$$=\color{red}{\frac{1}{2}}$$
Note here that the LHS is rational... the product of irrational numbers can trivially yield a rational number (take $\sqrt{2}\sqrt{2} = 2$) as can related operations
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Best way to expand $(2+x-x^2)^6$
I've completed part $(a)$ and gotten:
$64+192y+240y^2+160y^3+...$
Using intuition I substituted $x-x^2$ for $y$ and started listing the values for :
$y, y^2 $ and $y^3,$ in terms of $x$.
$y=(x-x^2)\\y^2=(x-x^2)^2 = x^2-2x^3+x^4;\\y^3 = (x-x^2)^3 = (x-x^2)(x^2-2x^3+x^4) = \;...$
Everything became complicated before I had even started to substitute these new-found values back into the polynomial devised from part $(a)$; for a 3 - out of 75 - mark question this seems extremely over-complicated. Am I doing it inefficiently / incorrectly; is there a pre-defined or easier method; or is it simple expansion and substitution that I'm too lazy to complete?
| Here is another proposal about how to calculate example 4b). We start with
Some considerations
*
*At first it was a good idea from OP to start with 4a.). Due to the similarity of the somewhat simpler expression
\begin{align*}
(2+y)^6=64+192y+240y^2+160y^3+\cdots\tag{1}
\end{align*}
compared with $(2+x-x^2)^6$ we can use the substitution
$$y:=x-x^2$$
and the result is
\begin{align*}
(2+x-x^2)^2=64+192(x-x^2)+240(x-x^2)^2+160(x-x^2)^3+\cdots\tag{2}
\end{align*}
*Since we only need an expansion with powers up to $x^3$ we don't need any terms $(x-x^2)^n$ with $n>3$.
*We also recall the binomial formulas $(a+b)^n$ for $n=2,3$
\begin{align*}
(a+b)^2&=a^2+2ab+b^2\\
(a+b)^3&=a^3+3a^2b+3ab^2+b^3
\end{align*}
which we will use when expanding (2).
Now we have all the ingredients to effectively expand (2) up to powers of $x^3$.
The calculation
We obtain from (1) with the substitution: $y:= x-x^2$
\begin{align*}
(2+x-x^2)^2&=64+192(x-x^2)+240(x-x^2)^2+160(x-x^2)^3+\cdots\\
&=64+192(x-x^2)+240(x^2-2x^3)+160(x^3)+\cdots\tag{3}\\
&=64+192x+(-192+240)x^2+(240\cdot(-2)+160)x^3+\cdots\tag{4}\\
&=64+192x+48x^2-320x^3+\cdots
\end{align*}
and we're done here.
Comment:
*
*In (3) we expand the binomial formula up to third powers of $x$; everything else is put to $+\cdots$.
*In (4) we collect terms with equal powers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the problem. Any help is appreciated.
Find the value of $$\frac{a^2}{a^4+a^2+1}$$ if $$\frac{a}{a^2+a+1}=\frac{1}{6}$$
| Hint. From the equation, one easily gets
$$
a^2=5a-1, \quad a^4=(5a-1)^2=25a^2-10a+1=115a-24
$$ giving in the first expression
$$
\frac{a^2}{a^4+a^2+1}=\frac{5a-1}{120a-24}=\frac{1 \times\color{red}{(5a-1)}}{24\times\color{red}{(5a-1)}}=\frac1{24}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
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Solution of $4 \cos x(\cos 2x+\cos 3x)+1=0$ Find the solution of the equation:
$$4 \cos x(\cos 2x+\cos 3x)+1=0$$
Applying trigonometric identity leads to
$$\cos (x) \cos \bigg(\frac{x}{2} \bigg) \cos \bigg(\frac{5x}{2} \bigg)=-\frac{1}{8}$$
But I can't understand what to do from here. Could some suggest how to proceed from here?
| Using the identities
$$
\cos(2x)=2\cos^2x-1\qquad \cos(3x)=4\cos^3x-3\cos x
$$
yields
$$
4\cos x(2\cos^2x-1+4\cos^3x-3\cos x)+1=0
$$
that is
$$
16\cos^4x+8\cos^3x-12\cos^2x-4\cos x+1=0
$$
Set $t=\cos x$. Our equation is
$$
\begin{array}{c}
16t^4+8t^3-12t^2-4t+1=0\\
(2t+1)(8t^3-6t+1)=0\\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
My solution: Consider
$$x^2 + y^2 + z^2 < 2(xy + yz + xz)$$
Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
Hence
$$(x+y+z)^2-2(xy+yz+xz) < 2(xy + yz + xz)$$
$$ (x+y+z)^2 - 4(xy+yz+xz) < 0 $$
As $x,y,z$ are all greater than zero as a side length of a triangle can not be negative. And because $(x+y+z)^2 >0$ for all real $x,y,z$ therefore the whole expression is less than zero. $Q.E.D$
Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :)
| Hint: In my personal experience, every inequality about sides of a triangle can be solved using the fact that there are positive real numbers $a,b,c$ such that $x=a+b,y=b+c,z=c+a$, i.e. the tangent points of the inner-circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with Change of Variable for the function $f(x,y)=e^{\frac{x}{2x+3y}}$ Let $D$ be the open triangle with the vertices $(0,0), (3,0), (0,2)$. For $f(x,y)=e^{ \frac{x}{2x+3y}}$ show that $f$ is integrable on $D$ and prove that $\iint_Df(x,y)dxdy=6\sqrt{e}-6$.
I was able to prove that $f$ is integrable on $D$, since $f$ is continuous everywhere but $(0,0)$ and around $(0,0)$, we have that $\frac{x}{2x+3y}<\frac{x}{2x}=\frac{1}{2}$, and therefore $f$ is bounded with a finite number of points where it's not continuous, and therefore, is integrable.
I also represented $D$ as $x\in (0,3)$ , $
y\in (0,-\frac{3x}{2}+3)$ since $y=-\frac{3x}{2}+2 $ is the hypertenuse of $D$.
Once I got to calculating the integral itself, I tried multiple changes of variables, such as polar, $u=x, v=\frac{1}{2x+3y}$, $u=x, v=2x+3y$, $u=\frac{1}{2x+3y}, v=-\frac{3x}{2}+3$, and none of these gave an integral that could be calculated using analytical tools only (no numerical tools. I also checked this with mathematica and all of these integrals require numerical tools to calculate).
What change of variables can be used here? Thanks!
| I think you had the upper limit for $y$ the wrong way around, unless you switched the vertices on the axes. If the vertices are on $(3,0)$ and $(0,2)$, the equation of the line joining them is $2x+3y=6$ so $y$ runs from $0$ to $-\tfrac{2}{3}x+2$.
If you let $u=x$ and $v=2x+3y$, then $u$ keeps the limits of $x$ and $v$ will go from $2u$ to $6$. The inverse relations are $x=u$ and $y=\tfrac{v}{3}-\tfrac{2u}{3}$, so the Jacobian is:
$$\begin{vmatrix}
1 & 0 \\
-\tfrac{2}{3} & \tfrac{1}{3}
\end{vmatrix} = \frac{1}{3}$$
The integral becomes:
$$\int_0^3 \int_0^{-\tfrac{2}{3}x+2} e^{\frac{x}{2x+3y}}\,\mbox{d}y \,\mbox{d}x = \frac{1}{3} \int_0^3 \int_{2u}^{6} e^{\frac{u}{v}}\,\mbox{d}v \,\mbox{d}u = (*)$$
Now $e^{\frac{u}{v}}$ doesn't have an elementary anti-derivative w.r.t. $v$, but you can change the order of integration. In the $uv$-plane, with $u$ from $0$ to $3$ and $v$ from $2u$ to $6$, the region is the triangle with vertices $(0,0)$, $(0,6)$ and $(3,6)$. Letting $v$ run fixed from $0$ to $6$ then gives limits for $u$ running from $0$ to $\tfrac{v}{2}$. The integral becomes easy to compute:
$$\begin{array}{rcl}
\displaystyle (*) = \frac{1}{3} \int_0^6 \int_{0}^{\tfrac{v}{2}} e^{\frac{u}{v}}\,\mbox{d}u \,\mbox{d}v
& = & \displaystyle \frac{1}{3} \int_0^6 \left[ ve^{\frac{u}{v}} \right]_{u=0}^{u=\tfrac{v}{2}} \,\mbox{d}v \\[8pt]
& = & \displaystyle \frac{1}{3} \int_0^6 \left( \sqrt{e}-1 \right)v \,\mbox{d}v \\[8pt]
& = & \displaystyle \frac{1}{3} \left( \sqrt{e}-1 \right) \left[ \frac{v^2}{2} \right]_{v=0}^{v=6} \\[8pt]
& = & \displaystyle 6\left( \sqrt{e}-1 \right)
\end{array}$$
| {
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"url": "https://math.stackexchange.com/questions/1804396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $(x+y)(y^2+z^2)(z^3+x^3) < \frac92$ for $x+y+z=2$ $x,y,z \geqslant 0$ and $x+y+z=2$, Prove
$$(x+y)(y^2+z^2)(z^3+x^3) < \frac92$$
While numerical method can solve this problem, I am more interested in classical solutions. I tried this problem for the past few months, using all kinds of AM-GM and CS, but still cannot solve it. I hope someone can help me out with this one.
| Let $f(x,y,z)=(x+y)(y^2+z^2)(z^3+x^3)$.
Assume that $y$ is largest of $x,y,z$. Then $$f(x,y,z)-f(x,z,y)=(y^2+z^2)(x + y) (x + z) (y - z) (x - y - z)\le0$$and therefore we may assume $x$ or $z$ is largest of $x, y, z$.
If $x$ is largest, then
$$f(x,y,z)-f(z,y,x)=(z^3+x^3)(z-x)(xy+xz-y^2+yz)\le0$$and therefore we may assume that $z$ is largest of $x,y,z$.
Now,$$f(0,x+y,z)-f(x,y,z)=(x+y)(((x+y)^2+z^2)z^3-(y^2+z^2)(z^3+x^3))\\=x (x+y)(-x^2 y^2 - x^2 z^2 + x z^3 + 2 y z^3)\ge0$$and we may assume that $x=0$.
Now it is just $y(y^2+(2-y)^2)(2-y)^3<\frac{9}{2}$, or$$2 y^6 - 16 y^5 + 52 y^4 - 88 y^3 + 80 y^2 - 32 y + 4.5>0$$which is true for all nonnegative $y$.
| {
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"source": "stackexchange",
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Let $n \in \mathbb{N}$. Proving that $13$ divides $(4^{2n+1} + 3^{n+2})$ Let $n \in \mathbb{N}$. Prove that $13 \mid (4^{2n+1} + 3^{n+2} ). $
Attempt: I wanted to show that $(4^{2n+1} + 3^{n+2} ) \mod 13 = 0. $ For the first term, I have $4^{2n+1} \mod 13 = (4^{2n} \cdot 4) \mod 13 = \bigg( ( 4^{2n} \mod 13) \cdot ( 4 \mod 13 ) \bigg) \mod 13. $ But still I don't know how to simplify the first term in the large bracket.
Any help/suggestions?
| Try induction. For $n=0$ the term $4^{2n+1} + 3^{n+2}$ becomes $4 + 9 = 13$, so that's OK.
Then suppose 13 divides $4^{2n+1} + 3^{n+2}$, then consider $4^{2(n+1)+1} + 3^{(n+1)+2} = 4^{2n+3} + 3^{n+3}$.
This equals $16 \cdot 4^{2n+1} + 3 \cdot 3^{n+2} = 13 \cdot 4^{2n+1} + 3\cdot( 4^{2n+1} + 3^{n+2})$, which is divisible by $13$ if the number in brackets is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808629",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot \sin^2 xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot (1-\cos^2 x)dx$$
$$I_{n} =I_{n-1}-\int_{0}^{\frac{\pi}{2}}\cos x\cdot \sin^{2n-2}\cdot \cos xdx$$
Now Using Integration by parts, We get $$I_{n} = I_{n-1}-\frac{I_{n}}{2n-1}\Rightarrow I_{n} = \frac{2n-1}{2n}I_{n-1}$$
Now Using Recursively, We get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}I_{n-2} =\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}I_{n-3}$$
So we get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \frac{2n-7}{2n-6}\cdot \cdot \cdot \cdot \cdot \cdot \cdot\cdot \frac{3}{2}I_{0}$$
and we get $\displaystyle I_{0} = \frac{\pi}{2}$
So we get $$I_{n} = \frac{(2n)!}{4^n\cdot n!\cdot n!}\cdot \frac{\pi}{2}$$
Now I did not understand How can I calculate value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Help Required, Thanks.
| $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{\sum_{n = 1}^{\infty}{1 \over 16^{n}}{2n \choose n}:\,?}$.
Lets $\ds{\mathrm{f}\pars{x} \equiv \sum_{n=0}^{\infty}{2n \choose n}x^{n}}$. Then,
\begin{align}
\mathrm{f}\pars{x} & =
1 + \sum_{n = 1}^{\infty}x^{n}\,{2n\pars{2n - 1}\pars{2n - 2}! \over
n\pars{n - 1}!\,n\pars{n - 1}!} =
1 + 2\sum_{n = 1}^{\infty}x^{n}\,{2n - 1 \over n}{2n - 2 \choose n - 1}
\\[3mm] & =
1 + 2\sum_{n = 0}^{\infty}x^{n + 1}\,{2n + 1 \over n + 1}{2n \choose n} =
1 + 4x\,\mathrm{f}\pars{x} - 2\sum_{n = 0}^{\infty}x^{n + 1}\,
{1 \over n + 1}{2n \choose n}
\\[3mm] & =
1 + 4x\,\mathrm{f}\pars{x} - 2\sum_{n = 0}^{\infty}x^{n + 1}\,
{2n \choose n}\int_{0}^{1}y^{\,n}\,\dd y =
1 + 4x\,\mathrm{f}\pars{x} - 2x\int_{0}^{1}\,\mathrm{f}\pars{xy}\,\dd y
\\[3mm] & =
1 + 4x\,\mathrm{f}\pars{x} - 2\int_{0}^{x}\,\mathrm{f}\pars{y}\,\dd y
\\[8mm]
\imp\quad\,\mathrm{f}'\pars{x} & = 4\,\mathrm{f}\pars{x} + 4x\,\mathrm{f}'\pars{x} - 2\,\mathrm{f}\pars{x}
\end{align}
$\ds{\mathrm{f}\pars{x}}$ satisfies:
$$
\mathrm{f}'\pars{x} - {2 \over 1 - 4x}\,\,\mathrm{f}\pars{x} = 0\,,
\qquad\,\mathrm{f}\pars{0} = 1
$$
Moreover,
$$
0 =\totald{\bracks{\root{1 - 4x}\,\mathrm{f}\pars{x}}}{x} =0\quad\imp
\root{1 - 4x}\,\mathrm{f}\pars{x} = \root{1 - 4 \times 0}\,\mathrm{f}\pars{0}
= 1
$$
such that
$$
{1 \over \root{1 - 4x}} = \sum_{n = 0}^{\infty}x^{n}{2n \choose n}
\quad\imp\quad
\color{#f00}{\sum_{n = 0}^{\infty}{1 \over 16^{n}}{2n \choose n}} =
\color{#f00}{{2 \over \root{3}}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Prove that $\phi \left(\frac{a^b-1}{c}\right)$ is divisible by $b$ What is the difference between the two questions below?
Prove that if $a,b,$ and $c$ are positive integers and $c \leq a$ where $\dfrac{a^b-1}{c}$ is an integer, then $\phi \left(\frac{a^b-1}{c}\right)$ is divisible by $b$.
Consider the integer $d = \frac{a^b-1}{c}$, where $a, b$, and $c$ are positive integers and $c \le a.$ Prove that the set $G$ of integers that are between $1$ and $d$ and relatively prime to $d$ (the number of such integers is denoted by $\phi(d)$) can be partitioned into $n$ subsets, each of which consists of $b$ elements.
I am confused how to go about proving this since $\dfrac{a^b-1}{c}$ is a weird looking fraction and we are taking the totient of it.
| Let $d=\frac{a^b-1}{c}$. Notice $\varphi(d)$ is the order of the multiplicative group of $\mathbb Z_d$.
What is the order of $a$ in $\mathbb Z_d^*$?
We have $a^b-1|a^b-1\implies \frac{a^b-1}{c}|a^b-1 \implies a^b \equiv 1 \bmod \frac{a^b-1}{c}$
Now notice $a^{b-1}-1\leq \frac{a^b}{c}-1< \frac{a^b-1}{c}$, so the order cannot be less than $b$.
So the order of $a$ is $b$ and so, by Lagrange's theorem $b|\varphi(d)$
| {
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How is $\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right) = n\times a^{n-1}$? In my book this is termed as a theorem and the proof given is as follows :-
$$\begin{align}
\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right)
&=\lim_{x \to a}\left(\frac{(x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^1*a^{n-2} + a^{n-1})}{x - a}\right) \\ &=(a^{n-1} + a*a^{n-2} + \cdots + a^{n-1}) \\ &=(a^{n-1} + a^{n-1} + \cdots + a^{n-1}) \\ &=(n*a^{n-1}).
\end{align}$$
Everything made sense to me except
$$x^n - a^n = (x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^1*a^{n-2} + a^{n-1})$$
Somebody please enlighten me on this topic.
| To adress your question. In the expression
$$
(x - a)\cdot(x^{n-1} + x^{n-2}\cdot a + x^{n-3}\cdot a^2 + x^{n-4}\cdot a^3 + \cdots + x^1\cdot a^{n-2} + a^{n-1})
$$ you may just expand it as
$$
x \cdot \left(x^{n-1} + x^{n-2}\cdot a + x^{n-3}\cdot a^2 + x^{n-4}\cdot a^3 + \cdots + x^1\cdot a^{n-2} + a^{n-1}\right)
$$ giving
$$\left(x^n + x^{n-1}\cdot a + x^{n-2}\cdot a^2 + x^{n-3}\cdot a^3 + \cdots + x^2\cdot a^{n-2} + x\cdot a^{n-1}\right)
$$
then substract
$$
a\cdot(x^{n-1} + x^{n-2}\cdot a + x^{n-3}\cdot a^2 + x^{n-4}\cdot a^3 + \cdots + x^1\cdot a^{n-2} + a^{n-1})
$$ that is substract
$$
(x^{n-1}\cdot a + x^{n-2}\cdot a^2 + x^{n-3}\cdot a^3 + x^{n-4}\cdot a^4 + \cdots + x^1\cdot a^{n-1} + a^n)
$$
then you can see that all terms cancel except
$$
x^n-a^n.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that: $97|2^{48}-1$ Show that: $97|2^{48}-1$
My work:
$$\begin{align}
2^{96}&\equiv{1}\pmod{97}\\
\implies (2^{48}-1)(2^{48}+1)&=97k\\
\implies (2^{24}-1)(2^{24}+1)(2^{48}+1) &=97k\\
\implies (2^{12}-1)(2^{12}+1)(2^{24}+1)(2^{48}+1)&=97k\\
\implies (2^6-1)(2^6+1)(2^{12}+1)(2^{24}+1)(2^{48}+1) &=97k
\end{align}$$
None of the terms on LHS seem to be divisible by 97!!
Direct calculation shows that: $97\mid 2^{24}+1$ , but how to find it mathematically (of course not using calculator)?
| Hint:
$2^{24}+1 = 16777217 = 172961 \times 97$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I thought maybe it's better to consider odd and even cases for $n$ but there's no unified rule here, because for example: $8^2\equiv{4}\pmod{10}$ and $8^4\equiv{6}\pmod{10}$. Any ideas??
| If $n$ is odd, then as in the answer by @Semiclassical, $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n\equiv5\pmod{10}.$
If $n$ is even, say $n=2m,$ then first observe $4^m\equiv 5+(-1)^m\pmod{10}$ and $9^m\equiv(-1)^m\pmod{10}.$
So we have
$\begin{align}1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n&\equiv2(1+2^n+3^n+4^n)+5^n\\&\equiv2(1+(5+(-1)^m)+((-1)^m)+(5+(-1)^{2m}))+5\\
&\equiv4(1+(-1)^m)+5\pmod{10}\end{align}$
Hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there is a unique positive integer $n$ such that $2^8 + 2^{11} + 2^n$ is a perfect square. I think I have almost got to the solution:
Let us start by considering two cases: $n\geq8$ and $n<8$.
1: $n\geq8$
$2^8+2^{11}+2^n$=$2^8(1+2^3+2^{n-8})=2^8(9+2^{n-8})$.......Factoring $2^8$ and simplifying terms inside the bracket.
Now since $2^8$ is already a perfect square,$(9+2^{n-8})$ has to be a perfect square as well for $2^8+2^{11}+2^n$.Therefore we can write the following equation:
$9+2^{n-8}=x^2$
=>$2^{n-8}=x^2-9$
=>$2^{n-8}=(x-3)(x+3)$
Hence, (x-3) and (x+3) are factors of a power of 2. The only factors of a power of 2,say $2^k$, is 2 and any $2^g$ where $g\leq k$. Therefore,
$x-3=2^y$.......equation 1
$x+3=2^z$.......equation 2 (Product of $2^y$ and $2^z$ is $2^{n-8}$)*
Subtracting equation 1 from 2 gives:
$2^z-2^y=(x+3)-(x-3)$
=>$2^z-2^y=6$
(let us say $y<z$)
$2^y(2^z-1)=6$. The factors of 6 that multiply to 6 are (1,6) and (2,3). So, $2^y$ and $2^z-1$ have to be equal to 1 and 6 or 2 and 3. Out of these, for $y,z$ to be integers, 2 and 3 can only fit the equation. Hence y=1;z-y=2 or z=3.From *,
$2^{n-8}=2^{y+z}=2^{3+1}=2^4$. Since $2^{n-8}=2^4$,$2^n=2^{12}$
If $z<y$,
$2^z(1-2^y)=6$. Again following the same rules we must have $(2^z,1-2^y)=(1,6)$ or (2,3). None of them satisfy the equation when $x$ and $y$ are integers.
Hence we can conclude that when $n\geq8$ , the only n that can make $2^8+2^{11}+2^n$ a perfect square is $n=12$.
2:$n<8$
Sub case 1: $n$ is odd.
Since $n$ is odd, $n$ could be expressed as $2l+1$, where $l$ is an integer. Therefore,
$2^8+2^{11}+2^n=2^8+2^{11}+2^{2l+1}=2^{2l+1}(2^{8-n}+2^{11-n}+1)=2*2^{2l}((2^{8-n}+2^{11-n}+1)$, Since there is a $2$ outside the bracket,$(2^{8-n}+2^{11-n}+1)$ must contain atleast one power of 2 and it is obvious that $(2^{8-n}+2^{11-n}+1)$ cannot be a power of 2.
In this problem, I was not able to prove that there is no $n\leq 8$ such that n is even which can be a solution to the problem . Can someone help me out with it? I arrived with three different solutions for proving that, only to realize their defects.
| For n < 8.
$2^8 + 2^{11}$ is $48^2$. That means $48^2 + 2^n$ is a perfect square. Since $2^n$ is even and $48^2$ is even, and even plus even is even, the square also has to be even. The next greatest even square after $48^2$ is $50^2$ which is $196$ greater than $48^2$. $2^7$ is $128$, which is less than $196$ which proves there is no $n$ in this case that works.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\sqrt{8+6i}$ in the form of $a+bi$ I need help with changing $\sqrt{8+6i}$ into complex number standard form.
I know the basics of complex number such as the value of $i$ and $i^2$, equality of complex number, conjugate and rationalizing method. This is my first encounter with such term and its only been a week since I learn about complex number. I would very appreciate a thorough explanation or just a calculation would suffice
Edit- Thank you guys, I've found both answer. And about the methods that Im not familiar with, It'll probably make sense to me after the next few class. I'll make sure to come back here and study them. Thanks again! You guys are great :D
| $$
8+6i = \sqrt{8^2+6^2} \cdot (\cos\theta + i\sin\theta) = 10(\cos\theta+i\sin\theta)
$$
where $\cos\theta = \dfrac 8 {\sqrt{8^2+6^2}} = \dfrac 8 {10} = \dfrac 4 5$ and $\sin\theta = \dfrac 6 {\sqrt{8^2+6^2}} = \dfrac 6 {10} = \dfrac 3 5$.
The square roots are
$$
\pm\sqrt{10} \left( \cos\frac\theta 2 + i \sin \frac\theta 2 \right).
$$
For the tangent half-angle formula we get
$$
\tan\frac \theta 2 = \frac{\sin\theta}{1+\cos\theta} = \frac{3/5}{1+ 4/5} = \frac 3 {5+4} = \frac 3 9 = \frac 1 3.
$$
So $\cos\dfrac\theta2 = \sqrt{\dfrac 1 {1+\tan^2\frac\theta 2}} = \dfrac 3 {\sqrt{10}}$ and $\sin\dfrac\theta2 = \dfrac1{\sqrt{10}}$.
Thus the square roots are
$$
\pm\sqrt{10} \left( \frac 3 {\sqrt{10}} + i \frac 1 {\sqrt{10}} \right) = \pm(3+i).
$$
It's easy to check this by multiplying: $(3+i)^2 = 8 + 6i$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 \\
\tan ^2 x & = 3
\end{align}
This is as far as I can get, and when I look at the Mark Scheme no other Trignometric Identities have been used. Thanks
| In line 4 you divided both side with $\tan x$ assuming that $\tan x\neq0$
($1$) So, if $\tan x\neq0$, you are on right track. $x=60^\circ$
($2$) But, if $\tan x=0$, you didn't consider this one. So, $\tan x=0\implies x=0^\circ$
| {
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Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$ Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$
Let the tangent planes be $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$
As the line $6x-3y-23=0=3z+2$ lies on both the planes,so put $z=\frac{-2}{3},x=\frac{3y+23}{6}$ in the both equations of the planes,
$A_1\frac{3y+23}{6}+B_1y-\frac{2C_1}{3}+D_1=0$ and $A_2\frac{3y+23}{6}+B_2y-\frac{2C_2}{3}+D_2=0$
I am stuck here.Please help.
The answer given in the book is $2x-y+4z-5=0$ and $4x-2y-z=16.$
| Given sphere:$$
x^2+y^2+z^2+2x-4y+6z-7=0.
$$
Required planes which intersect in the given line are given by$$
6x-3y-23+λ(3z+2)= 0. \tag{1}
$$
The plane (1) will be tangent to the given sphere when radius of the sphere is equal to the perpendicular distance of the plane from the center $(-1, 2, -3)$ of the sphere. This gives $λ = -\frac{1}{2}, 4$.
With these values of $λ$, equation (1) gives required tangent planes.
| {
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"url": "https://math.stackexchange.com/questions/1817414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine $\frac{f''(\frac{1}{2})}{f'(\frac{1}{2})}$ if $f(x) = \sum_{k=0}^{1000} \ {2015 \choose k}\ x^k(1-x)^{2015-k}$
Problem : Determine $\frac{f''(\frac{1}{2})}{f'(\frac{1}{2})}$ if $f(x) = \sum_{k=0}^{1000} \ {2015 \choose k}\ x^k(1-x)^{2015-k}$
Trying to simply brute force the problem, yields the following derivatives :
$$
\begin{equation}
\begin{split}
f(x) &= \sum_{k=0}^{1000} \ {2015 \choose k}\ x^k(1-x)^{2015-k} \\
f'(x) &= \sum_{k=0}^{1000} \ {2015 \choose k}\ \left[(kx^{k-1})(1-x)^{2015-k} - x^k(2015-k)(1-x)^{2014}\right]\\
f''(x) &= \sum_{k=0}^{1000} \ {2015 \choose k}\ \left[(k(k-1)x^{k-2})\right(1-x)^{2015-k} - 2kx^{k-1}(2015-k)(1-x)^{2014-k} + x^k(1-x)^{2014-k}\left[1+ (2015-k)(2014-k)(1-x)\right]]\\
\end{split}
\end{equation}
$$
But trying to evaluate $\frac{f''(\frac{1}{2})}{f'(\frac{1}{2})}$, given the above derivatives for $f'(x)$ and $f''(x)$ by conventional means is next to impossible.
Is there a simpler way to find a solution to this problem, or a method/trick to simplify the derivatives further to get them into a form that can be evaluated?
| Define $$f_{m,n}(x) = \sum_{k=0}^m \binom{n}{k} x^k (1-x)^{n-k}, \quad 0 < m \le n, \quad 0 < x < 1.$$ Then
$$f_{m,n}'(x) = \sum_{k=0}^m \binom{n}{k} k x^{k-1} (1-x)^{n-k} - (n-k) \binom{n}{k} x^k (1-x)^{n-k-1}.$$
Using the identities
$$k \binom{n}{k} = n \binom{n-1}{k-1}, \quad (n-k)\binom{n}{k} = n \binom{n-1}{k},$$
we obtain
$$\begin{align*} f_{m,n}'(x) &= n \sum_{k=1}^{m} \binom{n-1}{k-1} x^{k-1} (1-x)^{n-k} - n \sum_{k=0}^m \binom{n-1}{k} x^k (1-x)^{n-k-1} \\
&= n \sum_{k=0}^{m-1} \binom{n-1}{k} x^k (1-x)^{n-k-1} - n \sum_{k=0}^m \binom{n-1}{k} x^k (1-x)^{n-k-1} \\
&= -n \binom{n-1}{m} x^m (1-x)^{n-m-1},
\end{align*}$$
whenever such an expression is defined. Consequently,
$$\begin{align*} \frac{f_{m,n}''(x)}{f_{m,n}'(x)} &= \frac{d}{dx}\left[\log f_{m,n}'(x) \right] \\
&= \frac{d}{dx} \left[ \log \left(-n \binom{n-1}{m} \right) + m \log x + (n-m-1) \log (1-x) \right] \\
&= \frac{m}{x} - \frac{n-m-1}{1-x}.
\end{align*}$$
For $x = 1/2$, we obtain the special case
$$\frac{f_{m,n}''(1/2)}{f_{m,n}'(1/2)} = 2(1+2m-n),$$
and for the case $n = 2015$, $m = 1000$, we obtain the answer of $-28$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$(\sin^{-1} x)+ (\cos^{-1} x)^3$ How do I find the least and maximum value of $(\sin^{-1} x)+ (\cos^{-1} x)^3$ ?
I have tried the formula $(a+b)^3=a^3 + b^3 +3ab(a+b)$ , but seem to reach nowhere near ?
| $$\frac{d}{dx}\left(\arcsin x +\arccos^{3}x\right)$$
$$\frac{d}{dx}\left(\arcsin x\right) +\frac{d}{dx}\left(\arccos^{3}x\right)$$
$$\frac{1}{\cos(\arcsin x)}+3\arccos^2x\cdot\left(-\frac{1}{\sin(\arccos x)}\right)$$
$$\frac{1}{\sqrt{1-x^2}}+3\arccos^2x\cdot\left(-\frac{1}{\sqrt{1-x^2}}\right)$$
$$\frac{1-3\arccos^2 x}{\sqrt{1-x^2}}$$
This expression is equal to zero when $x=\cos\left(\frac{\sqrt{3}}{3}\right)$.
Plugging back into the original expression, we get a minimum value of $\frac{\pi}{2}-\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{9}$
Since there is no other place where the derivative equals zero, we conclude that the maximum must be at one of the two "endpoints" of the function. Trying both $x=1$ and $x=-1$, we see that the maximum occurs at $x=-1$, where $\left(\arcsin x +\arccos^{3}x\right)=\pi^3-\frac{\pi}{2}$
P.S. if you meant $\left(\arcsin x + \arccos x\right)^3$, that function is constant at $\frac{\pi^3}{8}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\frac{1}{1+x^2}\right) \cdot \frac{1}{x} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1+x^2-x^2}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x}+ \frac{1}{2}\int^{\infty}_{0} \frac{2x}{1+x^2} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 -\left[ \ln x -\frac{1}{2}\cdot \ln(1+x^2) \right]^\infty_0 $$
$$-2I=\left[ \frac{\ln x}{1+x^2}\right]^\infty_0 -\left[\ln \left (\frac{x}{\sqrt{1+x^2}} \right) \right]^\infty_0 $$
How can I evaluate the last limits ?
| A faster approach:
$$ J(a,b)=\int_{0}^{+\infty}\frac{x^b}{a^2+x^2}\,dx = a^{b-1} \frac{\pi}{2\cos\left(\frac{\pi b}{2}\right)}\qquad (a>0,b\in(-1,1))\tag{1}$$
is a consequence of the reflection formula for the $\Gamma$ function and Euler's beta function properties.
By considering:
$$ \lim_{b\to 1^-}\lim_{a\to 1^-}-\frac{1}{2}\frac{\partial^2 J}{\partial a \partial b} \tag{2}$$
we recover the value of our integral by differentiation under the integral sign. That gives:
$$ I=\int_{0}^{+\infty}\frac{x\log x}{(1+x^2)^2}\,dx = \lim_{b\to 1^-} -\frac{\pi}{8}\cdot \frac{2+\pi(b-1)\tan\left(\frac{\pi b}{2}\right)}{ \cos\left(\frac{\pi b}{2}\right)}=\color{red}{0}\tag{3}$$
that also follows from a symmetry argument:
$$ I = \int_{0}^{1}\frac{x\log x}{(1+x^2)^2}\,dx +\int_{0}^{1}\frac{\frac{1}{x}\log\left(\frac{1}{x}\right)}{x^2\left(1+\frac{1}{x^2}\right)^2}\,dx=\int_{0}^{1}0\,dx = \color{red}{0}.\tag{4} $$
| {
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"url": "https://math.stackexchange.com/questions/1822013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why my calculations aren't right? (Maclaurin series) Good evening to everyone!
I tried to calculate $ \cos\left( x- \frac{x^3}{3} + o(x^4)\right) $ using the MacLaurin series but instead of getting the final result equal to $1 - \frac{x^2}{2}+\frac{3x^4}{8} + o(x^4)$ I got this:
$$
\cos\left( x- \frac{x^3}{3} + o(x^4)\right) = 1-\frac{\left(x-\frac{x^3}{3}+o\left(x^4\right)\right)^2}{4}+o\left(x^5\right) = 1-\frac{x^2+\frac{x^6}{9}+o\left(x^8\right)-\frac{2x^4}{3}+2xo\left(x^4\right)-\frac{2x^3o\left(x^4\right)}{3}}{4}+o\left(x^5\right) = 1-\frac{x^2}{4}+\frac{x^6}{36}-\frac{x^4}{6}+o\left(x^4\right)
$$
Where for expanding what's between the parenthesis I'm using the formula $(a-b+c)^2 = a^2+b^2+c^2-2ab+2ac-2bc $.
I would really want to know where I'm doing wrong. Thanks for any possible answers.
| You are making two errors: (1) your series for $\cos x$ is wrong; (2) you are not treating the $x^4$ term correctly.
We have $\cos x=1-\frac{1}{2}x^2+\frac{1}{24}x^4+O(x^5)$.
When we replace $x$ by $x-\frac{1}{3}x^3+O(x^5)$ the $\frac{1}{2}x^2$ gives us $\frac{1}{2}(x^2-\frac{2}{3}x^4)+O(x^5)=\frac{1}{2}x^2-\frac{1}{3}x^4+O(x^5)$.
All we get from the $\frac{1}{24}x^4$ term after the replacement is $\frac{1}{24}x^4+O(x^5)$.
So we have finally $\cos x=1-\frac{1}{2}x^2+\frac{3}{8}x^4+O(x^5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822220",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $ I need help with this exercise from the book What is mathematics? An Elementary Approach to Ideas and Methods. Basically I need to proove:
$$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $$
$i)$ Particular cases
$ Q(1) = \frac{1}{2} ✓ $ ---------- $ P(1) = 2 - \frac{1+2}{2^1} = \frac{1}{2} ✓ $
$+ = 1 ✓ (=Q(1)+Q(2))$
$ Q(2) = \frac{2}{4}(=\frac{1}{2})$----$P(2) = 2 - \frac{2+2}{2^2} = 1 $ ✓
$ii)$ Hypothesis
$$F(k)=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{k}{2^k} = 2 - \frac{k+2}{2^k} $$
$iii)$ Proof
$P1 | F(k+1) = ( 2 -\frac{k+2}{2^k} ) + ( \frac{k+1}{2^{k+1}}) = 2 - \frac{k+3}{2^{k+1}} $
$P2 | \frac{2^{k+1}-k-2}{2^k} + \frac{k+1}{2^{k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
$P3 | \frac{2^{2k+2}-2^{k+1}K-2^{k+2}+2^kk+2k}{2^{2k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
$P4 | \frac{2(2^{k+1}-2^kk-2^{k+1}-2^{k-1}k+k)}{2^{k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
I get stuck here; if you could help me, that would be really kind.
Thanks in advance.
| You don't need any induction for this. This Series is simple AGP. Procedure for solving these series is assume
$$s=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}.....\frac{n-1}{2^{n-1}}+\frac{n}{2^n}$$ Now multiply whole series by $\frac{1}{2}$ and write the whole series as one term shifted $$\frac{s}{2}= \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}......\frac{n-1}{2^{n}}+\frac{n}{2^{n+1}}$$ Second series should be subtracted from first in such a way that terms which has same powers of $2$ comes togethar. Now subtract series $$\frac{s}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}......\frac{1}{2^{n}}-\frac{n}{2^{n+1}}$$ First n terms are of GP and can be written as $(1-\frac{1}{2^n})$, so $\frac{s}{2}=1-\frac{n+2}{2^{n+1}}$ . Now multiply with $2$ so as to get the correct expression for s $$s=2-\frac{n+2}{2^{n}}$$ Hence proved without induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Derivative of $x^y=y^x$ defines: $y=y(x)$ I need to find the derivative. given that:
$$x^y=y^x$$
defines:
$$y=y(x)$$
Thank you!
| Method 1
\begin{align*}
y^{x} &= x^{y} \\
x\ln y &= y\ln x \\
\ln y+\frac{xy'}{y} &= y'\ln x+\frac{y}{x} \\
\left( \frac{x}{y}-\ln x \right)y' &= \frac{y}{x}-\ln y \\
y' &= \frac{y(y-x\ln y)}{x(x-y\ln x)}
\end{align*}
Method 2
Let $y=(t+1)x$, then
\begin{align*}
[(t+1)x]^{x} &= x^{(t+1)x} \\
(t+1)^{x}x^{x} &= x^{tx} x^{x} \\
(t+1)^{x} &= x^{tx} \\
t+1 &= x^{t} \\
x &= (t+1)^{1/t} \\
y &= (t+1)^{(t+1)/t} \\
\dot{x} &= (t+1)^{1/t-1} \times \frac{t-(t+1)\ln (t+1)}{t^{2}} \\
\dot{y} &= (t+1)^{1/t+1} \times \frac{t-\ln (t+1)}{t^{2}} \\
\frac{dy}{dx} &= (t+1)^{2} \frac{t-\ln (t+1)}{t-(t+1)\ln (t+1)} \\
\end{align*}
Further points:
We omit the trivial case $y(x)\equiv x$ here.
| {
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How do I find the domain/range of functions algebraically? I've been having trouble when trying to find the domain/range of functions algebraically. Here is an example:
$P(x)=\frac{1}{3+\sqrt{x+1}}$
Finding the domain:
$x+1\ge0$
$x\ge-1$
Therefore, $x \in [-1,+\infty)$
Finding the range: Let $y=P(x)=\frac{1}{3+\sqrt{x+1}}$
From isolating x we find:
$x=(\frac{1}{y} -3)^2-1$
Therefore:
$(\frac{1}{y} -3)^2-1\ge-1$
$(\frac{1}{y} -3)^2\ge0$
$\frac{1}{y} -3\ge0$ or $\frac{1}{y} -3\le0$
$y\le \frac{1}{3}$ or $y\ge \frac{1}{3}$
This doesn't make any sense! Intuitively I can see that when $x=-1$ then $f(x)=\frac{1}{3}$ and as x approaches $+\infty$ then $f(x)$ approaches zero (without ever reaching it). How do I find this solution algebraically? What are the "rules" for working with inequalities w/ exponents and radicals (both positive and negative)? How do I find the range for other functions such as $g(x)=3+\sqrt{16-(x-3)^2}$ and $h(x)=\frac{12x-9}{6-9x}$ algebraically? A thorough explanation would be appreciated (also, feel free to point out errors in my work- there are obviously many).
| I don't fully follow what you are doing to determine the range. In any case, when you have:
$(1/y -3)^2\ge0$
The LHS is a square and thus always positive, this inequality is satisfied for all $y$...
In the formula:
$$y = \frac{1}{3+\sqrt{x+1}}$$
the range of the monotonically increasing part $\sqrt{x+1}$ is (clearly) $[0,+\infty)$, which means the denominator is monotonically decreasing with a maximum in $x=-1$, namely $y = 1/3$. For $x \to \infty$, $y \to 0$ but since $y \ne 0$ for all $x$, the range is: $0 < y \le \tfrac{1}{3}$.
Alternatively:
$$0 \le \sqrt{x+1} < +\infty$$
$$3 \le 3+\sqrt{x+1} < +\infty$$
$$ \frac{1}{3} \ge \frac{1}{3+\sqrt{x+1}} > \frac{1}{+\infty}$$
So:
$$ 0 < \frac{1}{3+\sqrt{x+1}} \le \frac{1}{3}$$
For the domain of:
$$g(x)=3+\sqrt{16-(x-3)^2}$$
You need:
$$16-(x-3)^2 \ge 0 \iff (x-3)^2 \le 16 \iff |x-3| \le 4 \iff -1 \le x \le 7 $$
For the range (given the domain as above):
$$0 \le 16-(x-3)^2 \le 16$$
$$0 \le \sqrt{16-(x-3)^2} \le 4$$
$$3 \le 3+ \sqrt{16-(x-3)^2} \le 7$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$ Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here?
Thx
| A hint to get you started:
$$3^{x+2}\cdot 4^{-(x+3)}+3^{x+4}\cdot 4^{-(x+3)}=4^{-(x+3)}\cdot\left(3^{x+2}+3^{x+4}\right)=\frac{1}{4^{x+3}}\cdot\left(3^{x+3-1}+3^{x+3+1} \right)$$
Now try to factor out a term with $3^{(\ldots)}$ and see if you can apply some exponent laws.
| {
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How to find number of solutions of an equation? Given $n$, how to count the number of solutions to the equation $$x + 2y + 2z = n$$ where $x, y, z, n$ are non-negative integers?
| We can restate this as:
$$x=n-2y-2z=n-2(y+z)$$
$y$ and $z$ are non-negative integers, so $y+z \geq 0$. We also have $x$ is a non-negative integer, so $n-2(y+z) \geq 0$, or $y+z \leq \lfloor \frac{n}{2} \rfloor$. Thus, we need to figure out the number of positive integers $y, z$ that satisfy:
$$0 \leq y+z \leq \left\lfloor \frac{n}{2} \right\rfloor$$
Now, if $y+z=0$, we have only one solution: $0+0$
If $y+z=1$, we have two solutions: $0+1=1+0$
If $y+z=2$, we have three solutions: $0+2=1+1=2+0$
From this pattern, we find that if $y+z=m$, we have $m+1$ solutions.
We need to sum $m+1$ from $m=0$ to $m=\lfloor \frac{n}{2} \rfloor$ since those are the possible values for $y+z$, so our answer is:
$$\sum_{m=0}^{\lfloor \frac{n}{2} \rfloor} m+1$$
We can say $k=m+1$ to change this sum to:
$$\sum_{k=1}^{\lfloor \frac{n}{2} \rfloor+1} k$$
Using the sum of consecutive positive integers identity, we get:
$$\frac{\left(\lfloor \frac{n}{2} \rfloor+1\right)\left(\lfloor \frac{n}{2} \rfloor+1+1\right)}{2}={\lfloor \frac{n}{2} \rfloor+2 \choose 2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$
Rewrite $(1)$ as
$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$
then
$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$
Simplified to
$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$
Then to
$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$
Any hints on what to do next?
Re-edit (Hint from Marco)
$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$
$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$
$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$
| $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{I \equiv \int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1}\,,\qquad
J \equiv \int_{0}^{\infty}{\dd x \over x^{8} + x^{4} + 1}}$
Note that
$\ds{x^{8} + x^{4} + 1 = \pars{x^{4} + x^{2} + 1}\pars{x^{4} - x^{2} + 1}}$
such that
\begin{align}
I - J & = \int_{0}^{\infty}{x^{4} - x^{2} \over x^{8} + x^{4} + 1}\,\dd x\
\stackrel{x\ \to\ 1/x}{=}\
\int_{\infty}^{0}{1/x^{4} - 1/x^{2} \over 1/x^{8} + 1/x^{4} + 1}
\,{\dd x \over -x^{2}} =
\int_{0}^{\infty}{x^{2} - x^{4} \over x^{8} + x^{4} + 1}\,\dd x
\\[3mm] & = J - I\quad\imp\quad
\fbox{$\ds{\quad\color{#f00}{I} = \color{#f00}{J}\quad}$}
\end{align}
The problem is reduced to evaluate $\ds{\underline{just\ one}}$ of the above integrals: For example, $\ds{\color{#f00}{I}}$.
\begin{align}
\color{#f00}{I} & =
\int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1}\
\stackrel{x\ \to\ 1/x}{=}\
\int_{0}^{\infty}{\dd x \over 1/x^{2} + 1 + x^{2}} =
\int_{0}^{\infty}{\dd x \over \pars{x - 1/x}^{2} + 3}\tag{1}
\\[3mm] & \mbox{Similarly,}
\\[3mm]
\color{#f00}{I} & =
\int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1} =
\int_{0}^{\infty}{1 \over x^{2} + 1 + 1/x^{2}}\,{\dd x \over x^{2}} =
\int_{0}^{\infty}{1 \over \pars{x - 1/x}^{2} + 3}
\,\dd\pars{-\,{ 1\over x}}\tag{2}
\end{align}
With $\pars{1}$ and $\pars{2}$:
\begin{align}
\color{#f00}{I} & = \color{#f00}{J} =
\half\int_{x = 0}^{x \to \infty}{1 \over \pars{x - 1/x}^{2} + 3}
\,\dd\pars{x - {1 \over x}}\
\stackrel{\pars{x - 1/x}\ \to x}{=}\
\half\int_{-\infty}^{\infty}{\dd x \over x^{2} + 3}
\\[3mm] \stackrel{x/\root{3}\ \to\ x}{=}\ &\
{1 \over \root{3}}\
\underbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}_{\ds{=\ {\pi \over 2}}}\ =\
\color{#f00}{\pi \over 2\root{3}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 5
} |
How to prove this Taylor expansion of $\frac{1}{(1+x)^2}=-1\times\displaystyle\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$? I came across this series of the Taylor Expansion-
$$\frac{1}{(1+x)^2}=1 - 2x + 3x^2 -4x^3 + \dots.=-1\times\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$$
But I have no idea how to prove this...
Thanks for any help!
| Here isn't a VERFICATION of hte fact, but a full derivation, as if you had to come up with this question all by yourself to ask someone else.
Recall that taylor's theorem states:
Given an smooth function $f(x)$ around a point $a$ (that is the function is infinitely differentiable around a, so $f(a)$, $f'(a)$, $f''(a)$ all exist.
Then it is the case that around this point
$$ f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \frac{1}{3!}f'''(a)(x-a)^3 + ... $$
So given:
$$ f(x) = \frac{1}{(1+x)^2}$$
Observe that
$$ f'(x) = -\frac{2}{(1+x)^3} $$
$$ f''(x) = \frac{2 \times 3}{(1+x)^4}$$
$$ f'''(x) = -\frac{2 \times 3 \times 4}{(1+x)^4}$$
And catching a pattern we prove in general that if:
$$ f^{(n)}(x) = (-1)^n \frac{(n+1)!}{(1+x)^{n+2}}$$
Then we verify by differentiating this that:
$$ f^{(n+1)}(x) = (-1)^{n+1} \frac{(n+2)!}{(1+x)^{n+3}}$$
So now we derive the series from taylors formula,
$$ f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \frac{1}{3!}f'''(a)(x-a)^3 + ... $$
Letting $a = 0$
$$ f(x) = f(a) + f'(a)x + \frac{1}{2!}f''(a)x^2 + \frac{1}{3!}f'''(a)x^3 + ... $$
Now substituting:
$$ \frac{1}{(1+x)^2} = 1-\frac{2}{(1+0)^3}x - \frac{1}{2!}\frac{3!}{(1+0)^4}x^2 + \frac{1}{3!} \frac{4!}{(1+0)^5}x^3 + ...$$
And simplifying $(1+0) = 1, \frac{(n+1)!}{n!} = (n+1)$ we have
$$ \frac{1}{(1+x)^2} = 1-2x + 3x^2 - 4x^3 + ...$$
As desired
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Determine $f(x)$ which satisfies given condition Suppose $f(x)$ is real valued function of degree $6$ satisfying the following conditions:
$1.$ $f(x)$ has minimum at $x=0$ and $x=2$
$2.$ $f(x)$ has maximum at x=1
$3.$ $lim(x \to 0)$ $\frac{ln(\Delta)}{x}=2$ where $\Delta$ is
\begin{vmatrix}
\frac{f(x)}{x} & 1 & 0 &\\
0 & \frac{1}{x} & 1 &\\
1 & 0 & \frac{1}{x} &\\
\notag
\end{vmatrix}
Determine $f(x)$
If we assume $f(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$, set $f'(x)=0$ at given critical points even then we will get only three equations.
One equation can be formed by using the fact that $\Delta$ must be equal to $1$ at $x=0$ as limit is finite.
One more equation can be formed by using $lim(x \to 0)$ $\frac{ln(\Delta)}{x}=2$ but still the solution is short of two equations (As we have to find 7 variables) and method it too length as well.
Could someone suggest a better approach?
| For the limit to exist we need
\begin{align*}
&\lim_{x\rightarrow 0} \ln \Delta(0) = 0\\
\implies &\lim_{x\rightarrow 0} \Delta(0) = 1\\
\implies &\lim_{x\rightarrow 0} f(x)/x^3+1 = 1\\
\implies &\lim_{x\rightarrow 0} f(x)/x^3 = 0.
\end{align*}
Since $f(x)$ is continuous this is only possible if $f(0) = 0$. By L'hopital you also get $f'(0) = f''(0) = f'''(0) = 0$, so $f(x)$ is divisible by $x^4$.
The derivative has zeros at $x=0,1,2$, so $f'(x) = g(x)x(x-1)(x-2)$ with $g(x)$ having degree 2. Since $f(x)$ is divisible by $x^4$, $f'(x)$ will be divisible by $x^3$, so $g(x)=cx^2$.
So
\begin{align*}
f'(x) &= cx^3(x-1)(x-2) = c(x^5-3x^4+2x^3)\\
f(x) &= c((1/6)x^6 - (3/5)x^5 + (1/2)x^4)
\end{align*}
Now solve for $c$ to make the limit work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A very curious rational fraction that converges. What is the value? Is there any closed form for the following limit?
Define the sequence
$$ \begin{cases}
a_{n+1} = b_n+2a_n + 14\\
b_{n+1} = 9b_n+ 2a_n+70
\end{cases}$$
with initial values $a_0 = b_0 = 1$. Then $\lim_{n\to\infty} \frac{a_n}{b_n} = ? $
The limit is approximately $0.1376$. My math teacher Carlos Ivorra says that this limit have a closed form involving the sine of an angle. What is the closed form for is limit?
NOTE: I have found this (and another series of converging sequences) by the use of an ancient method for calculating sines recently rediscovered. I'll give the details soon as a more general question.
| The requested limit is:
$$ \frac{4 \sqrt{57} - 20}{4 \sqrt{57} + 44} \approx 0.1374586 $$
This is
$$ \frac{ \sqrt{57} - 5}{ \sqrt{57} + 11} $$
and rationalizing the denominator gives
$$ \frac{ \sqrt{57} - 7}{ 4} $$
$$ a_{n+2} = 11 a_{n+1} - 16 a_n - 42 $$
$$ b_{n+2} = 11 b_{n+1} - 16 b_n - 42 $$
The separate linear recurrences are the result of the Cayley-Hamilton Theorem applied to the matrix
$$
\left(
\begin{array}{rr}
2 & 1 \\
2 & 9
\end{array}
\right)
$$
although I wrote everything out in detail because I was not sure what the constant terms $14,70$ would do.
Hmmm. Good thing I was careful, it was not necessary that the 42's come out the same. Given the matrix system $X_{n+1} = A X_n + B,$ where $\tau = \operatorname{trace} A$ and $\delta = \det A,$ we get
$$ X_{n+2} = \tau X_{n+1} - \delta X_n + (A - (\tau - 1)I) B. $$
There is no reason to expect the two components of $(A - (\tau - 1)I) B$ to come out the same, it was arranged for this particular problem. Indeed, here
$$
(A - (\tau - 1)I)^{-1} = \frac{1}{6}
\left(
\begin{array}{rr}
1 & 1 \\
2 & 8
\end{array}
\right)
$$
so to get the two constants the same it was required to take the constant vector $B$ as a scalar multiple of
$$
\left(
\begin{array}{rr}
1 & 1 \\
2 & 8
\end{array}
\right)
\left(
\begin{array}{r}
1 \\
1
\end{array}
\right) =
\left(
\begin{array}{r}
2 \\
10
\end{array}
\right),
$$
and they multiplied this by $7.$
$$ a_n = \left( 4 - \frac{20}{\sqrt{57}} \right) \left( \frac{11 + \sqrt {57}}{2} \right)^n + \left( 4 + \frac{20}{\sqrt{57}} \right) \left( \frac{11 - \sqrt {57}}{2} \right)^n - 7 $$
$$ b_n = \left( 4 + \frac{44}{\sqrt{57}} \right) \left( \frac{11 + \sqrt {57}}{2} \right)^n + \left( 4 - \frac{44}{\sqrt{57}} \right) \left( \frac{11 - \sqrt {57}}{2} \right)^n - 7 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 0
} |
Partial fraction integration problem I'm trying to solve this integral by partial fraction:
$$
\int \frac{2x-6} {(x-2)^2(x^2+4)} dx \
$$
i think i have to write the expression like
$$
2\int \frac{x-3} {(x-2)^3(x+2)} dx \
$$
Then i don't know how i should resolve the partial fraction!
| Careful!
$$
\int \frac{2x-6} {(x-2)^2(x^2+4)} dx \
$$
i think i have to write the expression like
$$
2\int \frac{x-3} {(x-2)^3(x+2)} dx \
$$
You seem to have replaced $x^2+4$ by $(x-2)(x+2)$, but these are not the same! You might be confusing with $x^2\color{red}{-}4=(x-2)(x+2)$...
You can take out the (constant) factor $2$, but you don't have to. For the partial fraction decomposition, you're looking for numbers $A$, $B$, $C$ and $D$ such that:
$$\frac{2x-6} {(x-2)^2(x^2+4)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4} $$
Once you have that, the integral splits into these three terms and these are all easy to integrate.
I can elaborate on the partial fraction decomposition; unless you can take it from here?
For the partial fraction decomposition:
$$\begin{array}{rl}
\displaystyle \frac{2x-6} {(x-2)^2(x^2+4)}
& \displaystyle = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4} \\[7pt]
& \displaystyle = \frac{A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2}{(x-2)^2(x^2+4)} \end{array}$$
Now equating the numerators:
$$2x-6 = A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2$$
Method 1
You can expand the RHS and group per power of $x$ and then identify the corresponding coefficients left and right; this gives you four linear equations in the 4 unknown variables.
$$2x-6 = (A+C)x^3+(-2A+B-4C+D)x^2+(4A+4C-4D)x+(-8A+4B+4D)$$
Which gives the system:
$$\left\{\begin{array}{rcl}
A+C & = & 0 \\
-2A+B-4C+D & = & 0 \\
4A+4C-4D & =& 2 \\
-8A+4B+4D & = & -6
\end{array} \right. \quad \Rightarrow \quad
\left\{\begin{array}{rcl}
A & = & \tfrac{3}{8} \\
B & = & -\tfrac{1}{4} \\
C & =& -\tfrac{3}{8} \\
D & = & -\tfrac{1}{2}
\end{array} \right.$$
Method 2
It is indeed also possible to simplify by choosing handy values for $x$ to substitute into
$$2x-6 = A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2$$
For example:
*
*Substitution of $x=2$ yields $-2=8B$ so $B = -\tfrac{1}{4}$.
*Substitution of $x=0$ yields $-6=-8A+4B+4D$ so $8A-4D=5$.
*(...)
If you know complex numbers, substitution of $x = 2i$ gives you $C$ and $D$ immediately, then $A$ follows from $8A-4D=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that
$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$
Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$
Let see (substitution of $y=x^2$)
$$\int_{-\infty}^{\infty}{x\over (x^4-x^2+1)^2}dx={1\over 2}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$
Substituion of $y=x^3$
$$\int_{-\infty}^{\infty}{x^3\over (x^4-x^2+1)^2}dx={1\over 4}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$
As for $\int_{-\infty}^{\infty}{x^2\over (x^4-x^2+1)^2}dx$ and $\int_{-\infty}^{\infty}{x^4\over (x^4-x^2+1)^2}dx$ are difficult to find a suitable substitution. This is the point where I am shrugged with to find a suitable substitution To lead me to a particular standard integral. Need some help, thank.
standard integral of the form
$$\int{1\over (ax^2+bx+c)^2}dx={2ax+b\over (4ac-b^2)(ax^2+bx+c)}+{2a\over 4ac-b^2}\int{1\over ax^2+bx+c}dx$$
And
$$\int{1\over ax^2+bx+c}dx={2\over \sqrt{4ac-b^2}}\tan^{-1}{2ax+b\over \sqrt{4ac-b^2}}$$
| Trick: the integral over $\mathbb{R}$ of a non-vanishing meromorphic function, $O\left(\frac{1}{\|z\|^2}\right)$ at infinity, is just $2\pi i$ times the sum of the residues for the poles in the upper half-plane. In our case such poles are located at $z=e^{\pi i/6}$ and $z=e^{5\pi i/6}$ (roots of $z^2-iz-1$) and the sum of the residues is
$$ -\frac{i}{2}\left(1+\pi+\pi^2\right). $$
After that, it is simple math. A worked example of the same technique in a similar (but apparently much harder) problem can be found here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$. Quadratic equation question, as specified in the title.
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.
I gather that $α + β = -\frac{b}{a} = \frac{6}{1} = 6$ and that $αβ = \frac{c}{a} = \frac{7}{1} = 7$. Do I need to convert $α + \frac{1}{β}$ and $β + \frac{1}{α}$ into a format whereby I can sub in the values for adding together or multiplying $α$ and $β$ ? If so, how ?
| $$(x-(\alpha + 1/\beta))(x - (\beta + 1/\alpha)) = x^2 - (\alpha + \beta + 1/\alpha + 1/\beta) x + (\alpha + 1/\beta)(\beta + 1/\alpha $$
You know $\alpha + \beta = 6$ and $\alpha \beta = 7$.
$$ \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha \beta} = \dfrac{6}{7}$$
$$\left( \alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right) = \alpha \beta + 2 + \frac{1}{\alpha \beta} = 7 + 2 + \frac{1}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $18$ which is clearly not even close. It says prove shouldn't it be always true?
| Proof by induction on $n$;
Set your base induction for $n = 1$; which gives you $18 = 18.$
Suppose your equation is true for $n$. We are interested to prove its correctness for $n+1.$ So
$3 + 3 \times 5 + \cdots + 3 \times 5^n + 3 \times 5^{n+1} = \frac{3 \times 5^{n+1} - 3}{4} + 3 \times 5^{n+1} = \frac{3 \times 5^{n+1} - 3 + 12 \times 5^{n+1} }{4} = \frac{3 ( 5^{n+1} + 4 \times 5^{n+1}) - 3}{4} = \frac{3 ( 5^{n+2} - 1)}{4}$
by using the induction steps. And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Feasible point of a system of linear inequalities Let $P$ denote $(x,y,z)\in \mathbb R^3$, which satisfies the inequalities:
$$-2x+y+z\leq 4$$ $$x \geq 1$$ $$y\geq2$$ $$ z \geq 3 $$ $$x-2y+z \leq 1$$ $$ 2x+2y-z \leq 5$$
How do I find an interior point in $P$?
Is there a specific method, or should I just try some random combinations and then logically find an interior point?
| Label the planes and their intersection points as follows
$\begin{array}{llllllll}
&\Pi_1:&x=1 & A & B & C & & & W & X\\
&\Pi_2:&y=2 & A & B & & D & & W & & Y\\
&\Pi_3:&z=3 & A & & C & D & & & X & Y & Z\\
&\Pi_4:&-2x+y+z=4 & & B & C & & E & & & Y & Z\\
&\Pi_5:&x-2y+z=1 & & B & & D & E & & X & & Z\\
&\Pi_6:&2x+2y-z=5 & & & C & D & E & W\\
\end{array}$
where $A=(1,2,3),B=(1,2,4),C=(1,3,3),D=(2,2,3),E=(2,3,5)$ are points which also satisfy all the inequalities and $W=(1,2,1),X=(1,\frac{3}{2},3),Y=(\frac{1}{2},2,3),Z=(0,1,3)$ fail to satisfy all the inequalities.
The five points $A,B,C,D,E$ are the vertices of a triangular bipyramid. The two pyramids are $ABCD$ and $BCDE$ with common base $BCD$ which is internal and not a face. The six faces are the planes $\Pi_i$.
The graphic below has the $z$-axis vertical, with $E$ the apex at the top ($z=5$), $B$ the vertex halfway down ($z=4$) and $\Pi_3=ACD$ the triangular base (at $z=3$). It has two hidden triangular faces - the base and the back $\Pi_6=CDE$. The light blue vertical face is $\Pi_1=ABC$, and the purple vertical face is $\Pi_2=ABD$. The other two visible faces are $\Pi_4,\Pi_5$.
The centroid of the polyhedron is $(1.4,2.4,3.6)$. It obviously satisfies the three inequalities related to $\Pi_1,\Pi_2,\Pi_3$. For $\Pi_4$ we have $-2.8+2.4+3.6=3.2<4$; for $\Pi_5$ we have $1.4-4.8+3.6=0.2<1$; and for $\Pi_6$ we have $2.8+4.8-3.6=4<5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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How to solve the limit of this sequence $\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$ $$\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$$
I have tried to split the subset into telescopic series but got no result.
I also have tried to use the squeeze theorem by putting the $a_n$ between $\frac{1}{(2n-1)(2n+1)}$ and $\frac{1}{(4n-1)(4n+1)}$ but it doesn't work.
| Alternatively one may recall the series representation for the digamma function
$$\begin{equation}
\psi(x+1) = -\gamma - \sum_{n=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant.
Then by partial fraction decomposition we have
$$
\begin{align}
\frac1{(6n-3)(6n+2)} &= \frac15\left(\frac{1}{6n-3}-\frac{1}{6n+2}\right)\\\\
&=\frac1{30}\left[\left(\frac{1}{n-1/2}-\frac1n\right)-\left(\frac{1}{n+1/3}-\frac1n\right)\right]
\end{align}
$$ then summing from $n=1$ to $+\infty$, using $(1)$, we get
$$
\sum_{n=1}^{\infty}\frac1{(6n-3)(6n+2)} =\frac1{30}\left(\psi\left(\frac43\right)-\psi\left(\frac12\right)\right)
$$ equivalently
$$
\sum_{n=1}^{\infty}\frac1{(6n-3)(6n+2)} =\frac{1}{180} \left(18-\sqrt{3} \pi -9 \ln 3+24 \ln 2\right)
$$
where we have used special values of the digamma function.
| {
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"url": "https://math.stackexchange.com/questions/1839102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove
$$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$
I rearranged it
$$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$
My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?
Thanks
| Hint:
Set $1/a=x$ etc.
and use $(x-y)^2\ge0$ for real $x,y$
| {
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For how many 3-digit prime numbers $\overline{abc}$ do we have: $b^2-4ac=9$?
For how many 3-digit prime numbers $\overline{abc}$ do we have: $b^2-4ac=9$?
The only analysis I did is:
$(b-3)(b+3)=4ac \implies\ b\geq3 $
$b=3\implies\ c=0\implies impossible!!$
So I deduced that $b\gt3$
Is there any better way for quickly solving this??
Or maybe a trick applies...
| $(b-3)(b+3) = 4ac$
Now, $4ac$ is an even number. Hence, $(b-3)(b+3)$ is also and even number. That means, b cannot be even number else, $(b-3)(b+3)$ will if odd.
Hence, $b > 3$ and b is odd. Thus, b can be 5, 7, 9.
Putting, the values for b:
When $b = 5$, we get $4ac = (5-3)(5+3) = 2 * 8 = 16$ or $ac = 4 = 2*2 = 1*4$. No prime numbers possible with these combinations.
When $b = 7$, we get $4ac = (7-3)(7+3) = 4.10 = 40$ or $ac = 10 = 2*5$. Again, no combinations possible.
When $b = 9$, we get $4ac = (9-3)(9+3) = 6.12 = 72$ or $ac = 18 = 2*9 = 3*6$. Again, no combinations possible.
Thus, no such 3 digit prime number exists.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = 0 $$
$$ \sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right) = 0 $$
Here one of the factors has to be $0$,
$$ \sin x = 0 \ \Rightarrow \ x = 0 \ or \ x = \pi $$
$$ \sin\left(\frac{x}{2}'\right) = 0 \ \Rightarrow \ x = 0 ;\ x \text{ can't be } \pi \text{ because of its range} $$
$$ \cos x = 0 \ \Rightarrow \ x = \frac{\pi}{2} \text{ or } \ x = \frac{3\pi}{2} $$
$$ \cos\left(\frac{3x}{2}\right) = 0 => x = \frac{\pi}{3} \text{ or } x = \pi $$
So the solutions are : $$ 0, \pi, \frac{\pi}{3} $$
I have seen other methods to solve this, so please don't post them. I'm really interested what's wrong with this one.
| $$\sin(2x)=\sin(x)\implies 2\cos(x)\sin(x)=\sin(x).$$
The elements of the form $k\pi$ are solutions (notice that if you divide the equation by $\sin(x)$ with not paying attention to this condition, you will loose solutions). Suppose $x\neq k\pi$. Then,
$$2\cos(x)\sin(x)=\sin(x)\implies 2\cos(x)=1\implies \cos(x)=\frac{1}{2}\implies x\in \left\{\pm\frac{\pi}{3}+2k\pi\mid k\in\mathbb Z\right\}.$$
Finally the solution are given by
$$\{k\pi\mid k\in\mathbb Z\}\cup\left\{\pm\frac{\pi}{3}+2k\pi\mid k\in\mathbb Z\right\}.$$
I let you take the solutions in $[0,2\pi[$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x\in \left(0,\frac{\pi}{4}\right)$ then $\frac{\cos x}{(\sin^2 x)(\cos x-\sin x)}>8$
If $\displaystyle x\in \left(0,\frac{\pi}{4}\right)\;,$ Then prove that $\displaystyle \frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$
$\bf{My\; Try::}$ Let $$f(x) = \frac{\cos x}{\sin^2 x(\cos x-\sin x)}=\frac{\sec^2 x}{\tan^2 x(1-\tan x)} = \frac{1+\tan^2 x}{\tan^2(1-\tan x)}$$
Now Put $\tan x= t \in (0,1)\;,$ Then $$h(t) = \frac{1+t^2}{t^2(1-t)}\;\;, 0<t<1 $$ where $h(t)=f(x)$.
Now How can i solve it after that, Help Required ,Thanks
| $\bf{My\; Solution::}$ Using $\bf{A.M\geq G.M}$
$$2\sqrt{\sin x(\cos x-\sin x)}\leq \cos x$$
so $$4\sin x(\cos x-\sin x)\leq \cos^2 x\Rightarrow \frac{\sin^2 x(\cos x-\sin x)}{\cos x}\leq \frac{\cos x\sin x}{4}$$
So $$\frac{\sin^2 x(\cos x-\sin x)}{\cos x}\leq \frac{\sin 2x}{8}< \frac{1}{8}$$
So $$\frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a}{b} + \frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}$ Prove that $\frac{a}{b} + \frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}$ with a,b,c > 0
| Hint:
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-3=\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)+\left(\dfrac{b}{c}+\dfrac{c}{a}-\dfrac{b}{a}-1\right)=\dfrac{(a-b)^2}{ab}+\dfrac{(a-c)(b-c)}{ac}\tag{1}$$
Similarly, $$\dfrac{a+c}{b+c}+\dfrac{b+c}{a+b}+\dfrac{a+b}{a+c}-3=\dfrac{(a-b)^2}{(b+c)(a+b)}+\dfrac{(a-c)(b-c)}{(b+c)(a+c)}\tag{2}$$
As $a,b,c >0$, it is obvious that $(1)$ is greater than $(2)$ as without loss of generality, we can set $c=\max{\{a,b,c\}}$ or $c=\min{\{a,b,c\}}$
| {
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Decreasing sequence numbers with first digit $9$
Find the sum of all positive integers whose digits (in base ten) form a strictly decreasing sequence with first digit $9$.
The method I thought of for solving this was very computational and it depended on a lot of casework. Is there a nicer way to solve this question?
Note that there are $\displaystyle\sum_{n=0}^{9} \binom{9}{n} = 2^9$ such numbers.
| For each digit $a$ from $0$ to $9$, let us count how many numbers there are in our sum with a digit of $a$ in the $10^n$ place. First, suppose $a<9$. A number with a digit of $a$ in the $10^n$ place has a subset of $\{0,1,\dots,a-1\}$ for its last $n$ digits, so there are $\binom{a}{n}$ choices for the last $n$ digits. The preceding digits (omitting the initial $9$) can form any subset of $\{8,7,\dots,a+1\}$, so there are $2^{8-a}$ choices for the preceding digits. So in total, all of the digits of $a$ in our numbers contribute $\sum_n 2^{8-a}\binom{a}{n}\cdot a10^n$ to the sum. By the binomial theorem, this is equal to $$2^{8-a}a(10+1)^a=2^8a\left(\frac{11}{2}\right)^a.$$
For $a=9$, we have no choice of preceding digits, so we just have $\binom{9}{n}$ choices. So the sum of the $9$ digits contributes $$\sum_n \binom{9}{n}9\cdot 10^n=9\cdot 11^9.$$
So in total, the sum is $$9\cdot 11^9+2^8\sum_{a=0}^8 a\left(\frac{11}{2}\right)^a.$$
Let us write $x=\frac{11}{2}$ and simplify the sum in the second term:
$$\sum_{a=0}^8 ax^a=\sum_{b=1}^8\sum_{c=b}^8x^c=\sum_{b=1}^8\frac{x^9-x^b}{x-1}=\frac{8x^9-\frac{x^9-x}{x-1}}{x-1}.$$
Putting it all together, the original sum is
$$9\cdot 11^9+2^8\cdot\frac{8x^9-\frac{x^9-x}{x-1}}{x-1}$$
for $x=\frac{11}{2}$. According to Wolfram Alpha, this evaluates to $23259261861$. That is, assuming I haven't made some algebra mistake somewhere.
| {
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Find $\sin \theta $ in the equation $8\sin\theta = 4 + \cos\theta$ Find $\sin\theta$ in the following trigonometric equation
$8\sin\theta = 4 + \cos\theta$
My try ->
$8\sin\theta = 4 + \cos\theta$
[Squaring Both the Sides]
=> $64\sin^{2}\theta = 16 + 8\cos\theta + \cos^{2}\theta$
=> $64\sin^{2}\theta - \cos^{2}\theta= 16 + 8\cos\theta $
[Adding on both the sides]
=> $64\sin^{2}\theta + 64\cos^{2}\theta= 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $64 = 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $48 = 8\cos\theta + 65\cos^{2}\theta$
=> $48 = \cos\theta(65\cos\theta + 8)$
I can't figure out what to do next !
| $$8\sin\theta=4+\cos\theta$$
$$8\sin\theta-\cos\theta=4$$
$$\sqrt{65}\sin\alpha\sin\theta-\sqrt{65}\cos\alpha\cos\theta=4$$
where $\alpha=\tan^{-1}8$
$$-\sqrt{65}\cos(\theta+\alpha)=4$$
Can you continue from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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implicit derivative for equation $y^5 = x^8$ I'm having a bit of trouble finding the second implicit derivative for the equation $y^5 = x^8$.
I have the first derivative, which is $8x^7 / 5y^4$, but I'm having trouble after that. I did the quotient rule and got $280x^6y^4 - 160x^7y^3 / (20y^3)^2$. I'm not quite sure what to do after this. Any help would be appreciated. Thanks!
| $${ y }^{ 5 }-x^{ 8 }=0\\ \\ 5{ y }^{ 4 }\frac { dy }{ dx } -8{ x }^{ 7 }=0\\ \frac { dy }{ dx } =\frac { 8{ x }^{ 7 } }{ 5{ y }^{ 4 } } \\ \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\frac { d }{ dx } \left( \frac { 8{ x }^{ 7 } }{ 5{ y }^{ 4 } } \right) =\frac { 56{ x }^{ 6 }\cdot 5{ y }^{ 4 }-8{ x }^{ 7 }\cdot 20{ y }^{ 3 }\frac { dy }{ dx } }{ { \left( 5{ y }^{ 4 } \right) }^{ 2 } } =\frac { 280{ x }^{ 6 }{ y }^{ 4 }-160{ x }^{ 7 }{ y }^{ 3 }\frac { 8{ x }^{ 7 } }{ 5{ y }^{ 4 } } }{ 25{ y }^{ 8 } } \\ =\frac { 280{ x }^{ 6 }{ y }^{ 4 }-256{ x }^{ 14 }{ y }^{ -4 } }{ 25{ y }^{ 8 } } $$
| {
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Possible assistance with a monstrous pair of integrals In my study of ODEs I have recently encountered this monster of an integral
the sum of two integrals:
$ \int_{0}^{2\pi} \frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx + \int_{0}^{2\pi} \frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx $
Even Wolfram is having trouble with this, I was hoping someone could please tell me if I can do it analytically or numerically any way at all? It really is scary looking. Of course $\omega$ and $A$ are parameters with $A$ between 0 and 1.
Edits:
The first one (with the cosine) is zero as the great answers point out but the one with the product of two sines is a mystery.
I feel I owe an explanation: the relation to ODE comes from a previous problem I stated here dealing with the Melnikov integral here the answer changes variables from t to x.
| It's simpler than it appears at first :
Let $\quad f(x)=\frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}$
$$\int_0^{2\pi} f(x)dx = \int_0^{\pi} f(x)dx +\int_{\pi}^{2\pi} f(x)dx $$
Let $\quad x=2\pi-t \quad$ It is easy to prove that $\quad f(x)=-f(t)$
because $\sin(x)=\sin(-t)=-\sin(t)$ and $\sin^2\left(\frac{x}{2}\right)=\sin^2\left(\pi-\frac{t}{2}\right)=\sin^2\left(\frac{t}{2}\right)$
$\int_{\pi}^{2\pi} f(x)dx =\int_{\pi}^{0} \left(-f(t)\right)(-dt) =-\int_{0}^{\pi} f(t)dt $
$$\int_0^{2\pi} f(x)dx = \int_0^{\pi} f(x)dx -\int_{0}^{\pi} f(t)dt=0 $$
$$\int_0^{2\pi}\frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx=0$$
With the same method and $g(x)=\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx$
$g(x)=g(t) \quad\to\quad \int_0^{2\pi}g(x)dx=\int_0^{\pi}g(x)dx+\int_0^{\pi}g(t)dt=2\int_0^{\pi}g(x)dx\neq 0$
$$\int_0^{2\pi}\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx \neq 0$$
In the equation above, the symbol $\neq 0$ means "generally not equal to $0$ ". It doesn't mean "always not equal to $0$ ".
| {
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Finding the integral $\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3}dx$ with substitution - how to think?
Find
$$\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3}dx$$
In the above question, I was literally stumped, and wasn't able to solve it for a long time. Turns out that you had to divide the numerator and the denominator by x15, and then we could substitute. Now this got my thinking, how can we understand where to divide what? I mean obviously, in this question - one big hint would be the numerical coefficients of the numerator, but other than that is there any logical way to proceed or is it basically a hit and try?
Link to the solution
| Generally these type of questions we simply put $\displaystyle x = \frac{1}{t}$ and then use normal substution method
Now let $$I = \int \frac{2x^{12}+5x^9}{(x^5+x^3+1)^3}dx$$
Put $\displaystyle x= \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So $$I = -\int\frac{2+5t^3}{\left(t^5+t^2+1\right)^3}\cdot \frac{t^{15}}{t^{12}}\cdot \frac{1}{t^2}dt = -\int\frac{2t+5t^4}{\left(1+t^2+t^{5}\right)^3}dt$$
Now using normal substution method, Put $(1+t^2+t^5)=u\;,$ Then $(2t+5t^4)dt=du$
So we get $$I = -\int\frac{1}{u^3}du = \frac{1}{2u^2}+\mathcal{C} = \frac{1}{2(1+t^2+t^5)^2}+\mathcal{C}$$
So $$I = \int \frac{2x^{12}+5x^9}{(x^5+x^3+1)^3}dx = \frac{x^{10}}{2(x^5+x^3+1)^2}+\mathcal{C}$$
$\bf{Bonus\; Question}::$ Evaluation of $$\int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx$$
| {
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Finding the generating function of a recurrence relation in dependence of a variable Given this inhomogeneous linear recurrence relation of 2nd order :
$F_n = F_{n-2} + a$ for $n \geq 2$
with $F_1 = 1$ and $F_0 = 0$
How do I find the generating function of this recurrence relation in dependence of the variable a? I tried solving it but the solution that I got ($\frac{x + a \cdot x^2}{1-x^2} + \frac{a}{1-x}$) doesn't seem to be right. I hope someone can help.
| It’s easy enough to derive a closed form directly from the recurrences:
$$\begin{align*}
F_{2n}&=an\\
F_{2n+1}&=an+1
\end{align*}$$
Thus,
$$\begin{align*}
\sum_{n\ge 0}F_{2n+1}x^{2n+1}&=\sum_{n\ge 0}(an+1)x^{2n+1}\\
&=x\sum_{n\ge 0}anx^{2n}+\sum_{n\ge 0}x^{2n+1}\\
&=x\sum_{n\ge 0}F_{2n}x^{2n}+\frac{x}{1-x^2}\;,
\end{align*}$$
so
$$\begin{align*}
\sum_{n\ge 0}F_nx^n&=\sum_{n\ge 0}F_{2n}x^{2n}+\sum_{n\ge 0}F_{2n+1}x^{2n+1}\\
&=(1+x)\sum_{n\ge 0}anx^{2n}+\frac{x}{1-x^2}\\
&=a(1+x)\sum_{n\ge 0}nx^{2n}+\frac{x}{1-x^2}\\
&=\frac{a}2(1+x)x\sum_{n\ge 0}2nx^{2n-1}+\frac{x}{1-x^2}\\
&=\frac{a}2(1+x)x\frac{d}{dx}\left(\sum_{n\ge 0}x^{2n}\right)+\frac{x}{1-x^2}\\
&=\frac{a}2(1+x)x\frac{d}{dx}\left(\frac1{1-x^2}\right)+\frac{x}{1-x^2}\\
&=\frac{ax^2(1+x)}{(1-x^2)^2}+\frac{x}{1-x^2}\\
&=\frac{ax^2}{(1-x)(1-x^2)}+\frac{x}{1-x^2}\\
&=\frac{(a-1)x^2+x}{(1-x)(1-x^2)}\;.
\end{align*}$$
| {
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Find the modulus of $|z-5|/|1-3z|$ when z is given If $z = 3-2i$ then find $$\frac { \left| z-5 \right| }{ \left| 1-3z \right| } $$
I've substituted z by $|z|^2/z$ conjugate but still cant figure out what to do,
Thanks in advance
| $$\frac { \left| z-5 \right| }{ \left| 1-3z \right| } =\frac { \left| 3-2i-5 \right| }{ \left| 1-9+6i \right| } =\frac { \left| -2-2i \right| }{ \left| -8+6i \right| } =\frac { \sqrt { { 2 }^{ 2 }+{ 2 }^{ 2 } } }{ \sqrt { { \left( -8 \right) }^{ 2 }+{ 6 }^{ 2 } } } =\frac { 2\sqrt { 2 } }{ 10 } =\frac { \sqrt { 2 } }{ 5 } $$
| {
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Another formula for the angle bisector in a triangle I have seen in an old geometry textbook that the formula for the length of the angle bisector at $A$ in $\triangle\mathit{ABC}$ is
\begin{equation*}
m_{a} = \sqrt{bc \left[1 - \left(\frac{a}{b + c}\right)^{2}\right]} ,
\end{equation*}
and I have seen in a much older geometry textbook that the formula for the length of the same angle bisector is
\begin{equation*}
m_{a} = \frac{2}{b + c} \sqrt{bcs(s - a)} .
\end{equation*}
($s$ denotes the semiperimeter of the triangle.)
I did not see such formulas in Euclid's Elements. Was either formula discovered by the ancient Greeks? May someone furnish a demonstration of either of them without using Stewart's Theorem and without using the Inscribed Angle Theorem?
| We may prove it by avoiding the usual path, and proving other interesting things along the way.
Let $I$ be the incenter and $I_A$ the $A$-excenter. $\widehat{IBI_A}=\widehat{ICI_A}=\frac{\pi}{2}$, hence $IBI_A C$ is a cyclic quadrilateral. By Van Obel's theorem we have
$$ \frac{AI}{IL_A}=\frac{b+c}{a}, $$
hence it is enough to find $IA^2$. Let $C'$ be the symmetric of $C$ with respect to $AI$. We have:
$$ AI\cdot AI_A = \text{pow}_A\left(\Gamma_{I_A B C}\right)=AB\cdot AC'=bc.$$
The problem boils down to finding $II_A=AI_A-AI$, i.e. the diameter of the circumcircle of $I_A BC$. The midpoint of $II_A$ is also the midpoint of the $BC$-arc in the circumcircle of $BCI_A$, hence the previous diameter just depends on the length of $BC$, i.e. $a$, and the angle $\widehat{BNC}=\pi-\widehat{A}$. Putting everything together, we get that $IA$ is a root of
$$ x^2+x\frac{a}{\cos\frac{A}{2}}-bc, $$
but $IA\cos\frac{A}{2}=\frac{b+c-a}{2}$ and
$$ \frac{a}{\cos^2\frac{A}{2}}=\frac{2a}{1+\cos A}=\frac{4abc}{(b+c-a)(b+c+a)},$$
hence:
$$ IA^2 = bc-\frac{2abc}{a+b+c} = \color{red}{bc-4rR}.$$
Now we may exploit the parallel axis theorem to compute the unpleasant squared distance $IG^2$ in a very slick way:
$$\begin{eqnarray*} \sum_{cyc}IA^2 = ab+ac+bc-12rR &=&3IG^2+GA^2+GB^2+GC^2\\ &=& 3IG^2+\frac{3}{4}(a^2+b^2+c^2)\end{eqnarray*}$$
leads to:
$$ IG^2 = ab+ac+bc-\frac{2abc}{a+b+c}-\frac{a^2+b^2+c^2}{4}$$
and we may prove $IO^2=R^2-2Rr$ in a similar way.
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Solving a system of linear congruences
Find all positive integer solutions to \begin{align*}x &\equiv -1 \pmod{n} \\ x&\equiv 1 \pmod{n-1}. \end{align*}
I rewrote the system as $x = nk_1-1$ and $x = (n-1)k_2+1$. Thus, we have $nk_1-1 = (n-1)k_2+1$ and so $n(k_1-k_2) = 2-k_2 \implies n = \frac{2-k_2}{k_1-k_2}$. How do I solve it from here?
| The two congruences are equivalent to $$\begin{cases}x=nu-1\\x=(n-1)v+1\end{cases}$$
It follows the diophantine equation $$nu-(n-1)v=2$$ which admits the particular solution $(u,v)=(2,2)$ hence the general solution is $$\begin{cases}u=(n-1)t+2\\v=nt+2\end{cases}$$ Finally one has
$$x=n((n-1)t+2)-1$$ where $t$ is an arbitrary integer.
(The first congruence is clearly verified; the second gives $x\equiv 2n-1\equiv 1\pmod{n-1}$ because of $2n-1=2(n-1)+1$)
| {
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Find the probability of getting two sixes in $5$ throws of a die.
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial?
My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for the second and third throw. Again the probability of getting a $6$ is fourth roll is $\frac{1}{6}$. So the probability of ending the game in the fifth roll is $\frac{5^3}{6^3}\times\frac{1}{6^2}=\frac{125}{6^5}$.
But the answer is not correct. Where is my mistake? Help please.
| So both the fourth and the fifth rolls need to be sixes:
$$P(4^{th},5^{th}\mbox{ rolls are sixes}) = \frac{1}{6^2}$$
There are following possible combinations of rolls for the game to not end until the fifth roll:
$$(XXXOO), (XOXOO),(OXXOO)$$
where $X,O$ represent non-six and six, respectively.
Thus, the probability is:
$$P(\mbox{end in fifth trial}) = \frac{1}{6^2}\left( \frac{5^3}{6^3} +2 \cdot \frac{5^2}{6^2} \cdot \frac{1}{6} \right) = \frac{175}{6^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
If $\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)=l$, then what is the value of $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$? If the function
$$\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)$$ exists and has a value equal to $l$ then what will be the value of $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$
| Let $$f(x)=\frac{1}{\sqrt{1+x}}-\frac{1+ax}{1+bx}$$ and let $$g(x)=x^3.$$ Apply L'Hospital's rule.
First derivatives
We have $$f'(x)=-\frac{1}{2(1+x)^{3/2}}-\frac{a}{1+bx}-\frac{b(1+ax)}{(1+bx)^2}$$ so that $f'(0)=b-a-1/2$. Also $g'(x)=3x^2$ so that $g'(0)=0$. Thus since there is a finite limit $l$ it must be that $f'(0)=0$ i.e. $b=a+1/2$. Substituting this into $f'(x)$ gives $$f'(x)=\frac{1}{2(1+x)^{3/2}}-\frac{2}{[2+(1+2a)x]^2}.$$
Second derivatives
Differentiating again, we have $$f''(x)=\frac{3}{4(x+1)^{5/2}}-\frac{4(2a+1)}{[2+(1+2a)x]^3},$$ so that $f''(0)=1/4-a$. Since $g''(x)=6x$ so that $g''(0)=0$, we need $f''(0)=0$ which gives $a=1/4$. Substituting this value gives $$f''(x)=\frac{3}{4(x+1)^{5/2}}-\frac{48}{(3x+4)^3}.$$
Third derivatives
Differentiating again, we have $$f'''(x)=-\frac{15}{8(x+1)^{7/2}}+\frac{432}{(3x+4)^4}.$$ Thus $f'''(0)=-3/16$. Finally, since $g'''(x)=6$, the limit is
$$l=\frac{f'''(0)}{g'''(0)}=\frac{(-3/16)}{6}=-\frac{1}{32}.$$
Conclusion
Thus $$\frac{1}{a}-\frac{2}{l}+\frac{3}{b}=4+64+4=72.$$
(Probably there is a faster way...)
| {
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"url": "https://math.stackexchange.com/questions/1860434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The $25$th digit of $100!$ I want to find The $25$th digit of $100!$.
My attempt:It is easy to know it has $24$ zeroes.Because:
$\lfloor {\frac{100}{5}} \rfloor+\lfloor {\frac{100}{25}} \rfloor =24$
By getting the fist digits(after deleting all $5$ factors and $24$,$2$ factors)and multiplying them to each other we get the answer $4$ but I want an easier way.
| If we exclude the numbers that are divisible by 5, we see a cycle that repeats.
$4! = 24\\
9!/6!\equiv 24 \mod 100$
$100! = \frac {100!}{5^{20} 20!} (5^{20} 20!) = \frac {100!}{5^{20} 20!}\frac {20!}{5^4 4!} (4!)(5^{24})$
$(24^{25})(5^{24}) = (12^{25})(10^{24})(2)$
$12^{25} \equiv 12^5 \mod 100\equiv 32 \mod 100$
The last $2$ non-zero digits of 100! are $64.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
solutions of the equation $x^3-y^3=z!-18$ What are the solutions of the equation $x^3-y^3=z!-18$? Here $x,y,z$ are non-negative integers. I have tried brute force but is there a better method?
| You can even solve this if $x,y,z\in\mathbb Z$ can also be negative. Note that $z!$ is never defined when $z<0$.
By Fermat's Little Theorem for all $a\in\mathbb Z$ we have $a^6\equiv 0,1\pmod{7}$ and so $a^3\equiv -1,0,1\pmod{7}$.
If $z\ge 7$, then $$x^3-y^3\equiv 0-18\equiv 3\pmod{7},$$ impossible, because $$x^3-y^3\equiv -2,-1,0,1,2\pmod{7}$$
Therefore $0\le z\le 6$. Then you can analyse each $7$ (or $6$, since $0!=1!$) case by using factorization:
$$x^3-y^3=(x-y)\left(x^2+xy+y^2\right)=t$$
for an integer $t$. One thing that will help is that $$x^2+xy+y^2=\frac{1}{4}(2x+y)^2+\frac{3}{4}y^2\ge 0$$
You'll solve systems of equations. E.g., if $z=4$, then $$(x-y)\left(x^2+xy+y^2\right)=6$$
and the cases are $x-y=6$, $x^2+xy+y^2=1$
or $x-y=3$, $x^2+xy+y^2=2$
or $x-y=2$, $x^2+xy+y^2=3$
or $x-y=1$, $x^2+xy+y^2=6$.
Which is $4$ systems of equations. E.g., in the first one, $y=x-6$ and so $x^2+x(x-6)+(x-6)^2=1$ is a simple quadratic equation in terms of $x$.
Another method here is noticing increase of differences between consecutive cubes. $2^3-1^3>6$ and $1^3-(-1)^3<6$ and $(-1)^3-(-2)^3>6$.
From the other answer you can tell that $z\not\in\{3,4,5\}$ by noticing that if $z\ge 3$, then $3\mid x^3-y^3$, but by Fermat's Little Theorem then $3\mid x-y$, and also $x^2+xy+y^2\equiv 3x^2\equiv 0\pmod{3}$, so $9\mid z!$, so $z\ge 6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Show that $a_n = [n \sqrt{2}]$ contains an infinite number of integer powers of $2$
Show that the sequence $\{a_n\}_{n \geq 1}$ defined by $a_n = [n \sqrt{2}]$ contains an infinite number of integer powers of $2$. ($[x]$ is the integer part of $x$.)
I tried listing out the first few values, but I didn't see a pattern: $1,2,4,5,7,8,9,11,\ldots.$ Should we do a proof by contradiction?
| Let $1/\sqrt{2} = \sum_{j=1}^\infty d_j 2^{-j}$ be the base-2 expansion of $1/\sqrt{2}$, where each $d_j$ is $0$ or $1$. Since $1/\sqrt{2}$ is irrational, there are infinitely many $0$'s and infinitely many $1$'s.
Let $x_N = \sum_{j=1}^N d_j 2^{-j}$ and $n_N = 1 + 2^N x_N = 1 + \sum_{j=1}^N d_j 2^{N-j}$ which is a positive integer. If $d_{N+1} = 1$ we have
$$1/\sqrt{2} - 2^{-N-1} < x_N + 2^{-N-1} < 1/\sqrt{2}$$ and then
$$2^N < \sqrt{2}\; n_N = \sqrt{2} (x_N + 1/2^N)2^N < 2^N+\sqrt{2}/2 < 2^N + 1$$
so that $2^N = \lfloor \sqrt{2} n_N \rfloor$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Analytical expressions for extreme values of $f(x):=\log(2)\left(\sum_\limits{k=-\infty}^\infty 2^{k+x}e^{-2^{k+x}}\right)-1$ The function $f(x):=\log(2)\left(\sum_\limits{k=-\infty}^\infty 2^{k+x}e^{-2^{k+x}}\right)-1$ is a periodic function. Numerical optimization shows that the minimum and maximum of $f$ are approximately $-9.885\cdot10^{-6}$ and $9.885\cdot10^{-6}$, respectively. Are there any analytical expressions for these values?
| The Fourier series of the function $f(x)$, which has period equal to 1, is
\begin{equation}
f(x) = \Re\!\left(\sum_{l=1}^{\infty}
a_l \exp(2\pi lx \rm{i})\right)
\end{equation}
with coefficients
\begin{multline}
a_l = 2\int_0^1 f(x) e^{-2\pi lx \rm{i}} dx
=
2\log(2)\int_0^1 \sum_{k=-\infty}^\infty 2^{k+x}e^{-2^{k+x}}e^{-2\pi lx \rm{i}}dx\\
=
2\log(2)\sum_{k=-\infty}^\infty \int_{k}^{k+1} 2^{x}e^{-2^{x}}e^{-2\pi lx \rm{i}}dx
=
2\log(2) \int_{-\infty}^{\infty} 2^x e^{-2^{x}}
e^{-2\pi lx \rm{i}} dx.
\end{multline}
The variable transformation $y = 2^x$ yields
\begin{equation}
a_l
=
2\int_{0}^{\infty} e^{-y}
y^{-\frac{2\pi l \rm{i}}{\log(2)}}dy
=
2\Gamma\!\left(1-\frac{2\pi l \rm{i}}{\log(2)}\right)
\end{equation}
where $\Gamma$ denotes the gamma function. Using the identity $\left|\Gamma(1+ \rm{i}x)\right| =
\sqrt{\frac{\pi x}{\sinh(\pi x)}}
$
we are able to write for the absolute values of the coefficients
\begin{equation}
|a_l| =
2\sqrt{\frac{bl}{\sinh(bl)}}
\quad
\text{with}
\
b:=\frac{2\pi ^2}{\log 2}.
\end{equation}
In particular, the amplitude of the first harmonic is
$|a_1|
\approx
9.884\cdot 10^{-6}$.
Next we show that the deviation of $f(x)$ from the first harmonic is small. It is obvious that the deviation must be smaller than $\sum_{l=2}^{\infty}
|a_l|$. The ratio of subsequent coefficients is given by
\begin{equation}
\frac{|a_{l+1}|}{|a_{l}|}
=
\sqrt{\frac{l+1}{l}}
\sqrt{\frac{
\sinh(b l)
}{
\sinh(b (l+1))
}}
=
\sqrt{\frac{l+1}{l}}
\sqrt{\frac{
1
}{
\cosh(b)
+
\frac{
\sinh(b)
}
{
\tanh(b l)
}
}}.
\end{equation}
For $l\geq 2$ we have $\sqrt{\frac{l+1}{l}}\leq \sqrt{\frac{3}{2}}$. Together with $\tanh(x)\leq 1$ we obtain
\begin{equation}
\frac{|a_{l+1}|}{
|a_{l}|
}
\leq
\sqrt{\frac{3}{2}}
\sqrt{\frac{
1
}{
\cosh(b)
+
\sinh(b)
}}
\\
=
\sqrt{\frac{
3
}{
2 e^b
}}.
\end{equation}
Consequently, $|a_{l}| \leq
|a_{2}|
\left(\sqrt{\frac{
3
}{
2 e^b
}}\right)^{l-2}$ for $l\geq 2$ and
\begin{equation}
\sum_{l=2}^{\infty}
|a_l|
\leq
|a_2|
\sum_{l=0}^{\infty}
\left(
\sqrt{\frac{
3
}{
2e^b
}}
\right)^l
=
2\sqrt{
\frac{2b}{\sinh(2b)}}
\frac{1}{1-
\sqrt{\frac{
3
}{
2 e^b
}}
}
\approx
9.154 \cdot 10^{-12}.
\end{equation}
Therefore the maximum and minimum of $f(x)$ are very close to $\pm|a_1|$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Circles in complex plane.
Find the real value of a for which there is at least one complex number satisfying $|z+4i|=\sqrt{a^2-12a+28}$ and $|z-4\sqrt{3}|\lt a$.
My solutions:-
*
*Graphical solution:-
$|z+4i|=\sqrt{a^2-12a+28}$ represents a circle with center at $A\equiv(0,-4)$ and radius $r_1=\sqrt{a^2-12a+28}$ and similarly, $|z-4\sqrt{3}|\lt a$ represents all the points inside the circle with center at $B\equiv(4\sqrt{3},0)$ and radius $r_2=a$. So, the distance between the centers of the circle is $\sqrt{(4\sqrt{3})^2+(4^2)}=8$
Now consider the graphs of $\sqrt{a^2-12a+28}$ and $a$. The graph of $\sqrt{a^2-12a+28}$ which is a hyperbola with its centre at $(6,0)$ and vertexes being $(6\pm 2\sqrt2,0)$
It is clear from the plot that $a\in (0,6-2\sqrt2]\cup [6+2\sqrt2,\infty)$ for $\sqrt{a^2-12a+28}\ge0$ and $a\gt 0$
Now, lets do some case study.
Case 1:- When $a \ge 6+2\sqrt2$
In this case the radius of the circle represented by $|z-4\sqrt3|=a$ is greater than the distance between the centers of the circles, i.e $r_2\gt AB$ as $r_2\gt 8$. So, the circle $|z-4\sqrt3|=a$ either encloses the circle $|z+4i|=\sqrt{a^2-12a+28}$(or the point $-4i$, which is the case when $a=6+2\sqrt2$) fully or encloses a portion of it and in both the cases we find that we obtain a number common to both the regions. So, $a\in(6+2\sqrt2,\infty)$
Case 2:- $0\le a\le 6-2\sqrt2$
In this case we see that due to the bounding of $a$, we see that $(r_1+r_2)\lt 8$, so there is no intersection of the wanted regions, so there is no solution in this region.
So the required interval of $a$ is $\boxed{a \in [6+2\sqrt2, \infty)}$
*
*Algebraic Solution(or whatever you wanna call it):-
Lets consider all the possible circles that can be drawn for the given circles, they are as given in the figure:-
From the above drawn circles we can consider the following cases.
Case 1:- From Figure-3 (I have not numbered it so consider it in the left to right manner) we can get the following condition $$AB\lt r_1+r_2 \implies 8 \lt \sqrt{a^2-12a+28} + a \implies a\gt 9$$ This case deals with the limiting condition of both the circles touching.
Case 2:- Now to not let the Figure-5 take place we have to consider the case $$AB\gt r_1-r_2 \implies 8 \gt \sqrt{a^2-12a+28} -a \implies a\gt -\frac{9}{7}$$
Now, I don't know what to conclude from this, so please do tell me that
Case 3:- Consider Figure-4(don't know why I drew the last figure as it IMO represents the same case as Figure 4), the case we get is
$$AB\lt r_2-r_1 \implies 8\lt a-\sqrt{a^2-12a+28} \implies a\gt 9$$
My deal with the question:-
$a\gt 9$, $a\gt 9$ everywhere not a single $a\ge 6+2\sqrt2$ to see.(It's a joke)
So as you can see that from the Algebraic solution I got $a\gt 9$ and from the Graphical approach to the question I get $a\ge 6+2\sqrt2$. And the book I am solving also gives the answer interval as $a\in (9,\infty)$. So what is wrong with the solutions and do point out the errors. As, always more elegant solutions are welcome.
In my opinion the answer should be $a\in [6+2\sqrt2, \infty)$, which is evident from the graph figure below.
| The answer should be $a\ge 6+2\sqrt 2$.
($a\gt 9$ is not correct since, for $a=9$, $z=-3i$ satisfies the conditions.)
In your Algebraic Solution, it seems that you have some errors.
Case 1:-
$$8 \lt \sqrt{a^2-12a+28} + a \implies a\gt 9$$
Case 2:-
$$8 \gt \sqrt{a^2-12a+28} -a \implies a\gt -\frac{9}{7}$$
Case 3:-
$$8\lt a-\sqrt{a^2-12a+28} \implies a\gt 9$$
These should be incorrect. Under $0\lt a\le 6-2\sqrt 2$ or $a\ge 6+2\sqrt 2$, I got the followings :
$$8 \lt \sqrt{a^2-12a+28} + a \iff a\ge 6+2\sqrt 2$$
(To solve $8-a\lt \sqrt{a^2-12a+28}$, separate it into two cases : If $8-a\lt 0$, then the inequality holds(the LHS is negative, the RHS is non-negative). If $8-a\ge 0$, then since the both sides are non-negative, squaring gives $a\gt 9$. There is no $a$ such that $a\le 8$ and $a\gt 9$. Therefore, $a\gt 8$. Considering $0\lt a\le 6-2\sqrt 2$ or $a\ge 6+2\sqrt 2$, the solution is $a\ge 6+2\sqrt 2$.)
$$8 \gt \sqrt{a^2-12a+28} -a \iff 0\lt a\le 6-2\sqrt 2\quad\text{or}\quad a\ge 6+2\sqrt 2$$
$$8\lt a-\sqrt{a^2-12a+28}\iff 6+2\sqrt 2\le a\lt 9$$
In the following, I will write a solution in the similar way as your Algebraic Solution.
We consider the two circles
$$C_1 : x^2+(y+4)^2=a^2-12a+28\qquad\text{and}\qquad C_2 : (x-4\sqrt 3)^2+y^2=a^2$$
where $0\lt a\le 6-2\sqrt 2$ or $a\ge 6+2\sqrt 2$.
The distance between the centers is $8$.
Case 1 : The two cirlces have no intersection points, and no circle is inside the other : $8\gt \sqrt{a^2-12a+28}+a\iff 0\lt a\le 6-2\sqrt 2$. This case has to be eliminated.
Case 2 : The two circles have the only one intersection point, and no circle is inside the other : $8=\sqrt{a^2-12a+28}+a$. There is no such $a$.
Case 3 : The two circles have two intersection points : $|\sqrt{a^2-12a+28}-a|\lt 8\lt \sqrt{a^2-12a+28}+a\iff a\gt 9$. This case is sufficient.
Case 4 : The two circles have the only one intersection point, and one circle is inside the other : $8=|\sqrt{a^2-12a+28}-a|$.
Case 4-1 : The case when $(4\sqrt 3,0)$ is inside $C_1$. There is no such $a$.
Case 4-2 : The case when $(0,-4)$ is inside $C_2$ : $a=9$. This case is sufficient.
Case 5 : The two circles have no intersection points, and one circle is inside the other : $8\lt |\sqrt{a^2-12a+28}-a|$
Case 5-1 : The case when $(4\sqrt 3,0)$ is inside $C_1$ : There is no such $a$.
Case 5-2 : The case when $(0,-4)$ is inside $C_2$ : $8\lt a\lt 9$. This case is sufficient.
Therefore, the answer is $a\ge 6+2\sqrt 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tangent and Circle in Complex Plane
Question:-
Three points represented by the complex numbers $a,b$ and $c$ lie on a circle with center $O$ and radius $r$. The tangent at $c$ cuts the chord joining the points $a$ and $b$ at $z$. Show that $$z=\dfrac{a^{-1}+b^{-1}-2c^{-1}}{a^{-1}b^{-1}-c^{-2}}$$
Attempt at a solution:- To simplify our problem let $O$ be the origin, then the equation of circle becomes $|z|=r$.
Now, the equation of chord passing through $a$ and $b$ can be given by the following determinant
$$\begin{vmatrix}
z & \overline{z} & 1 \\
a & \overline{a} & 1 \\
b & \overline{b} & 1 \\
\end{vmatrix}= 0$$
which simplifies to $$z(\overline{a}-\overline{b})-\overline{z}(a-b)+(\overline{a}b-a\overline{b})=0 \tag{1}$$
Now, for the equation of the tangent through $c$, I used the cartesian equation of tangent to a circle $xx_1+yy_1=r^2$ from which I got $$z\overline{c}+\overline{z}c=2r^2\tag{2}$$
Now, from equation $(1)$, we get
$$\overline{z}=\dfrac{z\left(\overline{a}-\overline{b}\right)+\left(a\overline{b}-\overline{a}b\right)}{(a-b)}$$
Putting this in equation $(2)$, we get $$z=\dfrac{2r^2(a-b)+\left(a\overline{b}-\overline{a}b\right)c}{\left(a\overline{c}+\overline{a}c\right)-\left(b\overline{c}+\overline{b}c\right)}$$
After this I am not able to get to anything of much value, so your help would be appreciated. And as always, more solutions are welcomed.
| Consider $O$ to be the origin, then $|a|=|b|=|c|=r$
We know that the equation of a line passing through points $z_1$ and $z_2$ is represented by
$$\begin{vmatrix}
z & \overline{z} & 1 \\
z_1 & \overline{z_1} & 1 \\
z_2 & \overline{z_2} & 1 \\
\end{vmatrix}= 0$$
Now as per the question $a, b$ and $c$, lie on a circle, now lets consider a dummy point $d$ also located on the circumfrence of the circle. Then the equation of chord passing through $a$ and $b$ is represented as
$$\begin{vmatrix}
z & \overline{z} & 1 \\
a & \overline{a} & 1 \\
b & \overline{b} & 1 \\
\end{vmatrix}= 0$$
which on simplification gives $$z(\overline{a}-\overline{b})-\overline{z}(a-b)+(\overline{a}b-a\overline{b})=0 \tag{1}$$
Similarly, equation of the chord passing through $c$ and $d$ can be given by
$$z(\overline{c}-\overline{d})-\overline{z}(c-d)+(\overline{c}d-c\overline{d})=0 \tag{2}$$
On eliminating $\overline{z}$ from equations $(1)$ and $(2)$, we get
$$z=\dfrac{(a-b)(\overline{c}d-c\overline{d})+(\overline{a}b-a\overline{b})(c-d)}{(\overline{a}-\overline{b})(c-d)+(\overline{c}-\overline{d})(a-b)}$$
Now, as $|a|^2=|b|^2=|c|^2=r^2$, so for every $\overline{z}, z\in\{a,b,c\}$, we can write $\overline{z}=\dfrac{r^2}{z}$
On simplification we get $$z=\dfrac{a^{-1}+b^{-1}-c^{-1}-d^{-1}}{a^{-1}b^{-1}-c^{-1}d^{-1}}$$
Now as the question stated that there is tangent at $c$ not a chord so we substitute $d=c$ to get the answer
$$z=\dfrac{a^{-1}+b^{-1}-2c^{-1}}{a^{-1}b^{-1}-c^{-2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\sum_{r=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3r+1}$
Show that $$\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\ldots=\dfrac{1}{3}\left[ 2^{n-2} + 2\cos{\dfrac{(n-2)\pi}{3}}\right]$$
My solution:-
$$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\ldots=\sum_{r=0}^{n}{\binom{n}{r}x^r} \\ \therefore x^2(1+x)^n=\binom{n}{0}x^2+\binom{n}{1}x^3+\binom{n}{2}x^4+\binom{n}{3}x^5+\ldots=\sum_{r=0}^{n}{\binom{n}{r}x^{r+2}}$$
In the above Binomial Expansion on substituting $x=1,\omega,\omega^2$, $\omega$ being a complex cube root of unity, we get the following three equations
$$\tag{1}(1)^2(1+1)^n=\sum_{r=0}^{n}{\binom{n}{r}}=2^n$$
$$(\omega)^2(1+\omega)^n=\sum_{r=0}^{n}{\binom{n}{r}\omega^{r+2}}=(-1)^n(\omega)^{2n+2} \tag{2}$$
$$(\omega)^4(1+\omega^2)^n=(\omega)(1+\omega^2)^n=\sum_{r=0}^{n}{\binom{n}{r}\omega^{2r+4}}=(-1)^n(\omega)^{n+1} \tag{3}$$
On adding $(1),(2) \text{ and }(3)$, we get
$$\dfrac{1}{3}\left(2^n+(-1)^n(\omega^{n+1}+\omega^{2n+2})\right)=\sum_{r=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3r+1}$$
Now, as $\omega=-e^{i(\pi/3)}$ ($\omega$ being the cube root of unity)
$$\begin{aligned}
\therefore (\omega^{n+1} +\omega^{2n+2})
&= \left(\left(-e^{i(\pi/3)}\right)^{n+1}+\left(-e^{-i(\pi/3)}\right)^{n+1}\right) \\
&=(-1)^{n+1}\left(e^{i(\pi(n+1)/3)}+e^{-i(\pi(n+1)/3)}\right) \\
&=(-1)^n\left(2\cos{\left(\dfrac{\pi(n+1)}{3}\right)}\right)
\end{aligned}$$
Now, substituting the value of $(\omega^{n+1}+\omega^{2n+2})$ back into $(4)$, we get
$$2^n+(-1)^n(\omega^{n+1}+\omega^{2n+2})=2^n-2\cos{\left(\dfrac{\pi(n+1)}{3}\right)}=\sum_{r=0}^{n}{\binom{n}{3r+1}}$$
$$\therefore \sum_{r=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3r+1}=\boxed{\dfrac{1}{3}\left(2^n-2\cos{\left(\dfrac{\pi(n+1)}{3}\right)}\right)}$$
So, where did I go wrong, or is it that the book has provided the wrong answer.
| $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\sum_{k = 0}^{\left\lfloor\pars{n - 1}/3\right\rfloor}
\,\,{n \choose 3k + 1}} & =
\sum_{k = 0}^{\infty}{n \choose n - 3k - 1} =
\sum_{k = 0}^{\infty}\oint_{\verts{z}\ =\ 1^{\color{#f00}{-}}}
{\pars{1 + z}^{n} \over z^{n - 3k}}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{\color{#f00}{-}}}
{\pars{1 + z}^{n} \over z^{n}}\sum_{k = 0}^{\infty}\pars{z^{3}}^{k}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z}\ =\ 1^{\color{#f00}{-}}}\,\,\,
{\pars{1 + z}^{n} \over z^{n}\pars{1 - z^{3}}}\,{\dd z \over 2\pi\ic}
\\[5mm] &\ \stackrel{z\ \mapsto\ 1/z}{=}\
\oint_{\verts{z}\ =\ 1^{\color{#f00}{+}}}\,\,\,
{z\pars{1 + z}^{n} \over z^{3} - 1}\,{\dd z \over 2\pi\ic} =
\sum_{p}{p\,\pars{1 + p}^{n} \over 3p^{2}}
\end{align}
$\ds{p}$ are the roots of $\ds{z^{3} - 1 = 0}$. Namely,
$\ds{p \in \braces{\expo{-2\pi\ic/3},1,\expo{2\pi\ic/3}}}$.
Then $\ds{~\pars{\mbox{note that}\ p^{2} = {p^{3} \over p} = {1 \over p}}~}$,
\begin{align}
\color{#f00}{\sum_{k = 0}^{\left\lfloor\pars{n - 1}/3\right\rfloor}
\,\,{n \choose 3k + 1}} & =
{1 \over 3}\sum_{p}p^{2}\pars{1 + p}^{n}
\\[5mm] & =
{1 \over 3}\,2^{n} + {2 \over 3}\,
\Re\bracks{\expo{4\pi\ic/3}\pars{1 + \expo{2\pi\ic/3}}^{n}}
\\[5mm] & =
{1 \over 3}\,2^{n} + {2 \over 3}\,
\Re\braces{\expo{\pars{n + 4}\pi\ic/3}\,\,\bracks{2\cos\pars{\pi \over 3}}^{n}}
\\[5mm] & =
\color{#f00}{{1 \over 3}\braces{2^{n} - 2\cos\pars{\bracks{n + 1}\pi \over 3}}}
\end{align}
Note that $\ds{2\cos\pars{\pi \over 3} = 1}$ and
$\ds{\expo{\pars{n + 4}\pi\ic/3} = -\expo{\pars{n + 1}\pi\ic/3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Sum to infinity of trignometry inverse: $\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)$ If we have to find the value of the following (1)
$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)
$$
I know that
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2 \right)
$$
I tried it lot and got a result but then stuck! (2)
(1) http://i.stack.imgur.com/26hA4.jpg
(2) http://i.stack.imgur.com/g2vBb.jpg
| We already have plenty of slick answers through creative telescoping, so I will go for the overkill.
By crude estimations we have $\sum_{r\geq 3}\frac{4}{r^2+3}\leq\frac{\pi}{2}$, hence:
$$ \sum_{r\geq 3}\arctan\left(\frac{4}{r^2+3}\right) = \text{Arg}\prod_{r\geq 3}\frac{r^2+3+4i}{r^2}\tag{1}=\text{Arg}\prod_{r\geq 3}\left(1+\frac{(2+i)^2}{r^2}\right)$$
but due to the Weierstrass product for the $\sinh$ function:
$$\prod_{r\geq 1}\left(1+\frac{(2+i)^2}{r^2}\right)=\frac{\sinh(2\pi+\pi i)}{\pi(2+i)}=-\frac{\sinh(2\pi)}{\pi(2+i)}\tag{2}$$
and that leads to:
$$ \sum_{r\geq 1}\arctan\left(\frac{4}{r^2+3}\right) = \color{red}{\pi-\arctan\frac{1}{2}}.\tag{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Integer solutions to $x^3+y^3+z^3 = x+y+z = 8$
Find all integers $x,y,z$ that satisfy $$x^3+y^3+z^3 = x+y+z = 8$$
Let $a = y+z, b = x+z, c = x+y$. Then $8 = x^3+y^3+z^3 = (x+y+z)^3-3abc$ and therefore $abc = 168$ and $a+b+c = 16$. Then do I just use the prime factorization of $168$?
| From $\sqrt[3]{abc}\approx 5.5 > 5\tfrac13=\frac{a+b+c}{3}$, we know that $a,b,c$ cannot all be positive (for in that case the AM-GM inequality states a "$\le $").
As $abc>0$, exactly two of $a,b,c$ must be negative.
Also, either all three numbers are even or exactly two are odd. In the first case, $\frac a2\frac b2\frac c2=21$, so all three integers $\frac a2,\frac b2,\frac c2$ are odd; but then $8=\frac a2+\frac b2+\frac c2$ would be odd as well. We conclude that exactly two of $a,b,c$ are odd, say $a,b$ are odd, $c$ is even. Wlog., $|a|\le |b|$. Then $c$ is an odd multiple of $8$, hence $a+b\equiv 8\pmod{16}$. The only ways to combine numbers from $\pm1,\pm3,\pm7,\pm21$ meeting this are $(a,b)=(\pm1,\pm7)$, or $(a,b)=(\pm3,\pm21)$; but the latter is not compatible with $ab\mid 168$. As two of $a,b,c$ must be negative, we conclude $$a=-1,\quad b=-7,\quad c=16-a-b=24$$
and from this
$$ x=\tfrac{b+c-a}2=9,\quad y=15,\quad z=-16.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving an equation involving an integral: $\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$ Determine a pair of number $a$ and $b$ for which
$$\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$$
I tried putting $x$ as $1-x$ as the integral wouldn't change but could not move forward from there so can you please suggest me what should I do next.
| Let $$I = \int_{0}^{1}\frac{ax+b}{(x^2+3x+2)^2}dx = \frac{a}{2}\int_{0}^{1}\underbrace{\frac{2x+3}{(x^2+3x+2)^2}dx}_{J}+\left(b-\frac{3a}{2}\right)\underbrace{\int_{0}^{1}\frac{1}{(x^2+3x+2)^2}dx}_{K}$$
So $$J = \int_{0}^{1}\frac{2x+3}{(x^2+3x+2)^2}dx = -\left[\frac{1}{x^2+3x+2}\right]_{0}^{1} = \left(\frac{1}{2}-\frac{1}{6}\right) = \frac{1}{3}$$
Similarly $$K = \int_{0}^{1}\frac{1}{(x^2+3x+2)^2}dx=\int_{0}^{1}\left[\frac{1}{x+1}-\frac{1}{x+2}\right]^2dx$$
So $$K = \int_{0}^{1}\frac{1}{(x+1)^2}dx+\int_{0}^{1}\frac{1}{(x+2)^2}dx-2\int_{0}^{1}\left[\frac{1}{x+1}-\frac{1}{x+2}\right]dx$$
So $$K = -\left[\frac{1}{x+1}\right]_{0}^{1}-\left[\frac{1}{x+2}\right]_{0}^{1}-2\left[\ln(x+1)-\ln(x+2)\right]_{0}^{1}$$
So $$K = \frac{1}{2}+\frac{1}{6}+2\left[\ln(4)-\ln(3)\right]=\frac{2}{3}+2\left[\ln\left|\frac{4}{3}\right|\right]$$
So $$I = \frac{a}{6}+\left(b-\frac{3a}{2}\right)\cdot \left(\frac{2}{3}+2\ln \left\|\frac{4}{3}\right\|\right) = \frac{5}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Algebraic Expressions with Fractions can someone review this and see if i've done it correctly please.
$$\frac{\frac {3x} {y}}{\frac {2x}{7}}
$$
$$= \frac{3x}{y} . \frac{7}{2x}$$
$$= \frac{21x}{2xy}
$$
$$= \frac {21}{2y}
$$
Thank you for your time.
| Your answer and method look correct (but as pointed out in the comments, you need to add the restriction $x\ne 0$). I would have liked to have seen more steps though to demonstrate understanding.
$$\frac{\frac{3x}{y}}{\frac{2x}{7}}$$
$$= \color{red}{\frac{\frac{3x}{y}}{\frac{2x}{7}}\cdot\frac{\frac{7}{2x}}{\frac{7}{2x}}}$$
$$= \color{red}{\frac{\frac{3x}{y}\cdot\frac{7}{2x}}{\frac{2x}{7}\cdot\frac{7}{2x}}}$$
$$= \color{red}{\frac{\frac{3x}{y}\cdot\frac{7}{2x}}{\frac{14x}{14x}}}$$
$$= \color{red}{\frac{\frac{3x}{y}\cdot\frac{7}{2x}}{1}}$$
$$= \frac{3x}{y}\cdot\frac{7}{2x}$$
$$= \frac{21x}{2xy}$$
$$= \frac{21}{2y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integration of the following trignometry We have to find the integration of the following ,
I tried but got stuck , can anyone help me
| Use Weierstrass substitution
By setting $t= \tan \left( \frac x2 \right) $
The integral is equivalent to:
$$2\int \frac{(1+t^2)^2}{(3-t^2)^3} dt$$
Decompose the fraction into:
$$ 2 \int \left( \frac 1{3-t^2} - \frac 8{(3-t^2)^2} + \frac{16}{(3-t^2)^3} \right)dt$$
Since $\frac 1{3-t^2} = \frac 1{2\sqrt 3} \left( \frac 1{\sqrt 3 - t} + \frac 1{\sqrt 3 +t} \right)$
(Or you can use inverse hyperbolic tangent, which is the same function)
Now try substituting $t=\sqrt 3 \sin (v)$ for the other two fractions:
$$\int \frac 1{(3- 3\sin^2 v)^2} \sqrt 3 \operatorname{cos} v dv= \frac{\sqrt{3}}{9}\int \sec^3 v dv$$
The third fraction requires integrating $\sec^5 x$, these two can be solved by integration by parts.
The remaining work is tedious, but not hard. Hope this answer helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to factorize the polynomial $a^6+8a^3+27$?
I would like to factorize $a^6+8a^3+27$.
I got different answers but one of the answers is
$$(a^2-a+3)(a^4+a^3-2a^2+3a+9)$$
Can someone tell me how to get this answer? Thanks.
| One may write
$$
\begin{align}
a^6+8a^3+27&=a^3\left(a^3+\frac{27}{a^3}+8 \right)
\\&=a^3\left(\left(a+\frac3a\right)^3-9\left(a+\frac3a\right)+9-1 \right)
\\&=a^3\left(\left[\left(a+\frac3a\right)^3-1\right]-9\left[\left(a+\frac3a\right)-1\right] \right)
\\&=a^3\left(a+\frac3a-1\right)\left(\left(a+\frac3a\right)^2+\left(a+\frac3a\right)-8 \right)
\\&=a \cdot \left(a+\frac3a-1\right)\cdot a^2 \cdot\left(\left(a+\frac3a\right)^2+\left(a+\frac3a\right)-8 \right)
\\&=\left(a^2-a+3\right)(a^4+a^3-2a^2+3a+9).
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Finding the value of an expression using substitution How to find the value of $x^{4000} + \frac{1}{x^{4000}}$ if $x+\frac{1}{x}=1$.
I think binomial theorem will be useful in this.
Here's my proceedings:
$$x+\frac{1}{x}=1$$
Raising both sides to the power of 4000
$$(x+\frac{1}{x})^{4000} = 1^{4000}$$
$${4000 \choose 0} x^{4000} + {4000 \choose 1} x^{3999} \frac{1}{x} + {4000 \choose 2} x^{3998} \frac{1}{x^2} + ... +{4000 \choose 4000} \frac{1}{x^{4000}} = 1$$
$$x^{4000} + \frac{1}{x^{4000}} = 1 - ({4000 \choose 1} x^{3999} \frac{1}{x} + {4000 \choose 2} x^{3998} \frac{1}{x^2} + ... + {4000 \choose 3999} x \frac{1}{x^{3999}})$$
$$x^{4000} + \frac{1}{x^{4000}} = 1- \sum\limits_{r=1}^{3999} {4000 \choose r} x^{4000-r} \frac{1}{x^r}$$
Now how to proceed further to get to my answer?
| HINT:
$$x^2-x+1=0\implies x^3+1=(x+1)(x^2-x+1)=0\implies x^3=-1$$
Now $4000=3(2n+1)+1$
Now $x^{3(2n+1))}=(x^3)^{2n+1}\cdot x=(-1)^{2n+1}\cdot x=-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Spivak's Calculus - Chapter 1 Question 1.5 - Proof by Induction In Spivak's Calculus Fourth Edition, Chapter 1 Question 1.5 is as follows:
Prove $x^n - y^n = (x - y)(x^{n-1}+x^{n-2}y+ \cdots + xy^{n-2}+y^{n-1})$ using only the following properties:
It's easy enough for me to use P9 to expand the right-hand side:
$$
\begin{array} { c l }
(x - y)(x^{n-1}+x^{n-2}y+ \cdots + xy^{n-2}+y^{n-1}) & \text{Given} \\
x^{n-1}(x - y)+x^{n-2}y(x - y)+ \cdots + xy^{n-2}(x - y)+y^{n-1}(x - y) & \text{P9} \\
x^n - yx^{n-1} + x^{n-1}y - x^{n-2}y^2 + \cdots + xy^{n-1} - xy^{n-2}y^2 + xy^{n-1} - y^n & \text{P9}
\end{array}
$$
Finally, all the terms except the first and last cancel out by P3.
I have a feeling this last step should be proved by induction. If so--how would one write that out?
( Also, I've noticed that a lot of people write the application of P9 to polynomials differently than I do when they're working on Spivak's Calculus. Why do they do that? What would it look like here? )
| If you're planning to do a proof by induction, the first step is to write down your inductive hypothesis, typically expressed as a proposition like "$P(n): (x^n - y^n) = (x-y) \sum_{i=0}^{n-1} x^i y ^ {n-1-i}$.
You then establish the truth of $P(0)$ or $P(1)$ or some other useful starting point. Then you say "Suppose that $P(1), P(2), \ldots, P(k)$ are all true. We will show $P(k+1)$, which states that [write out details here]." And that's the point at which YOUR proof started. But there's not much you can do because you don't have $P(1)$ and $P(2)$ and so on written down. In fact, you probably only need $P(n-1)$ and a few axioms to prove $P(n)$.
But "and so on" arguments like yours are exactly hidden inductions, so your proof is really not a solution to the problem given.
Much as I love Spivak's Calculus, this is one of the pedagogically worst problems in the book in my opinion. Part fo the difficulty is that powers haven't been defined, and the statement isn't true when either $x$ or $y$ is zero and $n = 0$, because $0^0$ is undefined. Setting all that aside, and agreeing that the claim applies to all nonzero $x$ and $y$, and that for these, $x^0 = 1$, let's go ahead and write out a proof.
Before I can do that proof, I need a lemma, also proved by induction:
For any $n \ge 1$, and any nonzero $x$ and $y$, and any $s$, $$
s\sum_{i=0}^n x^i y^{n-1-i} = \sum_{i=0}^n s (x^i y^{n-1-i}).
$$
I might even write out a proof of this down below, but for now, let's run with it.
To show: for every positive integer $n$, and for all nonzero $x$ and $y$ the statement
$P(n): (x^n - y^n) = (x-y) \sum_{i=0}^{n-1} x^i y ^ {n-1-i}$
is true.
Proof:
Step one: base case.
For $n = 1$, the statement asserts that
$(x^1 - y^1) = (x-y)(x^0y^0)$. We must show this is true.
Since for any nonzero number $a$, $a^0 = 1$, we can rewrite $P(1)$ as
\begin{align}
(x^1 - y^1) &= (x-y)(1\cdot 1)
\end{align}
To prove this statement, let's start with the right had side and convert it to the left hand side:
\begin{align}
(x-y)(1 \cdot 1) &= (x-y) \cdot 1, & \text{because of P6, applied to $a = 1$} \\
&= (x-y), & \text{because of P6, applied to $a = 1$}.
\end{align}
Thus the right hand side is equal to the left hand side, and we've established $P(1)$.
Step two: inductive step (following @BrebiusB.Brebs's approach)
We assume for some $k \ge 1$ that $P(k)$ is true, i.e., that
$$x^k - y^k = (x-y) \sum_{i=0}^{k-1} x^i y ^ {k-1-i}$$
for all nonzero $x$ and $y$.
Using this hypotheses, we will prove $P(k+1)$ is true, i.e., that
$$x^{k+1} - y^{k+1} = (x-y) \sum_{i=0}^{k} x^i y ^ {k-i}$$
for all nonzero $x$ and $y$.
To do this, let $x$ and $y$ be any two fixed nonzero values. Let's gradually alter the right hand side of the displayed equation above in an attempt to turn it into the left hand side. For each step on the left, I'll provide a reason on the right.
\begin{align}
(x-y) \sum_{i=0}^{k} x^i y ^ {k-i}
&= \sum_{i=0}^{k} (x-y) (x^i y ^ {k-i}), & \text{Lemma applied to $s = (x-y)$}\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i y ^ {k-i}) \right) + (x-y)(x^k y^{k-k}), & \text{The sum from 0 to $k$ is the sum from $0$ to $k-1$ plus the $k$th term; this is a hidden definition of the summation symbol}\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i y ^ {k-i}) \right) + (x-y)(x^k y^{0}), & \text{P3, definition of subtraction}\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i y ^ {k-i}) \right) + (x-y)(x^k \cdot 1), & \text{$y^0 = 1$ because $y$ is nonzero. }\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i y ^ {k-i}) \right) + (x-y)(x^k), & \text{P6 applied to last product }\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i (y \cdot y ^ {(k-i) - 1}) \right) + (x-y)(x^k), & \text{Property of exponents: $y^a \cdot y^b = y^(a+b)$, which also must be proved by induction! }\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i (y \cdot y ^ { ((k+ (-i)) + (- 1)}) \right) + (x-y)(x^k), & \text{Definition of subtraction, twice}\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i (y \cdot y ^ { ((k+ (-1)) + (- i)}) \right) + (x-y)(x^k), & \text{P1, applied within exponent}\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i (y \cdot y ^ { (k -1) - i)}) \right) + (x-y)(x^k), & \text{Definition of subtraction}\\
&= \left(\sum_{i=0}^{k-1} (x-y) (x^i y) ( y ^ { (k -1) - i)}) \right) + (x-y)(x^k), & \text{P5, applied within summation -- which is another hidden induction!}\\
&= \left(\sum_{i=0}^{k-1} (x-y) (y x^i) ( y ^ { (k -1) - i)}) \right) + (x-y)(x^k), & \text{P8, applied within summation -- which is another hidden induction!}\\
&= \left(\sum_{i=0}^{k-1} ((x-y) y) x^i ( y ^ { (k -1) - i)}) \right) + (x-y)(x^k), & \text{P5, again induction}\\
&= ((x-y)y)\left(\sum_{i=0}^{k-1} x^i ( y ^ { (k -1) - i)}) \right) + (x-y)(x^k), & \text{Lemma,applied to $s = (x-y)y$}\\
&= (y(x-y))\left(\sum_{i=0}^{k-1} x^i ( y ^ { (k -1) - i)}) \right) + (x-y)(x^k), & \text{P8}\\
&= y\cdot \left( (x-y))\sum_{i=0}^{k-1} x^i ( y ^ { (k -1) - i)}) \right) + (x-y)(x^k), & \text{P5}\\
&= y\cdot \left(x^k - y^k\right) + (x-y)(x^k), & \text{Inductive hypothesis, applied to middle summation}\\
&= (y\cdot x^k - y \cdot y^k) + (x\cdot x^k -y \cdot x^k), & \text{P9, twice ,with the definition of subtraction used as well. }\\
&= (y\cdot x^k + (- y) \cdot y^k) + (x\cdot x^k + (-y) \cdot x^k), & \text{Defn of subtraction, twice }\\
&= ((-y) \cdot y^k + y\cdot x^k ) + x^{k+1}) + (-y) \cdot x^k)), & \text{P4, definition of exponentiation }\\
&= ((-y) \cdot y^k + (y\cdot x^k + x^{k+1})) + (-y) \cdot x^k)), & \text{P1}\\
&= (((-y) \cdot y^k + x^{k+1}) +y\cdot x^k)) + (-y) \cdot x^k)), & \text{P1}\\
&= ((-y) \cdot y^k + x^{k+1}) +(y\cdot x^k + (-y) \cdot x^k), & \text{P1}\\
&= ((-y) \cdot y^k + x^{k+1}) +((y+ (-y)) \cdot x^k), & \text{P9}\\
&= ((-y) \cdot y^k + x^{k+1}) +(0 \cdot x^k), & \text{P3}\\
&= ((-y) \cdot y^k + x^{k+1}) +0, & \text{Lemma to be proved: $0 \cdot a = 0$ for every $a$}\\
&= (x^{k+1} + (-y) \cdot y^k ) , & \text{P4, P2}\\
&= (x^{k+1} + (-1\cdot y) \cdot y^k ) , & \text{Lemma to be proved: $-1 \cdot a = -a$ for any $a$. }\\
&= (x^{k+1} + (-1)\cdot (y \cdot y^k ) , & \text{P5}\\
&= (x^{k+1} + (-1)\cdot (y^{k+1} ) , & \text{Defn of exponent, P4 (to turn $ 1+k$ into $k+1$)}\\
&= (x^{k+1} + (-y^{k+1} ) , & \text{Most recent to-be-proved lemma again}\\
&= x^{k+1} -y^{k+1} , & \text{Def of subtraction}.
\end{align}
And with this (once we prove all those hidden lemmas!) we've proved that $P(k+1)$ is true.
(I apologize: I've got some spare parens dangling around in that mess, and I don't have the heart to go back and find/fix every one of them...)
Now, presumably, you see why I thin this is one of Spivak's worst exercises. The inductive definition of the summation symbol hasn't been given, and nor has the definition of exponentation for real numbers and nonnegative integer exponents, nor have properties like $a^{n+m} = a^n \cdot a^m$ been proved. It's just a plain old pain in the neck. On the other hand, it shows that you CAN do this stuff, and that once you've taken one algebraic manipulation and written it out this way, you (a) never want to do it again, and (b) are confident that you CAN do it agin if called on to do so.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| Let $$I = \int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx$$
$\displaystyle \bullet\;\; x\sin x+1\cdot \cos x = \sqrt{x^2+1}\left[\sin x\cdot \frac{x}{\sqrt{1+x^2}}+\cos x\cdot \frac{1}{\sqrt{1+x^2}}\right]$
$$=\sqrt{x^2+1}\sin \left(x+\alpha\right)$$
Where $\displaystyle \cot \alpha = x\Rightarrow \alpha = \cot^{-1}(x) = \frac{\pi}{2}-\tan^{-1}(x)$
$\displaystyle \bullet\;\; x\cos x-1\cdot \sin x = -\sqrt{x^2+1}\left[\sin x\cdot \frac{1}{\sqrt{1+x^2}}-\cos x\cdot \frac{x}{\sqrt{1+x^2}}\right]$
$$=-\sqrt{x^2+1}\sin \left(x-\beta\right)$$
Where $\tan \beta = x\Rightarrow \beta = \tan^{-1}(x)$
So $$I = -\int\frac{1}{1\cos(x- \tan^{-1}{x})\cdot \sin (x-\tan^{-1}(x))}\cdot \frac{1}{1+x^2}dx$$
Now Put $x-\tan^{-}(x)=t\;,$ Then $\displaystyle \frac{1}{1+x^2}dx = dt$
So $$I = -\int\frac{\sin^2 t+\cos^2 t}{\sin t \cdot \cos t }dt = -\int \tan t dt-\int \cot t dt$$
So $$I = \ln |\cos t|-\ln |\sin t|+\mathcal{C} = -\ln |\tan t|+\mathcal{C}$$
So $$I =-\ln \left|\tan \left(x-\tan^{-x}(x)\right)\right|+\mathcal{C} = -\ln \left|\frac{\tan x-x}{1+x\tan x}\right|+\mathcal{C}$$
So $$I = \ln \left|\frac{\cos x+x\sin x}{\sin x-x\cos x}\right|+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\{f(n) = \frac{1}{4n \tan \frac{\pi}{n}} \}$ is a monotonically increasing sequence Could anyone is able to give me a hint why $\{f(n) = \frac{1}{4n \tan \frac{\pi}{n}} \}$ is a monotonically increasing sequence?
| Function
$$f(x)=\frac{1}{4x\tan{\frac{\pi}{x}}}$$ is increasing, since $f'(x)>0$, which is easily proved:
$$f'(x)=\frac{-4\tan{\frac{\pi}{x}}+\frac{4\pi}{x\cos^2{\frac{\pi}{x}}}}{4x^2\tan^2{\frac{\pi}{x}}}>0$$
is equivalent with $$-4\tan{\frac{\pi}{x}}+\frac{4\pi}{x\cos^2{\frac{\pi}{x}}}>0,$$
which is again equivalent with inequality
$$-4\sin{\frac{\pi}{x}}\cos{\frac{\pi}{x}}+4\frac{\pi}{x}>0,$$
that holds, since $\sin{t}<t$ and $\cos{t}<1$ for every $t\in\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show $1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$, where $z=\cos x+i \sin x$ Let $z=\cos x+i \sin x$. Show that
$$1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$$
| Since
\begin{eqnarray}
\cos x&=&2\cos^2\frac{x}{2}-1\\
\sin x&=&2\sin\frac{x}{2}\cos\frac{x}{2},
\end{eqnarray}
we have
\begin{eqnarray}
1+z&=&1+\cos x+i\sin x\\
&=&1+\Big(2\cos^2\frac{x}{2}-1\Big)+i\Big(2\sin\frac{x}{2}\cos\frac{x}{2}\Big)\\
&=&2\cos^2\frac{x}{2}+2i\sin\frac{x}{2}\cos\frac{x}{2}\\
&=&2\cos\frac{x}{2}\Big(\cos\frac{x}{2}+i\sin\frac{x}{2}\Big)
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A tangent to the ellipse meets the $x$ and $y$ axes . If $O$ is the origin, find the minimum area of triangle $AOB$.
The question is that : A tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the $x$ and $y$ axes respectively at $A$ and $B$. If $O$ is the origin, find the minimum area of triangle $AOB$.
What I have attempted:
Because the equation of an ellipse is $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$$
We have that the $x$ intercept is $x=\frac{a^2}{x_1}$ and $y$ intercept is $y=\frac{b^2}{y_1}$
Hence the area of a triangle $(AOB)$ is given by $A=\frac{a^2b^2}{2x_1y_1}$ now I am stuck, how should I proceed?
| $A = \frac {a^2 b^2}{2xy}$
constrained by
$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$
$\frac{dA}{dx} = \frac {-2a^2b^2 (y + x y')}{(2xy)^2} = 0\\
y + x y' = 0\\
y' = -\frac yx$
Differentiating the constraint.
$\frac x{a^2} + \frac {y y'}{b^2} = 0\\
y' = -\frac {b^2x}{a^2y}\\
\frac {b^2x}{a^2y} = \frac yx\\
b^2 x^2 = a^2 y^2$
Again from the constraint:
$b^2 x^2 + a^2 y^2 = a^2 b^2\\
2a^2 y^2 = a^2 b^2\\
y^2 = \frac {b^2}{2}\\
y = \frac b{\sqrt {2}}\\
2xy = ab\\
A = \frac {a^2b^2}{2xy} = ab$
alternate
$x = a \cos t\\
y = b \sin t\\
A = \frac {ab}{2} \csc t \sec t\\
\frac{dA}{dt} = \frac {ab}{2} \sec t \csc t ( -\cot t + \tan t) = 0\\
\cot t = \tan t\\
t = \frac {\pi}{4}\\
A = ab$
That is much nicer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: Corrected the final integral and the limit from $45$ to $\pi/4$
| Hint:
note that
$$
x^6+1=(x^2+1)(x^2-\sqrt{3}+1)(x^2+\sqrt{3}+1)
$$
than use partial fraction decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Evaluation of this series $\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{4{{n}^{2}}-1}}=??$ I start first to use $u=n+1$ then
$$\sum_{u=2}^{\infty}
{\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}.$$
| Another way to do it is converting the series into a double integral:
Let $$I=\int_{0}^{1}\int_{0}^{1} \frac{x^2}{1+x^2y^2}dydx.$$ Because the region of integration implies $$0<x,y<1$$ we can convert the integrand into a geometric series as such:
$$\frac{x^2}{1+x^2y^2}=\sum_{n=0}^{\infty}x^2(-x^2y^2)^n=\sum_{n=0}^{\infty}(-1)^nx^{2n+2}y^{2n}.$$ We replace the integrand with the series and integrate term by term to get that
$$I=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+3)(2n+1)}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n+1)(2n-1)}$$ if you change the starting point of the summation.
Now we complete the proof by evaluating $$I=\int_{0}^{1}\int_{0}^{1} \frac{x^2}{1+x^2y^2}dydx.$$ directly.
Integrate with respect to $y$, we get $$I=\int_{0}^{1} x\arctan(x)dx.$$ Use the fact $\int_{0}^{1} \frac{1}{1+x^2y^2}dy=\frac{\arctan(x)}{x}$ to prove this. Now using integration by parts: $u=\arctan(x),du=\frac{dx}{1+x^2},v=\frac{x^2}{2},dv=xdx,$ we get $$I=\frac{\pi}{8}-\int_{0}^{1} \frac{x^2}{2(1+x^2)} dx.$$ To evaluate $$\int_{0}^{1} \frac{x^2}{2(1+x^2)} dx,$$ put $x=\tan(\theta),dx=\sec^2(\theta)d\theta$ to see that $$\int_{0}^{1} \frac{x^2}{2(1+x^2)} dx=\int_{0}^{\frac{\pi}{4}} \frac{\tan^2(\theta)}{2} d\theta=\int_{0}^{\frac{\pi}{4}} \frac{\sec^2(\theta)-1}{2} d\theta=\frac{1}{2}-\frac{\pi}{8}$$ Putting everything together , $$I=\frac{\pi}{4}-\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Do two nonsingular matrices exist such that this linear system can be solved? (I've posted this question on MathOverflow and was sent here. This is a copy of that submission)
I am trying to solve a system of linear equations. In this system I have two separate systems involving an unknown tension vector $\overrightarrow{T}$ in $\mathbb{R}^n$, which I want to cancel out of both equations by making them equal. In each system, $\overrightarrow{T}$ is multiplied by $n \times n$ matrices $C_{SU}$ and $S_{SU}$, defined as follows:
$$
C_{SU} =
\begin{bmatrix}
c_1 & -c_2 & 0 & \cdots & 0 \\
0 & c_2 & -c_3 & \cdots & 0 \\
0 & 0 & c_3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & c_n
\end{bmatrix}
$$
Where $c_k = \cos{\theta_k}$. $S_{SU}$ is defined similarly, replacing $\cos$ with $\sin$.
So to summarize, I have two linear systems of the following form:
$$
\overrightarrow{R_1} = C_{SU}\overrightarrow{T}\\
\overrightarrow{R_2} = S_{SU}\overrightarrow{T}
$$
Where $\overrightarrow{R_1}$ and $\overrightarrow{R_2}$ are some vectors in $\mathbb{R}^n$. I want to cancel $\overrightarrow{T}$ out of the equation by making both left hand sides equal. To do this, I intend to multiply $C_{SU}$ and $S_{SU}$ by some matrices $A$ and $B$ such that $A \cdot C_{SU} = B \cdot S_{SU}$. $A$ and $B$ should be $n \times n$ invertible matrices.
Do $A$ and $B$ exist, and if so, what are they?
| So, if I understood correctly your symbolism, we are
dealing with matrices of the type
$$
\begin{gathered}
\mathbf{C}\left( \mathbf{x} \right) = \left[ {\begin{array}{*{20}c}
{x_1 } & { - x_2 } & 0 & \cdots & 0 \\
0 & {x_2 } & { - x_3 } & \cdots & 0 \\
0 & 0 & {x_3 } & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & {x_n } \\
\end{array} } \right] = \left[ {\begin{array}{*{20}c}
1 & { - 1} & 0 & \cdots & 0 \\
0 & 1 & { - 1} & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1 \\
\end{array} } \right]\left[ {\begin{array}{*{20}c}
{x_1 } & 0 & 0 & \cdots & 0 \\
0 & {x_2 } & 0 & \cdots & 0 \\
0 & 0 & {x_3 } & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & {x_n } \\
\end{array} } \right] = \hfill \\
= \left( {\mathbf{I} - \mathbf{E}} \right)\;\mathbf{D}\left( \mathbf{x} \right) \hfill \\
\end{gathered}
$$
where:
$\mathbf{I}$ is the identity matrix, $\mathbf{E}$ is the matrix with elements $=1$ in the first upper diagonal and remaining null,
and $\mathbf{D}\left( \mathbf{x} \right)$ is the diagonal matrix with entries $\left( {x_1 ,\; \ldots ,\;x_n } \right)$.
Now, it is well known, and it is easily demonstrable, that
$$
\left( {\mathbf{I} - \mathbf{E}} \right)^{\, - \,\mathbf{1}} = \mathbf{S} = \left[ {\begin{array}{*{20}c}
1 & 1 & 1 & \cdots & 1 \\
0 & 1 & 1 & \cdots & 1 \\
0 & 0 & 1 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1 \\
\end{array} } \right]
$$
and of course:
$$
\mathbf{D}\left( \mathbf{x} \right)^{\, - \,\mathbf{1}} = \left[ {\begin{array}{*{20}c}
{1/x_1 } & 0 & 0 & \cdots & 0 \\
0 & {1/x_2 } & 0 & \cdots & 0 \\
0 & 0 & {1/x_3 } & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & {1/x_n } \\
\end{array} } \right]
$$
Concerning the determinants
$$
\begin{gathered}
\left| {\left( {\mathbf{I} - \mathbf{E}} \right)} \right| = \left| \mathbf{S} \right| = \;1 \hfill \\
\left| {\mathbf{D}\left( \mathbf{x} \right)} \right| = \prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {x_{\,k} } = \frac{1}
{{\left| {\mathbf{D}\left( \mathbf{x} \right)^{\, - \,\mathbf{1}} } \right|}} \hfill \\
\end{gathered}
$$
So the matrix $ \mathbf{C}\left( \mathbf{x} \right) $ is invertible if none of the $x_k$ is null:
$$
\mathbf{C}\left( \mathbf{x} \right)^{\, - \,\mathbf{1}} = \mathbf{D}\left( \mathbf{x} \right)^{\, - \,\mathbf{1}} \;\mathbf{S}
$$
Therefore the answer to your question is
$$
\begin{gathered}
\mathbf{A} = \mathbf{C}\left( \mathbf{c} \right)^{\, - \,\mathbf{1}} = \mathbf{D}\left( \mathbf{c} \right)^{\, - \,\mathbf{1}} \;\mathbf{S}\quad \left| {\;\cos \theta _k \ne 0} \right. \hfill \\
\mathbf{B} = \mathbf{C}\left( \mathbf{s} \right)^{\, - \,\mathbf{1}} = \mathbf{D}\left( \mathbf{s} \right)^{\, - \,\mathbf{1}} \;\mathbf{S}\quad \left| {\;\sin \theta _k \ne 0} \right. \hfill \\
\end{gathered}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Exponential generating functions balls problem I had a question about an exercise that I couldn't solve on my exam.
It was the following:
You have 3 boxes. 1 with an unlimited amount of identical red balls, 1 with an unlimited amount of identical green balls and 1 with an unlimited amount of identical blue balls. We now draw n balls in order out of these boxes and make an ordered list of the drawn balls.
Call:
*
*$A_n$ the number of possible lists of length n that contains at least 1 green ball
*$B_n$ the number of possible lists of length n that contains at least 2 blue balls
*$C_n$ the number of possible lists of length n that contains at least 1 green ball and at least 2 blue balls
*$D_n$ the number of possible lists of length n that contains at least 1 green ball or at least 2 blue balls
1) Find the exponential generating function of those situations with a simple expression
2) Find $A_n$, $B_n$, $C_n$ and $D_n$. The sum taht you find has to be 196.
How could I solve this problem?
Thanks in advance!
| To start off, we can find the exponential generating function where the coefficient of $\frac{x^r y^g z^b}{n!}$ is the number of lists containing $r$ red balls, $g$ green balls, and $b$ blue balls, where $r + g + b = n$. This is given by
$$
\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right)
\left(1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots \right)
\left(1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \right)
$$
which is just
$$
e^x e^y e^z = e^{x + y + z}.
$$
Plugging in $x = y = z$ (effectively, adding up the total number of balls in the exponent) gives the generating function for the total number of lists of size $n$, $e^{3x}$.
Then we modify the construction to get the answers to each part.
*
*For $A_n$, we want $y^1$ to be the smallest allowed factor of $y$, so we instead multiply
$$
\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right)
\left(y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots \right)
\left(1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \right)
$$
to get $e^x (e^y - 1) e^z$. Plugging in $x = y = z$,
$$
\sum_{n \ge 0} A_n \frac{x^n}{n!} = e^{2x}(e^x - 1).
$$
*For $B_n$, we want $z^2$ to be the smallest allowed factor of $z$, so we instead multiply
$$
\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right)
\left(y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots \right)
\left(\frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \right)
$$
to get $e^x e^y (e^z - z - 1)$. Plugging in $x = y = z$,
$$
\sum_{n \ge 0} B_n \frac{x^n}{n!} = e^{2x} (e^x - x - 1).
$$
*To get $C_n$, we use the power series in $y$ from $A_n$ and the power series in $z$ from $B_n$, to get $e^x (e^y - 1)(e^z - z - 1)$. Plugging in $x = y = z$,
$$
\sum_{n \ge 0} C_n \frac{x^n}{n!} = e^{x} (e^x - 1) (e^x - x - 1).
$$
*The easiest way to find $D_n$ is to notice that $D_n = A_n + B_n - C_n$. Hence, we have that
\begin{align*}
\sum_{n \ge 0} D_n \frac{x^n}{n!}
&= e^{2x}(e^x - 1) + e^{2x} (e^x - x - 1) - e^{x} (e^x - 1) (e^x - x - 1) \\
&= e^x\left(e^{2x} - x - 1 \right)
\end{align*}
All of the generating functions above can be found by a more direct method, if we notice that a list of $n$ balls is effectively the same as all the balls being labeled distinctly (say, by index in the list). Then $e^x$ is the generating function for 0 or more labeled balls of a specific color, $e^x - 1$ for one or more labeled balls of a color, and $e^x - x - 1$ for two or more labeled balls of a color. The generating functions for $A_n, B_n,$ and $C_n$ immediately follow.
For $D_n$, note that $e^{2x}$ counts collections of labeled balls of two colors, and we only want to exclude the case with exactly one blue and no greens and the case of none of each, so $e^{2x} - x - 1$ is the number of valid collections of blue and green balls, and then multiply by $e^x$ for red balls, to get $e^x(e^{2x} - x - 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral trigonometry $\int\sin3x\sin^2x\,dx$
$=\int \sin3x \frac{(1-cos2x)}{2}dx$
$=\frac{1}{2}(\int \sin 3x dx - \int \sin 3x \cos 2x dx)$
$I=\frac{-1}{2}\frac{1}{3}cos 3x-1/4(\int \sin 5x dx+ \int \sin x dx)$
So my question is how do i get from:
$\int \sin (3x)\cos (2x) dx$ to $\frac{1}{2}(\int \sin (5x) dx+ \int \sin (x) dx)$
Thanks for fast answer i solved it.
| by the formula $\sin { \left( x \right) \cos { \left( y \right) =\frac { 1 }{ 2 } \left[ \sin { \left( x+y \right) +\sin { \left( x-y \right) } } \right] } } $
in your case we have $$\\ \sin { \left( 3x \right) \cos { \left( 2x \right) =\frac { 1 }{ 2 } \left[ \sin { \left( 3x+2x \right) +\sin { \left( 3x-2x \right) } } \right] =\frac { 1 }{ 2 } \left[ \sin { \left( 5x \right) +\sin { \left( x \right) } } \right] } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How many different ways can a number be written as a sum of 1, 3, and 5? I had a programming question where we had to write code to output, given some number x, all the different ways x can be written as a sum of 1, 3 and 5. So for instance if x=6, then the answer is 4, as x=5+1=3+3=3+1+1+1=1+1+1+1+1+1.
This made me wonder how the same question could be done without using a computer, but I was unsure how one would go about doing this. Please try and explain how this could be done, preferably in the simplest way possible.
| The generating function approach is to note that if you write:
$$\begin{align}
f(x)&=\frac{1}{1-x}\frac{1}{1-x^3}\frac{1}{1-x^5}\\
&=(1+x+x^2+x^3+\cdots)(1+x^3+x^6+x^9+\cdots)(1+x^5+x^{10}+\cdots)\\
&=a_0+a_1x+a_2x^2+\cdots
\end{align}$$
Then $a_n$ counts the number of ways to partition $n$ into values in $1,3,5.$
Then we use the method of partial fractions to write:
$$f(x)=\frac{a}{(1-x)^3}+ \frac{b}{(1-x)^2}+\frac{c}{1-x}+ \frac{dx+e}{1+x+x^2}+\frac{fx^3+gx^2+hx+j}{1+x+x^2+x^3+x^4}$$
You can solve for $a,b,c,d,e,f,g,h,j$.
This is tedious, but you really only need to compute $a,b,c$, and there are tricks for doing so. Then you'll get a formula $c\binom{n+2}{2}+b\binom{n+1}{2}+a\binom{n}{2}$. As noted in comments, $a_n-\left(c\binom{n+2}{2}+b\binom{n+1}{2}+a\binom{n}{2}\right)$ will be periodic of period $15$, so you only have to figure out $a_0,\dots,a_{14}$ to figure out what the "correction" is.
I used Wolfram Alpha to find a,b,c.
It turns out the closed formula is the nearest integer to $\frac{(n+4)(n+5)}{30}=\frac{1}{15}\binom{n+5}{2}$.
I've checked this answer against Marcus's (now-deleted) answer, and it works up to $n=10000$, so I'm pretty sure my arithmetic was accurate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Find the value of the constants If $\displaystyle
\frac{\left( \frac{2x^2}{3a} \right)^{n-1}}
{\left( \frac{3x}{a} \right)^{n+1}} =
\left( \frac{x}{4} \right)^3$, determine the values of the constants $a$ and $n$
I could find the value of $a$, i.e, $\displaystyle \frac{\sqrt{x^6 \times 3^{2n}}}{2^n x^n 2^5 }$ and substituted the same to find the value of $n$, to no avail
The right values for $a$ is $\pm(3^6 2^{-11/2})$, and $n$ is equal to $6$
| Expand the LHS as
$$
\frac{{\left( {\frac{{2x^{\,2} }}
{{3a}}} \right)^{n - 1} }}
{{\left( {\frac{{3x}}
{a}} \right)^{n + 1} }} = \frac{{2^{n - 1} x^{2\left( {n - 1} \right)} }}
{{3^{n - 1} a^{n - 1} }}\;\frac{{a^{n + 1} }}
{{3^{n + 1} x^{n + 1} }} = \frac{{2^{n - 1} a^2 x^{n - 3} }}
{{3^{2n} }}\;
$$
The equating it to the RHS, you shall impose that the exponent
of $x$ be equal on both sides, as well as the multiplying coefficients.
Therefore:
$$
\begin{gathered}
\frac{{2^{n - 1} a^2 x^{n - 3} }}
{{3^{2n} }}\; = \frac{{x^3 }}
{{4^3 }}\quad \Rightarrow \quad \left\{ \begin{gathered}
x^{n - 3} = x^3 \hfill \\
\frac{{2^{n - 1} a^2 }}
{{3^{2n} }} = \frac{1}
{{4^3 }} \hfill \\
\end{gathered} \right. \hfill \\
\Rightarrow \quad \left\{ \begin{gathered}
n = 6 \hfill \\
\frac{{2^5 a^2 }}
{{3^{12} }} = \frac{1}
{{4^3 }} \hfill \\
\end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered}
n = 6 \hfill \\
a^2 = \frac{{3^{12} }}
{{2^5 4^3 }} = \frac{{3^{12} }}
{{2^{11} }} \hfill \\
\end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered}
n = 6 \hfill \\
a = \pm \sqrt 2 \frac{{3^6 }}
{{2^6 }} \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Two limits involving integrals: $\lim_{\varepsilon\to 0^+}\left(\int_0^{1-\varepsilon}\frac{\ln (1-x)}{x\ln^p x}dx-f_p(\varepsilon)\right)$, $p=1,2$. By applying the Taylor series expansion to $\ln x$, as $x \to 1$, one has the Laurent series expansion,
$$
\frac1{\ln x}=-\frac1{1-x}+\frac{1}{2}+O\left(1-x\right)
$$
then clearly
$$
\begin{align}
&\lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} \frac{\ln (1-x)}{x\ln x}\:dx=\infty
\\\\&\lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} \frac{\ln (1-x)}{x\ln^2 x}\:dx=-\infty.
\end{align}
$$
Thus I'm designing the following related limits.
Question. Find $f_1(\varepsilon)$, $f_2(\varepsilon)$ and find a closed form of $c_1$, $c_2$ such that
$$
\begin{align}
&\lim_{\varepsilon \to 0^+} \left(\int_0^{1-\varepsilon} \frac{\ln (1-x)}{x\ln x}\:dx-f_1(\varepsilon)\right)=c_1
\tag1
\\\\&
\lim_{\varepsilon \to 0^+} \left(\int_0^{1-\varepsilon} \frac{\ln (1-x)}{x\ln^2 x}\:dx-f_2(\varepsilon)\right)=c_2. \tag2
\end{align}
$$
Edit. A complete answer is now given below.
| We can try to use your idea for the other integral. Note that $$\frac{\log\left(1-x\right)}{x\log^{2}\left(x\right)}=\frac{\log\left(1-x\right)}{\left(1-x\right)^{2}}+\frac{\log\left(1-x\right)}{x}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)
$$ so we have $$\int_{0}^{1-\epsilon}\frac{\log\left(1-x\right)}{x\log^{2}\left(x\right)}dx=\frac{\log\left(\epsilon\right)+1}{\epsilon}-1
$$ $$ +\int_{0}^{1}\frac{\log\left(1-x\right)}{x}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx+O\left(\epsilon\right).\tag{1}
$$ Let us define $$F\left(s\right)=\int_{0}^{1}x^{s}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx
$$ so if we differentiate twice $$F''\left(s\right)=\int_{0}^{1}x^{s}\left(1-\frac{\log^{2}\left(x\right)x}{\left(1-x\right)^{2}}\right)dx=\frac{1}{s+1}-\int_{0}^{1}\frac{\log^{2}\left(x\right)x^{s+1}}{\left(1-x\right)^{2}}dx
$$ and we have that $$\int_{0}^{1}\frac{\log^{2}\left(x\right)x^{s+1}}{\left(1-x\right)^{2}}dx=\sum_{k\geq1}k\int_{0}^{1}\log^{2}\left(x\right)x^{s+k}dx
$$ $$ =2\sum_{k\geq1}\frac{k}{\left(s+k+1\right)^{3}}=2\psi^{\left(1\right)}\left(s+2\right)+\left(s+1\right)\psi^{\left(2\right)}\left(s+2\right)$$ so integrating twice and observing that $$F\left(0\right)=\int_{0}^{1}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx
$$ $$=\gamma+\int_{0}^{1}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}-\frac{1}{\log\left(x\right)}\right)dx=\gamma-\frac{1}{2}
$$ since $$\int\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}-\frac{1}{\log\left(x\right)}\right)dx=\frac{1}{x-1}-\frac{x}{\log\left(x\right)}+c
$$ and $$F'\left(0\right)=\int_{0}^{1}\left(\frac{1}{\log\left(x\right)}-\frac{\log\left(x\right)}{\left(1-x\right)^{2}}+\frac{\log\left(x\right)}{1-x}\right)dx=\gamma-\frac{\pi^{2}}{6}+1
$$ since $$\int\left(-\frac{1}{1-x}-\frac{\log\left(x\right)}{\left(1-x\right)^{2}}\right)dx=\frac{\log\left(x\right)}{x-1}+\log\left(x\right)+d
$$ we have $$F\left(s\right)=\left(s+1\right)\left(\log\left(s+1\right)-\psi^{\left(0\right)}\left(s+2\right)\right)+\frac{1}{2}.
$$ From $(1)$ we get $$\int_{0}^{1}\frac{\log\left(1-x\right)}{x}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx
$$ $$=-\sum_{n\geq1}\frac{1}{n}\int_{0}^{1}x^{n-1}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx
$$ $$=-\sum_{n\geq1}\frac{n\left(\log\left(n\right)-H_{n}+\gamma\right)+\frac{1}{2}}{n}=\sum_{n\geq1}\left(H_{n}-\log\left(n\right)-\gamma-\frac{1}{2n}\right)$$ and the closed form of the series can be found here, so
$$\int_{0}^{1-\epsilon}\frac{\log\left(1-x\right)}{x\log^{2}\left(x\right)}dx-\frac{1+\log\left(\epsilon\right)}{\epsilon}\stackrel{\epsilon\rightarrow0^{+}}{\rightarrow}\frac{\gamma-1-\log\left(2\pi\right)}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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} |
Volume of ellipsoid outside sphere I have the ellipsoid $\frac{x^2}{49} + y^2 + z^2 = 1$ and I want to calculate the sum of the volume of the parts of my ellipsoid that is outside of the sphere $x^2+y^2+z^2=1$
How to do this? I know the volume of my sphere, $\frac{4\pi}{3}$, and that I probably should set up some double- or triple integral and transform the coordinates to spherical coordinates and evaluate but I have to admit I'm stuck on how to set this up.
| Hints.
1) If $(x,y,z)$ satisfies $x^2+y^2+z^2\leq 1$ then
$$\frac{x^2}{49} + y^2 + z^2 \leq x^2+y^2+z^2\leq 1.$$
What does this inequality mean?
2) $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}=1$ is the equation of an ellipsoid centered at the origin of semi-principal axes of length $a$, $b$, $c$, and its volume is $\frac{4\pi (a\cdot b \cdot c)}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum and Maximum value of |z| This is a question that I came across today:
If $|z-(2/z)|=1$...(1) find the maximum and minimum value of |z|, where z represents a complex number.
This is my attempt at a solution:
Using the triangle inequality, we can write:
$||z|-|2/z||≤|z+2/z|≤|z|+|2/z|$
Let $|z|=r$ which implies that $|r-2/r|≤1≤r+2/r$ (From (1))
How must I proceed to find the value of |z|? Please help! Much thanks in advance :)
| Since $|z-2/z| = 1$, we have
$$\begin{aligned}
1 = |z- 2/z|^2 &= (z - 2/z)(\overline{z} - 2/\overline{z}) \\
&= |z|^2 - 2z/\overline{z} - 2\overline{z}/z + 4/|z|^2
\end{aligned}$$
Write $z = re^{i\theta}$. Then $\overline{z} = re^{-i\theta}$. Substituting into the above, we obtain
$$\begin{aligned}
1 &= r^2 - 2e^{i2\theta} - 2^{-i2\theta} + 4/r^2 \\
&= r^2 - 4\cos(2\theta) + 4/r^2 \\
\end{aligned}$$
We can rearrange this to get
$$r^2 + 4/r^2 \leq 4\cos(2\theta) + 1 \leq 5$$
or equivalently,
$$r^4 -5r^2 + 4 \leq 0$$
Factor the left hand side to obtain
$$(r-1)(r+1)(r-2)(r+2) \leq 0$$
As $r$ must be positive, this means that $r+1$ and $r+2$ are positive. Therefore, $r-1$ and $r-2$ must have opposite signs (or one of them is zero), which forces $1 \leq r \leq 2$.
You can check that $z=1$ and $z=2$ are solutions to the original equation, so $1$ and $2$ are the minimum and maximum values of $|z|$.
Here is a picture showing the set of solutions. From this, we might be tempted to speculate that we are looking at two circles of radius $1/2$, centered at $z=3/2$ and $z=-3/2$. But this is not the case. To see this, consider the point $z = 3/2 + i(1/2)$. This point is on the circle of radius $1/2$ centered at $z=3/2$. However, it does not satisfy the given equation:
$$\begin{aligned}
|z-2/z| &= |3/2 + i(1/2) - 2/(3/2 + i(1/2))| \\
&= |3/2 + i(1/2) - 6/5 + i(2/5)| \\
&= |3/10 + i(9/10)| \\
&= \sqrt{9/10} \neq 1
\end{aligned}$$
I don't think it's an ellipse, either, since these are of the form $|z-a| + |z-b| = c$. If there is a way to transform $|z-2/z| = 1$ into that form, it's not obvious to me (also, our figure has not one but two "ellipses").
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
When does this equation have positive roots? I am reading a paper where the author states (without proofing) that the equation
$$
{x}^2 \left(1-2 \sqrt{2}\, {x} \right) \sin ^2(\psi )-{y}^2
\left(1-2 \sqrt{2} \,{y} \right)=0
$$
has no positive roots for $y<3\sqrt{2}$ unless $\sin(\psi)>3\sqrt{2}\,y\sqrt{3(1-2\sqrt{2}\,y)}$
Could someone please explain why is this true?
| Suppose $y<\frac{1}{2\sqrt{2}}$ is fixed. Then define
\begin{align*}
f(x)&={x}^2 \left(1-2 \sqrt{2}\, {x} \right) \sin ^2(\psi )-{y}^2
\left(1-2 \sqrt{2} \,{y} \right)\\
f'(x)&=\left(2x -6 \sqrt{2}\, x^2 \right) \sin ^2(\psi )
\end{align*}
Since $f$ is concave down the maximum value occurs at $x_0=\frac{1}{3 \sqrt{2}}$. Plugging it in $f(x)$ gives us
\begin{align*}
f(x_0)&= \frac{\sin ^2(\psi )}{54}-{y}^2
\left(1-2 \sqrt{2} \,{y} \right)
\end{align*}
If $\sin(\psi)<3\sqrt{2}\,y\sqrt{3(1-2\sqrt{2}\,y)}$ then the maximum value of this function is negative hence there is no roots.
| {
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"url": "https://math.stackexchange.com/questions/1898973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ This is a problem that I tried to solve and didn't come up with any ideas
.?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$
All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2}}$ which evaluates to $1<0$Do you know what to do here ?
| This proof true when $\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.$
$$\frac { 1 }{ 2 } \cdot \frac { 3 }{ 4 } \cdots \frac { 2n-1 }{ 2n } \frac { 2n+1 }{ 2n+2 } <\frac { 1 }{ \sqrt { 2n+1 } } .\frac { 2n+1 }{ 2n+2 } =\frac { 1 }{ \sqrt { 2n+3 } } .\frac { \sqrt { 2n+3 } }{ \sqrt { 2n+1 } } \frac { 2n+1 }{ 2n+2 } =\\ =\frac { 1 }{ \sqrt { 2n+3 } } \sqrt { \frac { 4{ n }^{ 2 }+8n+3 }{ 4{ n }^{ 2 }+8n+4 } } <\frac { 1 }{ \sqrt { 2n+3 } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
| Intro: this is the business about taking the gcd of a polynomial and its derivative, in order to detect a repeated factor. In this case, the repeated factor just squares to give the original, so this is one of the simplest types. For this answer yesterday, How can one find the factorization $a^4 + 2a^3 + 3a^2 + 2a + 1 = (a^2 + a + 1)^2$ from scratch? I deliberately made up a more complicated situation. In that one, it was actually the cube of something.
$$ f = m^4 + 6 m^3 + 11 m^2 + 6 m + 1 $$
$$ f' = 4 m^3 + 18 m^2 + 22 m + 6 $$
$$ f'/2 = 2 m^3 + 9 m^2 + 11 m + 3 $$
$$ 2 f - m (2 m^3 + 9 m^2 + 11 m + 3) = 3 m^3 + 11 m^2 + 9 m + 2 $$
$$ 3(2 m^3 + 9 m^2 + 11 m + 3) - 2 (3 m^3 + 11 m^2 + 9 m + 2) = 5 m^2 + 15 m + 5. $$
With $$ 5 m^2 + 15 m + 5 = 5 (m^2 + 3m + 1) $$ we want
$$ \gcd_{\mathbb Q}(2 m^3 + 9 m^2 + 11 m + 3,m^2 + 3m + 1 ). $$
However,, $$ 2 m^3 + 9 m^2 + 11 m + 3 = (2m+3)(m^2 + 3m + 1) $$
so the GCD is $m^2 + 3m + 1$ itself. This must be a repeat factor of the original $f,$ and we check that, in fact, $f = (m^2 + 3m+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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How to determine that the solution to the following problem is unique? The question is:
[BMO2 2000 Q3] Find positive integers a and b such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20\sqrt[3]{6}$$
It suffices to find one solution to gain full marks, which I did by expanding and assuming that neither $a$ nor $b$ are perfect cubes, so that $2\sqrt[3]{ab} = 48$. The solution I get from this method is $(a, b) = (48, 288)$. However, from this two questions arise in my mind:
a) Is this is the only solution? I.e. are there $c$ and $d$ such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = (\sqrt[3]{c} + \sqrt[3]{d} - 1)^2$$
$\Rightarrow \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{c} + \sqrt[3]{d}$ or $\sqrt[3]{a} + \sqrt[3]{b} - 1 = -\sqrt[3]{c} - \sqrt[3]{d} + 1$
b) Can one solve the original problem in such a way that it is clear that the solutions found are unique?
EDIT: Thank you to gammatester for his remark that there is another solution: $(a, b) = (288, 48)$, but I am referring to distinct solutions which are not permutations of other solutions.
| I apologize, this solution is a bit hand-wavy and (it may in fact be wrong) as my skills/knowledge in this topic is rather weak. Perhaps someone else can later provide a more rigorous version of this argument if it is correct.
We have,
$$\sqrt[3]{a^2}+\sqrt[3]{b^2}-2(\sqrt[3]{a}+\sqrt[3]{b}) = 20\sqrt[3]{6}.$$
Consider the highest power of $6$ that divides $a$ and $b$ modulo $3$. I believe (likely using some argument about the linear independence of cuberoots of integers over $\mathbb{Q}$) that it can be shown that the highest power of $6$ (modulo $3$) that divides $a$ is $6^1$ and similarly $6^2$ for $b$. Such an argument would probably use the fact that the righthand side doesn't contain any $\sqrt[3]{6^2}$ terms.
Also, we have, like you said,
$$2\sqrt[3]{ab} = 48 = 2^4 \cdot 3$$
$$\implies 2^3 | \sqrt[3]{ab}$$
$$\implies 2^9 | ab.$$
That means that there are $9$ factors of $2$ between $a$ and $b$. Since $a$ contains $1$ factor of $6$ and $b$ contains $2$ factors of $6$, that means that there are a total of $6$ factors of $2$ left between $a$ and $b$ after we subtract out the factors of $2$ belonging to the $6$'s. In order to write $\sqrt[3]{a} = k\sqrt[3]{6}$ and $\sqrt[3]{b} = l\sqrt[3]{6^2}$ for some integers $k,l$; then those $6$ factors of $2$ must be distributed between $a$ and $b$ in multiples of $3$. There are three ways to do this:
*
*Give all six factors of $2$ to $a$
*Give all six factors of $2$ to $b$
*Give both $a$ and $b$ three factors of $2$ each.
This gives us three candidates for our choices of $a,b$:
*
*$a = 6\cdot 2^6$ and $b=6^2$
*$a = 6$ and $b=6^2\cdot 2^6$
*$a = 6\cdot 2^3$ and $b=6^2\cdot 2^3$
Of these three, only the last one solves the equation, hence the solution is unique.
As a side note, I've verified that $(48,288)$ is the only solution up to reordering in Mathematica:
Solve[(x^(1/3) + y^(1/3) - 1)^2 == 49 + 20*6^(1/3) && x > 0 && y > 0, {x, y}, Integers]
Hope this helps!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve $\frac{xdx-ydy}{xdy-ydx}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$
Solve $\dfrac{xdx-ydy}{xdy-ydx}=\sqrt{\dfrac{1+x^2-y^2}{x^2-y^2}}$
I tried doing this-
$\sqrt{x^2-y^2}d(x^2-y^2)=x^2\sqrt{1+x^2-y^2}d(\frac{x}{y})$ but it does not get better from here.
I even tried putting $\sqrt{x^2-y^2}$ as $u$ but that does not simplify either.
| I would suggest to you a more intricate change of variables:
\begin{align}
x &= z \cosh{s} = z \,\left(\frac{e^{s} + e^{-s}}{2}\right)\\
y &= z \sinh{s} = z \left(\frac{e^{s} - e^{-s}}{2}\right)
\end{align}
Then
\begin{align}
dx = \cosh(s) \, dz + z\,\sinh(s)\, ds \\
dy = \sinh(s) \, dz + z\,\cosh(s)\, ds
\end{align}
as well as
$$x^2-y^2 = z^2\big(\cosh^2(s) - \sinh^2(s)\big) = z^2$$ which means that
$$z = \sqrt{x^2 - y^2}$$
Now
\begin{align}
y\,dx = z \sinh(s) \cosh(s) \, dz + z^2\,\sinh^2(s)\, ds \\
x\,dy = z \cosh(s) \sinh(s) \, dz + z^2\,\cosh^2(s)\, ds
\end{align}
which leads to the difference
\begin{align}
x\,dy - y\,dx &= z \cosh(s) \sinh(s) \, dz + z^2\,\cosh^2(s)\, ds\\
& - z \sinh(s) \cosh(s) \, dz - z^2\,\sinh^2(s)\, ds \\
&= z^2\,\cosh^2(s)\, ds - z^2\,\sinh^2(s)\, ds \\
&= z^2\big(\cosh^2(s) - \sinh^2(s)\big) \, ds\\
&= z^2\, ds
\end{align}
Moreover,
\begin{align}
x\,dx - y\,dy &= \frac{1}{2}d \left(x^2-y^2\right)\\
&= \frac{1}{2} d(z^2)\\
&= z\,dz
\end{align}
Consequently
\begin{align}\frac{x\,dx - y\,dy}{ x\,dy - y\,dx } &= \frac{z\, dz}{z^2 \, ds}\\ &= \frac{1}{z} \frac{dz}{ds}\end{align}
and finaly the equations becomes
Moreover,
\begin{align}
\frac{1}{z} \frac{dz}{ds} = \frac{x\,dx - y\,dy}{ x\,dy - y\,dx } = \sqrt{\frac{1 + x^2-y^2}{x^2-y^2}} = \sqrt{\frac{1 + z^2}{z^2}} = \frac{\sqrt{1+z^2}}{z}
\end{align}
multiply both sides by $z$ and obtain the simple differential equation
$$ \frac{dz}{ds} = \sqrt{1+z^2}.$$
I think you can take it from here.
| {
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"url": "https://math.stackexchange.com/questions/1906473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
From $\lim_{n\to\infty}n\sum_{k=1}^\infty\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}$ to a multiple of $\zeta(3)$ This afternoon I tried do the specialisation of another problem due to Furdui. THat is PROBLEMA 103, La Gaceta de la RSME, Volumen 12, número 2 (2009), see the first identity in page 317 of the solution (in spanish).
Thus I know that one can state an identity from the following limit with a multiple of $\zeta(3)$ following the quoted identity, $$\lim_{n\to\infty}n\sum_{k=1}^\infty\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}.$$
My attempt to try show such identity was write $\frac{1}{1+n^2x^2}$ as $$\frac{-i}{2n}\frac{1}{x-\frac{i}{n}}+\frac{i}{2n}\frac{1}{x+\frac{i}{n}}.$$
THen after the change of variables $u=(x+k)^3$, I can write $$\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}=\int_{k^3}^{(k+1)^3}\frac{n/3du}{u^{5/3}((nu^{1/3}-nk)^2+1)}.$$
Also I've calculated with help of Wolfram Alpha previous indefinite integral, with the code
integrate x^(-5/3)/(n^2(x^(1/3)-k)^2+1) dx.
On the other hand I tried calculate the previous indefinite integral by parts
$$\int\frac{nx^{-5/3}dx}{(n^2(x^{1/3}-k)^2+1)}=\frac{\arctan n(x^{1/3}-k)}{x}+\int \frac{\arctan n(x^{1/3}-k)}{x^2}dx+\text{cte}.$$
I don't know how evaluate (the limits of integration and after take the series and the limit) previous calculations to get an identity involving $\zeta(3)$ without using Furdui's result. I don't know if there are mistakes in my calculations.
Question. What's is a right approach and set of calculations, to show that $$\lim_{n\to\infty}n\sum_{k=1}^\infty\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}$$
is related with $\zeta(3)$ (neccesarly thus agree with Furdui's identity) without using PROBLEMA 103? Thus you need take the limit, sum the series and compute the integral. Thanks in advance.
| Using partial fraction decomposition $$\frac{1}{(1+n^2x^2)(k+x)^3}$$ write $$\frac{n^2 \left(3 k^2 n^2-1\right)}{\left(k^2 n^2+1\right)^3 (k+x)}+\frac{2 k
n^2}{\left(k^2 n^2+1\right)^2 (k+x)^2}+\frac{1}{\left(k^2 n^2+1\right)
(k+x)^3}+\frac{\left(k^3 n^6-3 k n^4\right)+x \left(n^4-3 k^2 n^6\right)}{\left(k^2 n^2+1\right)^3
\left(n^2 x^2+1\right)}$$ which is not to bad (and easy to integrate).
So computing $$I_k=\int_0^1\frac{1}{(1+n^2x^2)(k+x)^3}\,dx$$ does not make any specific problem (the formulae are not reported here because of their length using MathJax).
Expanding as Taylor series for large values of $k$ leads to $$I_k=\frac{\tan ^{-1}(n)}{k^3 n}-\frac{3 \log \left(n^2+1\right)}{2 k^4 n^2}+\frac{6 \left(n-\tan ^{-1}(n)\right)}{k^5 n^3}
+O\left(\frac{1}{k^6}\right)$$ which makes $$n I_k=\frac{\tan ^{-1}(n)}{k^3 }-\frac{3 \log \left(n^2+1\right)}{2 k^4 n}+\frac{6 \left(n-\tan ^{-1}(n)\right)}{k^5 n^2}
+O\left(\frac{1}{k^6}\right)$$ $$\sum_{k=1}^\infty n I_k=\zeta (3) \tan ^{-1}(n)-\left( \frac{1}{60} \pi ^4 \log \left(n^2+1\right)+\frac{\pi ^6}{189}-6 \zeta (5)\right)\frac{1}{ n}+\cdots$$ and hence the limit and also how it is approached when $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
In a triangle $ABC$, $AB = a-b$ and $BC = 2\sqrt{ab}$, then find $\angle B$? Is this question solvable?
In $\Delta ABC$, $AB = a-b$ and $BC = 2\sqrt{ab}$, then $\angle B$ is
(a) $\: 60^{\circ}$
(b) $\: 30^{\circ}$
(c) $\: 90^{\circ}$
(d) $\: 45^{\circ}$
| HINT:
What do we know from $\triangle\text{ABC}$:
*
*$$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ=\pi\space\text{radians}$$
*For the length $\text{AB}$:
$$\text{AB}=\text{a}-\text{b}$$
*For the length $\text{BC}$:
$$\text{BC}=2\cdot\sqrt{\text{a}\cdot\text{b}}$$
Using the law of sinus and the law of cosines:
*
*$$\frac{\text{BC}}{\sin\left(\angle\text{A}\right)}=\frac{\text{AC}}{\sin\left(\angle\text{B}\right)}=\frac{\text{AB}}{\sin\left(\angle\text{C}\right)}$$
*$$
\begin{cases}
\text{BC}^2=\text{AC}^2+\text{AB}^2-2\cdot\text{AC}\cdot\text{AB}\cdot\cos\left(\angle\text{A}\right)\\
\text{AC}^2=\text{BC}^2+\text{AB}^2-2\cdot\text{BC}\cdot\text{AB}\cdot\cos\left(\angle\text{B}\right)\\
\text{AB}^2=\text{BC}^2+\text{AC}^2-2\cdot\text{BC}\cdot\text{AC}\cdot\cos\left(\angle\text{C}\right)
\end{cases}
$$
So, using your information (and $\left(2\cdot\sqrt{\text{a}\cdot\text{b}}\right)^2=4\cdot\text{a}\cdot\text{b}$):
*
*$$\frac{2\cdot\sqrt{\text{a}\cdot\text{b}}}{\sin\left(\angle\text{A}\right)}=\frac{\text{AC}}{\sin\left(\angle\text{B}\right)}=\frac{\text{a}-\text{b}}{\sin\left(\angle\text{C}\right)}$$
*$$
\begin{cases}
4\cdot\text{a}\cdot\text{b}=\text{AC}^2+\left(\text{a}-\text{b}\right)^2-2\cdot\text{AC}\cdot\left(\text{a}-\text{b}\right)\cdot\cos\left(\angle\text{A}\right)\\
\text{AC}^2=4\cdot\text{a}\cdot\text{b}+\left(\text{a}-\text{b}\right)^2-2\cdot\left(2\cdot\sqrt{\text{a}\cdot\text{b}}\right)\cdot\left(\text{a}-\text{b}\right)\cdot\cos\left(\angle\text{B}\right)\\
\left(\text{a}-\text{b}\right)^2=4\cdot\text{a}\cdot\text{b}+\text{AC}^2-2\cdot\left(2\cdot\sqrt{\text{a}\cdot\text{b}}\right)\cdot\text{AC}\cdot\cos\left(\angle\text{C}\right)
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Secondary school level mathematical induction
*
*It is given that
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$
Then, how to find the value of
$2^3+4^3+\cdots+30^3$?
Which direction should I aim at?
*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.
$$5^n-4^n=9m,\text{where $m$ is an integer.}$$
What I am thinking in the $n+1$ step is,
\begin{align}
& 5^{n+2}-4^{n+2} \\
= {} & 5^2(5^n-4^n)+5^24^n-4^{n+2} \\
= {} & 5^2(5^n-4^n)+4^n(5^2-4^2) \\
= {} & 5^29m+4^n9 \\
= {} & 9(5^2m+4^n)
\end{align}
Does this approach make sense?
*Show that $a+b$ is a factor of $a^n+b^n$ where $n$ is a positive odd number.
I am thinking this in the $n+1$ step.
$$a^{2n+1}+b^{2n+1}$$
But then I cannot get it further.
| 1) $2^3 + 4^3 + 6^3 + .... +30^3 = 2^3*1^3 + 2^3*2^3 + 2^3*3^3 + ... + 2^3*15^3 = 2^3(1^3 + 2^3 + .... + 25^3)$
2) Perfect. You did great. (Better than I did when I didn't realize it was only true for even numbers.)
3) If $n$ is odd then you don't want $a^{2n+1} + b^{2n+1}$ for the inductive step but either: $a^{n+2} + b^{n+2}$ (much like you did in 2) or $n = 2k+1;$ inductive step on $a^{2(k+1) + 1} + b^{2(k+1)+1}$. 2 1/2) or $n = 2k-1$ and $b^{2k+1} + b^{2k +1}$. Either way will work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$ Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$
First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything simpler. Also tried expressing $x^3$ as $\sqrt{x^6}$, but that doesn't really help either.
One would think that, as $x^6$ grows more quickly than $x^3$ by a factor of $x^3$, the contribution of the $x^3$ term to the term in the square root would be dwarfed by the contribution of the the $x^6$ term, so the overall behavior of the first term in the limit would "behave" like $x^3$, as x gets bigger and bigger, so I would think intuitively that the limit would evaluate to 0.
| Hmmm, no one has pointed out the obvious:
$\lim \sqrt{x^6 - 9x^3} - x^3 = \lim \sqrt{x^6 - 9x^3 + 36/4} - x^3$
$ = \lim \sqrt{(x^3 - 9/2)^2} - x^3 = \lim x^3 - 9/2 - x^3 = -9/2$
Intuition.... hmm .... I guess realizing the $x^3$ from $\sqrt {x^6 + stuff}$ was going to cancel the $-x^3$. So I want some $\lim \sqrt {Y_{x^3}^2} - x^3$ and figuring $Y_{x^3}^2$ must be.
=====
So why does $\lim \sqrt{x^6 -9x^3} = \lim\sqrt{x^6 -9x^3 + 36/4}$?
Let $\epsilon > 0$. We wish to solve for which $x$ we have $|\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}| = x^3 - \frac 92 - \sqrt{x^6-9x^3} < \epsilon$. If we can show that this can be solved for all $x > M$ for some $M$ we are done.
$(x^3 - \frac 92) - \epsilon < \sqrt{x^6 - 9x^3} $. For large enough $x$ we may assume this are both positive.
$x^6 - 9x^3 + 9 - 2\epsilon*(x^3 - \frac 92) + \epsilon^2 < x^6 - 9x^3$
$9 - 2\epsilon*(x^3 - \frac 92) + \epsilon^2 < 0$
$(x^3- \frac 92) > \frac{(9 + \epsilon^2)}{2\epsilon}$
So for any $x > \sqrt[3]{\frac 92 + \frac{(9 + \epsilon^2)}{2\epsilon}}$ this will be true.
So for all $\epsilon > 0$ if $x > M = \sqrt[3]{\frac 92 + \frac{(9 + \epsilon^2)}{2\epsilon}}$ we have $|\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}| < \epsilon$
So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}= 0$
So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4}= \lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3}$
So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3}-x^3 = \lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4} -x^3 = -\frac 92$
Or more generally...
If $c_x \rightarrow \infty$ and $f$ is continuous, then $\lim_{x\rightarrow \infty}f(c_x + h) = \lim_{x\rightarrow \infty}f(c_x(1 + h/c_x))=\lim_{x\rightarrow \infty}f(c_x(\lim_{x_\rightarrow \infty}(1 + h/c_x))=\lim_{x\rightarrow \infty}f(c_x*1)=\lim_{x\rightarrow \infty}f(c_x)$
Let $c_x = x^6 - 9x$, $h= 9=36/4$ $f(y) = \sqrt{y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 2
} |
Reduce a fraction Given the function
$f(x)=\frac{x^2-5}{x+\sqrt{5}}$
If I draw this function in maple, I will get a line. How can that be true? I should expect a line except in area of $x = -\sqrt{5}$, where $f(x) \rightarrow \infty$ or $f(x) \rightarrow -\infty $.
Of course Maple has factorized the numerator and reduced.
$\frac{x^2-5}{x+\sqrt{5}} = x-\sqrt{5},\ x\neq -\sqrt{5}$
My question is now, how can we ever reduce such a fraction with only condition $x\neq -\sqrt{5}$, when it is "$-\sqrt{5}$ and around it".
| Note that $x^2 - 5 = x^2 - \sqrt 5^2 = (x-\sqrt5)(x+\sqrt 5)$, so we have
$$
f(x) = \frac{x^2 - 5}{x + \sqrt 5} = \frac{(x-\sqrt 5)(x + \sqrt 5)}{x+\sqrt 5} = \frac{x-\sqrt 5}{1} = x-\sqrt 5
$$
However, this manipulation does not suddenly make $f$ defined when $x = -\sqrt 5$. The only thing it does is that it simplifies the process of studying how $f$ behaves very close to $x = -\sqrt 5$, namely, it behaves exactly like the function $g(x) = x-\sqrt 5$, which is defined on the whole of the real line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to factorise this expression $ x^2-y^2-x+y$ This part can be factorised as $x^2-y^2=(x+y)(x-y)$, How would the rest of the expression be factorised ?
:)
| Another method to factor $${ x }^{ 2 }-x+\frac { 1 }{ 4 } -{ y }^{ 2 }+y-\frac { 1 }{ 4 } ={ \left( x-\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( y-\frac { 1 }{ 2 } \right) }^{ 2 }=\left( x-\frac { 1 }{ 2 } +y-\frac { 1 }{ 2 } \right) \left( x-\frac { 1 }{ 2 } -y+\frac { 1 }{ 2 } \right) =\left( x-y \right) \left( x+y-1 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Set of values of $a$ for which function always increases If the set of all values of the parameter $a$ for which the function $$f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$$ increases for all $x \in R$ and has no critical point for all $x \in R$ is $(- \infty, m- \sqrt n) \cup (\sqrt n, \infty)$, then find the value of $m^2+n^2$ (where $m,n$ are prime numbers).
After finding $f'(x)$, we need to set $f'(x)>0$ but I am not having any clue how to extract condition on $a$? Could someone help me with this?
| Taking the derivative
$$f'(x) = 2\cos(2x) - 8(a+1)\cos x + (4a^2 + 8a -14)$$
Now both $\cos (2x)$ and $\cos x$ oscillate between $-1$ to $1$. Thus assuming $a \ge -1$,
$$-2 - 8(a+1) + 4a^2 + 8a - 14 \le f'(x) \le 2 + 8(a+1) + 4a^2 + 8a - 14$$
$$4a^2 - 24 \le f'(x) \le 4a^2 + 16a - 4$$
$$4(a^2 - 6) \le f'(x) \le 4(a^2 + 4a - 1)$$
Then $a > \sqrt{6}$ implies $0 < 4(a^2 - 6) \le f'(x)$.
On the other hand, if $a < 1$,
$$-2 + 8(a+1) + 4a^2 + 8a - 14 \le f'(x) \le 2 - 8(a+1) + 4a^2 + 8a - 14$$
$$4a^2 + 16a - 8 \le f'(x) \le 4a^2 - 20$$
$$4(a^2 + 4a - 2) \le f'(x) \le 4(a^2 - 5)$$
Then $a < -2 - \sqrt{6}$ implies $0 < 4(a^2 + 4a - 2) \le f'(x)$.
Therefore $a \in (-\infty,-2-\sqrt{6})\cup(\sqrt{6},\infty)$ makes the derivative positive for all $x$, which means the function is strictly increasing and has no critical points.
This seems to match what they are after except that $6$ isn't prime. Did I make an error somewhere?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Absolute value inequality with variable on both sides I am trying to solve the following inequality:
$$|3-5x| \le x$$
I am not familiar with inequalities including one absolute value with variables on both sides. I tried to solve it as follows:
$$-x \le 3-5x \le x$$
Then I solved for each side separately,as follows:
$$3-5x \le x$$
$$ x \ge (1/2)$$
$$-x \le 3-5x$$
$$x \le \frac34$$
I know my solution is incorrect and that it actually lies between $\frac12$ and $\frac34$, but I wanted to know what is wrong with my method and what is the appropriate approach to solving such inequalities.
Thanks,
| You need to break up the absolute value into its intervals:
$$
|x| = \begin{cases}
x & x > 0 \\
-x & x < 0 \\
0 & x = 0
\end{cases}
$$
Therefore for $|3-5x|$ you need to find the interval when it's less than zero and when it's greater than zero:
$$
|3-5x| = \begin{cases}
3 - 5x & 3 - 5x > 0 \rightarrow 3 > 5x \rightarrow x < \frac{3}{5}\\
5x - 3 & 3 - 5x < 0 \rightarrow 3 <5x \rightarrow x > \frac{3}{5} \\
0 & 3 - 5x = 0 \rightarrow 3 = 5x \rightarrow x = \frac{3}{5}
\end{cases}
$$
Now you solve the inequality in each case:
*
*$x < \frac{3}{5} \rightarrow |3 - 5x| = 3 - 5x$
$$
3 - 5x \leq x \\
3 \leq 6x \\
x \geq \frac{1}{2}
$$
*$x > \frac{3}{5} \rightarrow |3 - 5x| = 5x - 3$
$$
5x - 3 \leq x \\
4x \leq 3 \\
x \leq \frac{3}{4}
$$
*$x = \frac{3}{5} \rightarrow |3 - 5x| = 0$
$$
0 \leq x \\
x \geq 0
$$
Now you need to analyze each case:
*
*$x < \frac{3}{5} \wedge x \geq \frac{1}{2}$
It is true that $\frac{3}{5} = \frac{6}{10} \geq \frac{5}{10}$. Therefore this particular interval is true for $\frac{1}{2} \leq x < \frac{3}{5}$.
*$x > \frac{3}{5} \wedge x \leq \frac{3}{4}$
Since $\frac{3}{5} = \frac{12}{20}$ and $\frac{3}{4} = \frac{15}{20}$, this is true for $\frac{3}{5} < x \leq \frac{3}{4}$.
*$x = \frac{3}{5} \wedge x \geq 0$
$\frac{3}{5} > 0$ therefore this is trivially satisfied--thus $x = \frac{3}{5}$ is allowed.
When we combine these results, we find that $\frac{1}{2} \leq x \leq \frac{3}{4}$.
PS Edit:
It's probably easier to use:
$$
|x| = \begin{cases}
x & x \geq 0 \\
-x & x \leq 0 \\
\end{cases}
$$
In that case you get the two intervals:
$$
\frac{1}{2} \leq x \leq \frac{3}{5}
$$
and
$$
\frac{3}{5} \leq x \leq \frac{3}{4}
$$
Which clearly combines to give: $\frac{1}{2} \leq x \leq \frac{3}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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