Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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A hard integral $\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \mathrm{d}x$ This integtare is very hard for me，I do not known how to deal with it.
$$\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \mathrm{d}x$$
I use “mathematica” and obtain this
next the “maple” ：
the “maple” tell me it's a Elementary function
| Hint:
$$\dfrac{(x^4-1)\sqrt{x^4+1}}{x^8+1}=x\cdot\dfrac{\left(1-\dfrac1{x^4}\right)\sqrt{x^2+\dfrac1{x^2}}}{x^4+\dfrac1{x^4}}$$
Set $x^2=y$ to find
$$\int\dfrac{(x^4-1)\sqrt{x^4+1}}{x^8+1}dx=\dfrac12\int\dfrac{\left(1-\dfrac1{y^2}\right)\sqrt{y+\dfrac1y}}{\left(y+\dfrac1y\right)^2-2}dy$$
Now choose $\sqrt{y+\dfrac1y}=u\implies y+\dfrac1y=u^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$ The question is this:
If $a\ge b\ge c\ge 0$ and $a^2+b^2+c^2=3$, then prove that
$$abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$$
For my work on this inequality, I have proved already under constraints that it is true.
Proof for: $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0.$
$$
\sqrt{3}abc +
\sqrt{2}a -
\sqrt{3} -
\sqrt{2}c
\geqslant 0
$$
$$
a\left(
\sqrt{3}bc +
\sqrt{2}
\right)
+ (-1)\left(
\sqrt{3} +
\sqrt{2}c
\right) \geqslant 0
$$
$$
(1 + 1)(a\left(
\sqrt{3}bc +
\sqrt{2}
\right)
+ (-1)\left(
\sqrt{3} +
\sqrt{2}c
\right)) \geqslant 0
$$
By Chebyshev,
$$
(a - 1)
(\sqrt{3}bc +
\sqrt{2} +
\sqrt{3} +
\sqrt{2}c
)\geqslant0
$$
$$
a \geqslant 1
$$
Chebyshev Inequality requires the sequences to be monotonous. As $a+1>0$, we need to have the other sequence also in the same order, hence the condition: $\sqrt{3}bc + \sqrt{2} \geqslant\sqrt{3} + \sqrt{2}c$. The sequences are $(a,-1)$ and $(\sqrt{3}bc + \sqrt{2} ,\sqrt{3} + \sqrt{2}c)$.
I have tried another way but that was untrue. I have reached this far. The constraint $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0$ isn't true always. Try $(a,b,c) = (\sqrt{3},0,0)$.
Thanks for extensions or other solutions too are welcome!
| If $a\geq b\geq c\geq0$ then prove
$$3\sqrt3abc+\sqrt2\left(a-c\right)\left(a^2+b^2+c^2\right)\geq\left(a^2+b^2+c^2\right)^{\frac{3}2}.$$
Case 1: $c=0,$ it's obvious. Equality at $a=b\iff a=b=\sqrt{\frac{3}2}.$
Case 2: $c=1.$ If $a=1,$ then we are done. Equality at $a=b=c=1.$ If $a>1$ then consider on $[1,a]$ the function
$$f(b):=3\sqrt3ab+\sqrt2\left(a-1\right)\left(a^2+b^2+1\right)-\left(a^2+b^2+1\right)^{\frac{3}2}.$$
We have:
$$f'(b)=b\left(\frac{3\sqrt3a}b+2\sqrt2\left(a-1\right)-3\sqrt{a^2+b^2+1}\right)\implies$$
$f$ is pseudo-concave $\implies\min_{b\in[1,a]}{f(b)}\in\{f(1),f(a)\}.$ But
$$f(1)>0$$
and
$$f(a)>\sqrt3\left(2a^2+1\right)+\sqrt2\left(a-1\right)\left(2a^2+1\right)-\left(2a^2+1\right)^{\frac{3}2}>0.$$
We are done.
Edit: Let me give further details about $f(1)>0.$ We need to prove
$$3\sqrt3a+\sqrt2\left(a-1\right)\left(a^2+2\right)>\left(a^2+2\right)^{\frac{3}2}\iff$$
$$6\sqrt6a\left(a-1\right)\left(a^2+2\right)>a\left(a-1\right)^2\left(-a^3+2a^2+a+16\right)\iff$$
$$a^4-3a^3+3a^2+a^2\left(-2+6\sqrt6\right)-15a+16+12\sqrt6>0,$$
which is obviously true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Using a power series to approximate $\int_0^{.3} \frac{x^2}{1+x^7}dx$ to six decimal places Use a power series to approximate $\int_0^{.3} \frac{x^2}{1+x^7}dx$ to six decimal places
$\int_0^{.3} \frac{x^2}{1+x^7}dx$
Lets get this ready to substitute in a power series
$=\int_0^{.3} x^2 \frac{1}{1-(-x^7)}dx$
Lets substitute in a power series
$=\int_0^{.3} x^2 \sum_{n=0}^\infty (-1)^nx^{7n}dx$
Lets multiply in the $x^2$ term
$=\int_0^{.3} \sum_{n=0}^\infty (-1)^nx^{7n+2}dx$
Lets move the integral past the infinite sum
$=\sum_{n=0}^\infty \int_0^{.3} (-1)^nx^{7n+2}dx$
Now we integrate
$=\sum_{n=0}^\infty [(-1)^nx^{7n+3}\frac{1}{7n+3}]|_0^.3dx$
Now we evalute this, notice one of our points we evaluate at is $0$ so that whole term will be zero and thus we get:
$=\sum_{n=0}^\infty (-1)^n(.3)^{7n+3}\frac{1}{7n+3}$
This is an alternating series where $b_n = \frac{(.3)^{7n+3}}{7n+3}$
We can see that:
i) $b_{n+1} \leq b_n$
ii) $\lim_{n \rightarrow \infty} b_n = 0$
Thus we can use the alternating series estimation test.
Note that the second term of the sequences corresponds to when $n=1$
So we have:
$|R_1| \leq b_2 = \frac{(.3)^{14+3}}{14+3}=\frac{(.3)^{17}}{17} \cong 7.70(10^{-11})$
Wow... So the remainder after 2 terms is VERY small, and so we only need to add the first two terms together to get the series accurate up to 6 decimal places!! Note that adding up to $2$ terms means adding up when $n=0$ and $n=1$
$\sum_{n=0}^1 (-1)^n(.3)^{7n+3}\frac{1}{7n+3} = \frac{.3^3}{3} - \frac{3^{10}}{10} \cong .008999$
| Too long for comments.
This is a very interesting problem almost if you are coding; not knowing in advance how many terms have to be added for a given accuracy, at the level of each summation, you need an IF test and this is an expensive operation in terms of computer resources.
Mabing the problem mode general, consider that you want to compute
$$I=\int_0^t \frac {x^a}{1+x^b}\,dx \qquad \text{with}\qquad a\geq 0\qquad \text{and}\qquad b\geq 1$$ for an absolute error $\leq 10^{-k}$.
As you properly did, using the binomial expansion, we have
$$\frac {x^a}{1+x^b}=\sum_{n=0}^\infty (-1)^n x^{a+n b}$$ Writing the result as
$$I=\sum_{n=0}^p (-1)^n \frac{t^{a+b n+1}}{a+b n+1}+\sum_{n=p+1}^\infty (-1)^n \frac{t^{a+b n+1}}{a+b n+1}$$ since it is an alternatic series, we look for $p$ such that
$$R_p=\frac{t^{a+b (p+1)+1}}{a+b(p+1)+1} \leq 10^{-k}$$ There is an explicit solution which is
$$a+b (p+1)+1=-\frac{W\left(-10^k \log (t)\right)}{\log (t)}\implies p=\cdots$$ where $W(.)$ is Lambert function.
Applied to your problem $a=2$, $b=7$, $k=6$ and $t=\frac 3{10}$ , this gives
$$7p+10=\frac{W\left(10^6 \log \left(\frac{10}{3}\right)\right)}{\log
\left(\frac{10}{3}\right)}\approx 9.59664 \implies p=-0.058 \quad (!!)$$
So, as you properly showed, a single term should be sufficient. Effectively
$$R_1=\frac{\left(\frac{3}{10}\right)^{17}}{17}\approx 7.60 \times 10^{-11}\ll 10^{-6}$$
But, changing the problem to $k=60$ would give
$$7p+10=\frac{W\left(10^{60} \log \left(\frac{10}{3}\right)\right)}{\log
\left(\frac{10}{3}\right)}\approx 110.839 \implies p=14.4056 $$ So $\lceil p\rceil=15$. Ckecking
$$R_{14}=3.13\times 10^{-59} >10^{-60} \quad \text{and}\quad R_{15}=6.43\times 10^{-62}< 10^{-60}$$
In the linked page, you will find simple formulae for the approximation of $W(x)$ when $x$ is large.
| {
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"url": "https://math.stackexchange.com/questions/3878302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ if $x^2+y^2+z^2=1$ If $x^2+y^2+z^2=1$ what is the minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ for $x,y,z \gt 0$ ?
I would like to know if the minimum could be found using simpler ways.(like $AM \ge GM \ge HM$).
knowing $xy+yz+xz \geq \frac{-1}{2}$ for real $x,y,z$ might help.
| Comment
May be this idea can help you:
Let's start with a simpler form where in two dimension we have: $x^2+y^2=1$ corresponds to a circle $r=1$ and we want to find minimum of $\frac xy+\frac yx$. So we have to find minimum of:
$f(\alpha)=tan(\alpha)+ cotan(\alpha)$
taking derivative and equating to zero we get:
$\frac1{cos^2(\alpha)}-\frac1{sin^2(\alpha)}=0$
Which gives $\alpha=\frac{\pi}4$
That is $\frac xy+\frac yx$ is minimum when $x=y=\frac1{\sqrt2}$
Now we extend this idea to three dimension where $x^2+y^2+z^2=1$ corresponds to a sphere with radius 1 and we have to find minimum of $z\cdot tan(\alpha)+x\cdot tan(\beta)+ y\cdot tan(\gamma)$.Where $\alpha$. If we apply the idea we got from two dimension we have to have:
$x=y=z=\frac1{\sqrt3}$
$\alpha=\beta=\gamma=\frac{\pi}4$
Which gives:
$\frac{xy}z+\frac{yz}x+\frac{xz}y= 3\times\frac{(\frac1{\sqrt3})^2}{\frac 1{\sqrt3}}=\sqrt3$
I ckecked this with when $x=\frac1{\sqrt2}$, $y=\frac1{\sqrt3}$ and $z=\frac1{\sqrt6}$ which gives:
$1.98$ where $\sqrt3=1.73$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
finding integrating factor and solving ordinary differential equation Justify that a differential equation of the form: $$[y+xf(x^2+y^2)]dx+[yf(x^2+y^2)-x]dy=0$$ , where $f(x^2+y^2)$ is an arbitrary function of $(x^2+y^2)$, is not an exact differential equation and $1/(x^2+y^2)$ is an integrating factor for it. Hence, solve this differential equation for $$f(x^2+y^2)=(x^2+y^2)^2$$
I am unable to solve after making it an exact equation, having difficulty in the integration of this question.
| $$(y+xf(x^2+y^2))dx+(yf(x^2+y^2)-x)dy=0$$
$$f(x^2+y^2))(xdx+ydy)+ydx-xdy=0$$
$$\dfrac 12f(x^2+y^2))(d(x^2+y^2))+ydx-xdy=0$$
Divide by $x^2+y^2$:
$$\dfrac {f(x^2+y^2)d(x^2+y^2)}{2(x^2+y^2)}+\dfrac {ydx-xdy}{x^2+y^2}=0$$
Note that:
$$d \arctan (y/x)=\dfrac {xdy-ydx}{x^2+y^2}$$
$$\dfrac {f(x^2+y^2)d(x^2+y^2)}{2(x^2+y^2)}-d(\arctan (y/x)) =0$$
Integrate for the given function $f$.
| {
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"url": "https://math.stackexchange.com/questions/3886083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $(3x^2 \tan y- 2y^3/x^3 )dx + (4y^3 + x^3 \cos^2 y+ 3y^2/x^2 ) dy=0$ Somebody please help me to solve this Differential equation.
$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0 $
| Claim: This is an exact first order ordinary differential equation.
To verify this claim, we impose some standard notation, then perform a simple test. Let
\begin{align*}
M(x,y) &= 3x^2 \tan(y) - \frac{2y^3}{x^3} \text{,} \\
N(x,y) &= 4 y^3 + \frac{x^3}{\cos^2 y} + \frac{3y^2}{x^2}
\end{align*}
(Notice that, for a while, we will not treat $y$ as a dependent variable.)
We wish to find an $f(x,y)$ such that both
\begin{align*}
\frac{\partial}{\partial x} f(x,y) &= M(x,y) \\
\frac{\partial}{\partial y} f(x,y) &= N(x,y)
\end{align*}
If so, the equation is exact. We perform the test:
$$ \frac{\partial}{\partial y} M(x,y) \overset{?}= \frac{\partial}{\partial x} N(x,y) $$
(This test verifies $\frac{\partial^2}{\partial y \, \partial x} f(x,y) = \frac{\partial^2}{\partial x \, \partial y} f(x,y)$, which is true under mild assumptions about the various partial derivatives of $f$.)
So we calculate \begin{align*}
\frac{\partial}{\partial y} M(x,y) &= \frac{-6y^2}{x^3} + 3x^2 \sec^2(y) \text{,} \\
\frac{\partial}{\partial x} N(x,y) &= \frac{-6y^2}{x^3} + 3x^2 \sec^2(y) \text{.}
\end{align*}
These are equal, so the equation is exact.
Since the equation is exact, such an $f(x,y)$ exists and the general solution is
$$ f(x,y) + C \text{.} $$
We know two partial derivatives of $f$, so we can (partially) recover $f$ by integration.
\begin{align*}
f_1(x,y) &= \int M(x,y) \,\mathrm{d}x \\
&= \int 3x^2 \tan(y) - \frac{2y^3}{x^3} \,\mathrm{d}x \\
&= \frac{y^3}{x^2} + x^3 \tan(y) + C_1(y) \text{, and} \\
f_2(x,y) &= \int N(x,y) \,\mathrm{d}y \\
&= \int 4 y^3 + \frac{x^3}{\cos^2 y} + \frac{3y^2}{x^2} \,\mathrm{d}y \\
&= \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C_2(x) \text{.}
\end{align*}
Here, $C_1(y)$ is a function of integration, analogous to the constant of integration. Notice that if $f(x,y)$ has terms that depend only on $y$ or are constant, those terms are sent to zero in $\frac{\partial}{\partial x} f(x,y)$, so the general antiderivative must represent all possible $f$s differing only by a function of $y$ (and constants). Similarly, $C_2(x)$ is a function of integration.
Notice that we actually can use $f_2$ to find out about $C_1(y)$. $C_1(y) = y^4 + C$, where $C$ is a constant of integration. And reasoning the similarly, $C_2(x) = C$. (Some describe this process of merging the terms from the two solutions into a single solution that matches both partial derivatives. Perhaps a better way to describe this is that we are antidifferentiating both sides of the original equation, each with respect to the variable in the differential already present, and then matching the resulting functions of integration.) So we have
$$ f(x,y) = \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C \text{.} $$
Then $f(x,y) = C$ is the general solution, so,
$$ C = \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C $$
or, what is the same thing (because the range of "an arbitrary number minus another arbitrary number" is an arbitrary number),
$$ C = \frac{y^3}{x^2} + x^3 \tan(y) + y^4 $$
is the general solution to the given equation.
We can, of course, verify that this is the case using implicit differentiation. First, we stop pretending that $y$ is an independent variable. Next, we implicitly differentiate the claimed general solution
$$ 0 = \frac{x^2 (3y^2 y') - y^3 (2x)}{x^4} + 3x^2 \tan(y) + x^3 \sec^2(y) y' + 4y^3 y' \text{.} $$
Then, we solve for $y'$. \begin{align*}
\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y) &= \frac{x^2 (3y^2 y')}{x^4} + x^3 \sec^2(y) y' + 4y^3 y' \\
\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y) &= \left( \frac{x^2 (3y^2)}{x^4} + x^3 \sec^2(y) + 4y^3 \right) y' \\
\frac{\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y)}{\frac{x^2 (3y^2)}{x^4} + x^3 \sec^2(y) + 4y^3} &= y' \\
\frac{\frac{2 y^3 }{x^3} - 3x^2 \tan(y)}{\frac{3y^2}{x^2} + x^3 \sec^2(y) + 4y^3} &= \frac{\mathrm{d}y}{\mathrm{d}x} \\
\left(\frac{2 y^3 }{x^3} - 3x^2 \tan(y) \right)\mathrm{d}x &= \left(\frac{3y^2}{x^2} + x^3 \sec^2(y) + 4y^3 \right) \mathrm{d}y \text{,}
\end{align*}
which should be easy enough to see is the given equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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The Functional Equation $ ( x + y ) \big( f ( x ) - f ( y ) \big) = ( x - y ) f ( x + y ) $, need solution have answer Find all functions $ f: \mathbb R \to \mathbb R $ such that for all reals $ x $ and $ y $,
$$ ( x + y ) \big( f ( x ) - f ( y ) \big) = ( x - y ) f ( x + y ) \text . $$
I actually got the answer by guessing and checking, $ f ( x ) = a x ^ 2 + b x $, but I want to see the solution. My friend suggested surjectivity but I don't see how to continue.
$ f ( x ) = f \left( \frac x { f ( x ) ^ 2 } \right) $;
this is what I just got.
The answer is correct, just need a solution. Thanks!
| For any real numbers $ a $ and $ b $, if we define the function $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = f ( x ) - a x ^ 2 - b x $, then using the original functional equation, we'll have
$$ ( x + y ) \big( g ( x ) - g ( y ) \big) = ( x - y ) g ( x + y ) \text . \tag 0 \label 0 $$
In particular, setting $ a = \frac { f ( 1 ) + f ( - 1 ) } 2 $ and $ b = \frac { f ( 1 ) - f ( - 1 ) } 2 $, we will also have $ g ( 1 ) = g ( - 1 ) = 0 $. Letting $ y = 1 $ in \eqref{0}, we get
$$ ( x + 1 ) g ( x ) = ( x - 1 ) g ( x + 1 ) \text , \tag 1 \label 1 $$
while substituting $ x + 1 $ for $ x $ and $ - 1 $ for $ y $ in \eqref{0} we have
$$ x g ( x + 1 ) = ( x + 2 ) g ( x ) \text . \tag 2 \label 2 $$
\eqref{1} and \eqref{2} together show that
$$ 2 g ( x ) = x ( x + 1 ) g ( x ) - ( x - 1 ) ( x + 2 ) g ( x ) = x ( x - 1 ) g ( x + 1 ) - ( x - 1 ) x g ( x + 1 ) = 0 \text . $$
Therefore $ g $ is the constant zero function, and hence $ f ( x ) = a x ^ 2 + b x $ for all $ x \in \mathbb R $. It's straightforward to see that any function of this form satisfies the original functional equation, and thus they form the class of all the solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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$(a^4+b^2) \leq (a^2+b^2)^2$ for all $a, b \in \mathbb{R}_+$? Let $a, b \in \mathbb{R}_+$.
Question. Is valid that
$$(a^4+b^2) \leq (a^2+b^2)^2?\tag{1}$$
I tried what follows:
$$(a^2+b^2)^2=a^4+2a^2b^2+b^4 \geq a^4+2a^2b^2.$$
For this, can I to conclude that $(1)$ holds?
| $$\bigg( \left(\frac{1}{2} \right)^2+ \left(\frac{1}{2} \right)^2 \bigg)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}$$
but
$$\bigg( \left(\frac{1}{2}\right)^4+ \left(\frac{1}{2}\right)\bigg)^2= \left(\frac{1}{2}\right)^4+\frac{1}{4} > \frac{1}{4}= \bigg( \left(\frac{1}{2} \right)^2+ \left(\frac{1}{2}\right)^2 \bigg)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886957",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$ Prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$, where a, b, c >0.
I tried,
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} = \frac{a}{\sqrt{a}\sqrt{a+b}}+\frac{b}{\sqrt{b}\sqrt{b+c}}+\frac{c}{\sqrt{c}\sqrt{c+a}} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{a}\sqrt{a+b}+\sqrt{b}\sqrt{b+c}+\sqrt{c}\sqrt{c+a}}$.
Then $\sqrt{a}\sqrt{a+b} \le \sqrt{2}/2 (2a+(a+b))$.
This only gave me $\sqrt{2}/2 + \sqrt{2}(\frac{ab+bc+ca}{a+b+c})$.
| Hint: $\dfrac{a}{a+b} > \dfrac{a}{a+b+c}$
| {
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"question_score": "1",
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Completely factoring $a^3 + b^3 + c^3 - 3abc$? How do I completely factor $a^3 + b^3 + c^3 - 3abc$ over the complex numbers? The first thing I did is factor this into $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$. But I am not sure how to factor $a^2 + b^2 + c^2 - ab - bc - ac$ over the complex numbers. Can anyone help me?
| $a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$, where $\omega$ is the root of $x^2+x+1=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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} |
Proving $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$ This question is the last part of a problem leading to proof of Stirling's approximation. I've already proved that $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists and that $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$.
Hence, the question asks to assume $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists, and then asks to use $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$ to show $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$.
My attempt goes like this:
Since $\sqrt{x}$ is continuous, we can use $\sqrt{\underset{n\to \infty }{\text{lim}}f(n)}=\underset{n\to \
\infty }{\text{lim}}\sqrt{f(n)}$ to get $\underset{n\to \infty }{\text{lim}}\frac{2^{2 n} (n!)^2}{(2 n)! \
\sqrt{2 n+1}}=\sqrt{\frac{\pi }{2}}$.
Then we can eliminate $2^{2 n}n!$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{\sqrt{1+2 n} \left(n-\frac{1}{2}\right) \left(n-\frac{3}{2}\right) \cdots \
\frac{3}{2}\frac{1}{2}}$$
Then factor out $n^n$ and adjust $\sqrt{1+2n}$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{n^{n+\frac{1}{2}} \sqrt{2+\frac{1}{n}} \left(1-\frac{1}{2n}\right) \left(1-\frac{3}{2n}\right) \cdots \
\frac{3}{2n}\frac{1}{2n}}$$
The $\sqrt{2+\frac{1}{n}}$ factor will give a $\sqrt{2}$, so the remaining $\prod _k^n \left(1-\frac{2 k-1}{2 n}\right)$ must somehow relate to $\sqrt{2} e^{-n}$.
However, I'm not sure how to do this.
| Elaborating on Daniel Schepler's comment: if $L = \lim_{n \to \infty} \frac{n!}{n^{n+1/2} e^{-n}}$ then
$$\sqrt{\pi/2}
= \lim_{n\to \infty} \frac{2^{2n} (n!)^2}{(2n)!\sqrt{2n+1}}
= \lim_{n \to \infty} \frac{2^{2n}n^{2n+1} e^{-2n} L^2}{(2n)^{2n+1/2} e^{-2n} L \sqrt{2n+1}}
= L\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{2}\sqrt{2n+1}}
= \frac{L}{2}.$$
For what it's worth, I tried to continue your approach by approximating $\log \left(\prod_{k=1}^n (1 - \frac{2k-1}{2n})\right)$ with the integral $n \int_0^{1-1/(2n)} \log(1-x) \, dx$ but it was terribly messy and didn't seem tight enough to get the exact equivalence with $\sqrt{2} e^{-n}$. But I think the above approach is the intended approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Derive the formula for the Arc Length in polar coordinates I want to prove the formula:
$\frac{ds}{d\theta } =\sqrt{\rho^{2} +\left( \acute{\rho } \right)^{2} } $
One solution is to start with rectangular coordinates
$x=\rho .\cos \left( \theta \right) \ \ \ and\ \ \ \ y=\rho .\sin \left( \theta \right) \ \ \ ,\ \rho =f\left( \theta \right)$
$\frac{dx}{d\theta } =\acute{\rho } .\cos \left( \theta \right) -\rho .\ \sin \left( \theta \right) $
$\frac{dy}{d\theta } =\acute{\rho } .\sin \left( \theta \right) +\rho .\ \cos \left( \theta \right) $
$\frac{ds}{d\theta } =\sqrt{\left( \frac{dx}{d\theta } \right)^{2} +\left( \frac{dy}{d\theta } \right)^{2} } =\sqrt{\rho^{2} +\left( \acute{\rho } \right)^{2} } \ $
Now I want to reach the same result using only polar variables.
With reference to attached figure, the arc length could be approximated by a sum of very small line segments
$\left( \Delta s\right)^{2} =\rho^{2} +\left( \rho +\Delta \rho \right)^{2} -2\cdot \rho \cdot \left( \rho +\Delta \rho \right) \cdot \cos \left( \Delta \theta \right)$
$\left( \frac{\Delta s}{\Delta \theta } \right)^{2} =\left( \frac{\rho +\Delta \rho }{\Delta \theta } \right)^{2} +\left( \frac{\rho }{\Delta \theta } \right)^{2} -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right) \cdot \cos \left( \Delta \theta \right) }{\left( \Delta \theta \right)^{2} } $
$\left( \frac{\Delta s}{\Delta \theta } \right)^{2} =\left[ \frac{\rho +\Delta \rho }{\Delta \theta } -\frac{\rho }{\Delta \theta } \right]^{2} +\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right) }{\left( \Delta \theta \right)^{2} } -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right) \cdot \cos \left( \Delta \theta \right) }{\left( \Delta \theta \right)^{2} }$
$\left( \frac{\Delta s}{\Delta \theta } \right)^{2} =\left( \frac{\Delta \rho }{\Delta \theta } \right)^{2} +\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right) }{\left( \Delta \theta \right)^{2} } -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right) \cdot \cos \left( \Delta \theta \right) }{\left( \Delta \theta \right)^{2} }$
Now $\ \ \ \ \ \left( \lim_{\Delta \theta \rightarrow 0} \ \frac{\Delta \rho }{\Delta \theta } \right)^{2} =\left( \frac{d\rho }{d\theta } \right)^{2} =\left( \acute{\rho } \right)^{2} $
I couldn’t figure out how the remaining expression could be equal to $\rho^{2} $
Could this or a different approach lead to the same formula?
| Keep the $\Delta$ terms up to second order only,
\begin{align}
\cos (\Delta \theta) & \approx 1-\frac{(\Delta \theta)^2}{2} \\
(\Delta s)^2
& \approx
\rho^{2}+(\rho +\Delta \rho)^{2}-2\rho (\rho +\Delta \rho) \cos ( \Delta \theta ) \\
& \approx
[\rho^{2}-2\rho (\rho +\Delta \rho)+(\rho +\Delta \rho)^{2}]+
\rho (\rho +\Delta \rho) (\Delta \theta)^2 \\
& \approx
[\rho-(\rho+\Delta \rho)]^2+\rho^2 (\Delta \theta)^2+
\require{cancel} \cancel{\rho^2 (\Delta \rho)(\Delta \theta)^2} \\
& \approx (\Delta \rho)^2+\rho^2 (\Delta \theta)^2 \\
\end{align}
Take $\Delta \theta \to 0$, the metric becomes
$$ds^2=d\rho^2+\rho^2 d\theta^2$$
that is
$$
\left(
\frac{ds}{d\theta}
\right)^2=\rho^2+
\left(
\frac{d\rho}{d\theta}
\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$ for $c = \min(a, b, c)$ and $ab + bc + ca = 2$
Problem 1: Let $a, b, c \ge 0$ with $c = \min(a, b, c)$ and $ab + bc + ca = 2$. Prove that
$\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$.
Background: This problem was proposed by csav10@AoPS
(https://artofproblemsolving.com/community/q1h2321937p18543869)
which is a variant of the following:
Problem 2: Let $a, b, c \ge 0$ with $ab + bc + ca = 2$. Prove that $\sqrt{a+ab} + \sqrt{b+bc} + \sqrt{c+ca} \ge 3$.
(Proposed by KaiRain@AoPS. See https://artofproblemsolving.com/community/c6h1905412.
Also see: https://artofproblemsolving.com/community/q1h2308247p18537783.)
For Problem 2, yleo@AoPS and csav10@AoPS gave nice solutions, respectively.
Also, I solved Problem 2 by the Buffalo Way (BW) which is a computer solution.
For Problem 1, I have not yet solved it by the Buffalo Way (BW).
By the way, Equality case: $(a, b, c) = (2, 1, 0)$.
Any comments and solutions are welcome and appreciated.
Add the Buffalo Way solution (outline) for Problem 2
Fact 1: If $b \ge a \ge c \ge 0$, then
$$\sqrt{a + ab} + \sqrt{b + bc} + \sqrt{c + ca} \ge \sqrt{b + ba} + \sqrt{a + ac} + \sqrt{c + cb}.$$
From Fact 1, we assume that $a \ge b \ge c$.
After homogenization, we need to prove that
$$\sqrt{\frac{a \sqrt{\frac{ab+bc+ca}{2}} + ab}{\frac{ab+bc+ca}{2}}} +
\sqrt{\frac{b \sqrt{\frac{ab+bc+ca}{2}} + bc}{\frac{ab+bc+ca}{2}}} +
\sqrt{\frac{c \sqrt{\frac{ab+bc+ca}{2}} + ca}{\frac{ab+bc+ca}{2}}} \ge 3.$$
Denote
$$X = \frac{1}{4}\cdot\frac{a \sqrt{\frac{ab+bc+ca}{2}} + ab}{\frac{ab+bc+ca}{2}},
Y = \frac{b \sqrt{\frac{ab+bc+ca}{2}} + bc}{\frac{ab+bc+ca}{2}},
Z = \frac{c \sqrt{\frac{ab+bc+ca}{2}} + ca}{\frac{ab+bc+ca}{2}}.$$
We need to prove that $2\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge 3$.
By using $\sqrt{u} \ge \frac{2u}{1+u}$ for $u\ge 0$, it suffices to prove that
$$\frac{4X}{1+X} + \frac{2Y}{1+Y} + \frac{2Z}{1+Z} \ge 3.$$
After some manipulation, it suffices to prove $A\sqrt{\frac{ab+bc+ca}{2}} + B \ge 0$ where $A, B$ are some polynomials and $A \ge 0$.
It suffices to prove that $A^2\cdot \frac{ab+bc+ca}{2} \ge B^2$.
BW works.
| Partial answer
Let $f = \sqrt{a + ab} + \sqrt{b} + \sqrt{c} - 3$ and $g = ab + bc + ca -2$, so we are to show $f\ge 0$ under the condition $g =0$.
Start by noting that if $a=b=c=c_0$, we have $c_0 = \sqrt{2/3} \simeq 0.8165$ and $f = 0.0251$ so the condition is very tight.
Nevertheless, in this partial answer the inequality can be proved for small $c$, and for $c$ in the (larger) vicinity of $c_0$. Indications are given to extend these calculations.
Treat $c$ as a parameter. First, note that for $a\ge b$, we observe for the exchange of $a\leftrightarrow b$ that $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} - 3 \le \sqrt{b + ab} + \sqrt{a} + \sqrt{c} - 3$, hence we need to consider the case $a\ge b$ only.
Since $c$ is required to be the smallest variable, we consider $a\ge b \ge c$.
Case 1: small $c$.
We show that the equality condition $(a,b,c) = (2,1,0)$ is actually a minimum of $f$. A convenient way to do so is variational calculus, where $c$ is arbitrary but fixed. This gives
$
0 = dg = (b+c)da + (a+c)db$
and
$ 2 \sqrt{a + ab}\sqrt{b} \; df = \sqrt{b} (1+b) \; da + (a \sqrt b + \sqrt{a + ab}) \; db$
For the RHS to become zero we insert the $dg=0$-equation and get
$ d \tilde f = [\sqrt{b} (1+b) (a+c) - (a \sqrt b + \sqrt{a + ab}) (b+c)] \, da $.
For $c=0$, $ d \tilde f = 0$ gives
$ \sqrt{a}= \sqrt{1 + b} \sqrt{b}$. Now inserting, from $g=0$, the condition $a = 2/b$, gives $2 = b^2 + b^3$ which has the only positive solution $b=1$, from which follows again $a = 2$. This establishes that $(a,b,c) = (2,1,0)$ is actually a minimum of $f$.
Now in the vicinity of that minimum, i.e. for small $c > 0$, the $f$-equation itself guarantees the inequality. The reason is that $ d \tilde f = 0$ tells us that the value of $f$ will shift away in linear dependence with $c$, as $a$ and $b$ vary to obey $ d \tilde f = 0$. In contrast, in $f$ the additional term $\sqrt c$ increases $f$, where this term has an infinite growth rate at $c=0$ (illustrated also in the second figure below). Hence for small $c > 0$, $f$ will always move in positive direction with $c$.
This behavior can be utilized with the following sketch: use $d \tilde f = 0$ and $g=0$ to calculate the (small) change of $a$ and $b$ once $c$ moves away from $0$. With this change, calculate the new $f$ where the first terms change linearly and the last term goes with $\sqrt c$. Then, equate the "worst" change in the first terms with $\sqrt c$ to get an estimate for the highest $c$ where $f \ge 0$ holds.
Case 2: large $c$.
The largest $c$ that can be attained is $c_0 = \sqrt{2/3}$, as $c$ must be the smallest variable. In this case, $a=b=c=c_0$, $f = 0.0251$, and variations of $(a,b,c)$ under the condition $g=0$ change $f$ linearly with $c$. Hence, in some environment $f$ is guaranteed to stay positive.
Indeed, here we can estimate $f$. Method 1: As $a$ is the largest of the variables, the smallest $a$, for given $c$, that can be attained, is observed when $a=b$. Under $g=0$, this gives $a = \sqrt{2 + c^2} - c$. The smallest $b$ is $b=c$. For $a=b=c=c_0$ both estimates are tight, so for some range of $c < c_0$, we can ensure $f\ge 0 $ to hold. We have
$f \ge \hat f = \sqrt{(\sqrt{2 + c^2} - c)} \, \sqrt{1 + c} + 2\sqrt{c} - 3$ and this holds until $\hat f = 0$ which is attained for $c_{min} \simeq 0.793$.
Method 2: Inserting $a$ from $g =0$ gives $f = \sqrt{2-b c} \, \frac{\sqrt{1 + b}}{\sqrt{c + b}} + \sqrt{b} + \sqrt{c} - 3$. Now $\frac{\sqrt{1 + b}}{\sqrt{c + b}} \ge \frac{\sqrt{1 + b_{max}}}{\sqrt{c_{max} + b_{max}}}$ and, by AM-GM, $\sqrt{b} + \sqrt{c} \ge 2 \sqrt[4]{b c}$. Letting $bc = x$ this gives $$f \ge \hat f = \sqrt{2-x} \, \frac{\sqrt{1 + b_{max}}}{\sqrt{c_{max} + b_{max}}} + 2 \sqrt[4]{x} - 3$$ which is a function of $x$ only, with $b_{max} = \max\{\sqrt{2 + c^2} - c\} = \sqrt{2}$ and $c_{max} = c_0$. Again, all estimates are tight for $a=b=c=c_0$, which entails $x = \frac23$ and $f = 0.0251$. So $\hat f = 0$ can be used to calculate the smallest $x$ where the estimate holds, which is given for $\hat x \simeq 0.632$. Since $c = \hat x/b$, this makes it feasible to reduce $c$ while increasing $b$, however this is only possible until $b$ reaches its highest value. For given $c$, this value is $\sqrt{2 + c^2} - c$ which gives $c(\sqrt{2 + c^2} - c) = 0.632$ or $c_{min} = 0.737$, which is better than the bound with method 1.
Again, this can be improved, if better values for $b_{max}$ and $c_{max}$ are found. For example, with the result just obtained, one can iterate the method and reduce $c$ once again, where now it is established that $c_{max} = 0.737$ can be used. Also, the estimate $b_{max} = \max\{\sqrt{2 + c^2} - c\}$ needs the smallest value of $c$ which goes into consideration, where $c=0$ was the most conservative one. Here, once, from the extension of case 1, a highest value of $c$ is known for which the inequality holds, this value can be used.
Case 3: "small $c$" $ < c < 0.737$.
To be done. Indications are given above.
Here is a figure which shows what to expect, and consequently what to do:
In green is the location of the minimum, i.e. the solution of $df = 0$ under $g=0$. In red is the range for $b$, i.e. $c \le b \le \sqrt{2 + c^2} \, - c$. For $c=0$, the minimum is to be found at $b_{min}=1$, as was calculated above. As $c$ increases, $b_{min}$ gets smaller. For $c \simeq 0.345$, the location of $b_{min}$ reaches the lower limit of the range for $b$. Hence for the remaining $ 0.345 < c < \sqrt{2/3}$, the smallest $f$ is to be found at $b_{min} = c$. This makes it easy to show the inequality in this interval of $c$: we have
$f \ge f_{min}^* = \sqrt{2- c^2} \, \frac{\sqrt{1 + c}}{\sqrt{2 c}} + 2\sqrt{c} - 3$ and this is clearly positive, see the following plot (created with values from numeric evaluations). In that plot, $f_{min}$ is plotted over the first $c$-interval $\mbox{[0 $\,$ 0.345]}$ where the value of $b$ is taken as the free minimum, from the solution of $df = 0$ under $g=0$ (blue section). The green circle is there to mark the transition to the second $c$-interval [$0.345 \,$ $\sqrt{2/3}$] where $f_{min}^*$ is plotted (see above) which is taken at $b = c$ (orange section). Note the rise of $f_{min}$ with infinite rate at $c=0$ acording to the $\sqrt{c}$-term in $f$, as discussed above under case 1: small $c$.
Hence the agenda for analytic work is:
*
*establish for $c < 0.345$ that $f$ has a free minimum w.r.t. $b$, and that at that minimum, $f_{min}(c) > f_{min}(c=0) =0$.
*establish for $c > 0.345$ that $f(b,c) \ge f(c,c)$, and then $f(c,c) > 0$ can be shown easily (see above), which gives the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Sum of a non arithmetic nor geometric sequence Express the following sum as a common fraction:
$\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{97\cdot100}$
This is from a timed competition, so fastest answers are best.
I tried rearranging it to
$$\frac{7\cdot10\cdot13\cdot...\cdot100+1\cdot10\cdot13\cdot...\cdot100+...+1\cdot4\cdot7\cdot...\cdot 94}{1\cdot4\cdot7\cdot10\cdot13\cdot...\cdot100}$$but I don't know what to do after this. Any help?
| $$\frac{1}{n(n+3)}=\frac{1}{3}(\frac{1}{n}-\frac{1}{n+3})$$
$$\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{97\cdot100}=\frac{1}{3}(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100})=\frac{1}{3}(1-\frac{1}{100})=\frac{33}{100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $B=\sqrt[3]{x\sqrt[5]{x^4\sqrt[9]{x^{24}\sqrt[17]{x^{240}...}}}}$ What value does $\dfrac{B}{x}$ approach?
Source: Select Problems of "Lumbreras Editors"
If I divide I get that:
$$\dfrac{B}{x}=\sqrt[3]{\dfrac{x}{x}.\dfrac{1}{x^2}\sqrt[5]{x^4\sqrt[9]{x^{24}\sqrt[17]{x^{240}...}}}}$$
$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{\dfrac{x^4}{x^4} \dfrac{1}{x^6}\sqrt[9]{x^{24}\sqrt[17]{x^{240}...}}}}$$
$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{ 1\sqrt[9]{\dfrac{x^{24}}{x^{24}}\dfrac{1}{x^{30}}\sqrt[17]{x^{240}...}}}}$$
$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{ 1\sqrt[9]{1\sqrt[17]{\dfrac{x^{240}}{x^{240}}\dfrac{1}{x^{270}}...}}}}$$
Can I assume that $\dfrac{B}{x}$ is approximately 1?
| $B = x^\alpha$ where
$$
\begin{align}
\alpha &=
{\small \frac13}(1 +
{\small \frac15}(4 +
{\small \frac19}(24 +
{\small \frac{1}{17}}(240+\cdots ))))\\
&= \frac13 + \frac{4}{3\cdot 5} + \frac{24}{3\cdot 5 \cdot 9} + \frac{240}{3\cdot 5 \cdot 9\cdot 17} + \cdots\\
&= \frac{2}{2\cdot 3} + \frac{4}{3\cdot 5} + \frac{8}{5\cdot 9} + \frac{16}{9\cdot 17} + \cdots\\
&= \sum_{k=0}^\infty\frac{2^{k+1}}{(2^k+1)(2^{k+1}+1)}\\
&= \sum_{k=0}^\infty \left(\frac{2}{2^k+1} - \frac{2}{2^{k+1}+1}\right)\\
&= \frac{2}{2^0+1}\\ &= 1
\end{align}
$$
This means $\frac{B}{x} = \frac{x}{x} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3906425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $G$ is a finitely generated group under multiplication.
Consider the set of matrices $$G = \left \{\begin{pmatrix} s & b \\ 0 & 1 \end{pmatrix}\ \bigg |\ b \in \Bbb Z,\ s \in \left \{-1,+1 \right \} \right \}.$$ Then show that $G$ is a finitely generated group under multiplication.
It is easy to show that $G$ is a group under multiplication. What I think is that $$G = \left \langle \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix} \right \rangle.$$
The reason behind my claim is that the elements of $G$ are of the form $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ or $\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix},$ for some $n \in \Bbb Z.$ Now for any $n \in \Bbb Z$ we have $$\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} = {\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}}^n.$$ Also if $n$ is an odd integer then $$\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix} = {\begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}}^n;$$ and if $n$ is an even integer then $$\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix} = {\begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}}^{n+1} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} ;$$ which proves the claim.
Is my above reasoning correct at all? Can anybody please verify it?
Thanks in advance.
| Well, you need to consider what happens when the generators multiply:
$\begin{pmatrix} -1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}$ and
$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} -1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 0 & 1\end{pmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving the given system of equations
$$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$
I got $x^3 + y^3 = 6xy = (x+y)^2$ which I simplified to $x^2 - xy + y^2 = x + y$ and I'm not sure what to do now. Please help.
| Hint:
Let $x+y=a, xy=b$
$$\implies 6b=a^3-3ab\ \ \ \ (1)\text{ and }4b=a^2-2b\implies a^2=6b\ \ \ \ (2)$$
Clearly, $ab\ne0$
Replacing the value of $b$ in $(1)$,
$$a^2=a^3-3\cdot a\cdot\dfrac{a^2}6\iff2a^2=a^3\implies a=?\text{ as }a\ne0$$
So, we can find $b=\dfrac{a^2}6$
So, $x,y$ are the roots of $$t^2-at+b=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Is this a correct way to show convergence for $\sum_{k=1}^{\infty} x^k (arctan(\frac{1}{k}))^2$? Is this a correct way to show convergence for $\sum_{k=1}^{\infty} x^k (arctan(\frac{1}{k}))^2$?
Ratio test with $lim_{k->\infty} |\frac{a_{k+1}}{a_k}|: \sum_{n=0}^\infty (\frac {\frac{(\frac{1}{k+1})^{2n+1}}{2n+1}}{\frac{(\frac{1}{k})^{2n+1}}{2n+1}})^2=...=\sum_{n=0}^{\infty}(1-\frac{1}{k+1})^{4n+2}=\sum_{n=0}^{\infty}((1-\frac{1}{k+1})^n)^4(1-\frac{1}{k+1})^2$
According to the binomial theorem we have: $\lim_{n->\infty} \lim_{k->\infty}(1-\frac{1}{k+1})^n=\lim_{n->\infty} \lim_{k->\infty} 1-\frac{1}{k+1}+(-\frac{1}{k+1})^2)+...+(-\frac{1}{k+1})^n=\\lim_{n->\infty} 1+0+0^2+...+0^n=1$
So $R=1$ which means that it converges for $|x|<1$.
For $x=1$ we have exactly $\sum_{k=1}^{\infty} (arctan(\frac{1}{k}))^2$, which converges $\implies$ convergence for $x=-1$ also. So convergence $\forall x: |x|\leq 1$.
| Set up your limit, then apply L'Hopital's rule.
\begin{align*}
\lim_{k \rightarrow \infty} \left| \frac{a_{k+1}}{a_k} \right|
&= \lim_{k \rightarrow \infty} \left| \frac{x^{k+1} \arctan^2 \left( \frac{1}{k+1} \right)}{x^{k} \arctan^2 \left( \frac{1}{k} \right)} \right| \\
&= \lim_{k \rightarrow \infty} \left|x \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)} \right| \\
&= \lim_{k \rightarrow \infty} |x| \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)} & & \text{squares are automatically nonnegative} \\
&= |x| \lim_{k \rightarrow \infty} \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)} \\
&\overset{L'H}= |x| \lim_{k \rightarrow \infty} \frac{ \frac{1}{1 + \left( \frac{1}{k+1} \right)^2 }}{ \frac{1}{1 + \left( \frac{1}{k} \right)^2 }} \\
&= |x| \cdot 1 \\
&= |x| \text{.}
\end{align*}
Now the ratio test gives you a condition on the value of this limit to ensure convergence: $|x| < 1$.
Now to the endpoints. Suppose $x = 1$. Then we consider $\sum_{k=1}^\infty \arctan^2 \frac{1}{k}$. One way to proceed is to note $\arctan x < x$ for $x > 0$. So
$$\sum_{k=1}^\infty \arctan^2 \frac{1}{k} < \sum_{k=1}^\infty \frac{1}{k^2} \text{,} $$
which is a convergent series.
Now suppose $x = -1$. From the previous, we know the sum we obtain is absolutely convergent, hence convergent. Therefore, the sum converges for $x \in [-1,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Combinatorics problem - choosing $6$ cards out of $32$-card deck so that there are exactly three different suits (Inclusion-Exclusion) Suppose we have a deck of $32$ cards with $8$ cards of each of the four suits. In how ways we can choose six cards such that there are cards of exactly three different suits among the chosen cards?
I believe inclusion-exclusion principle is the way to solve it, where we first count the number of total ways to choose $6$ cards out of $32$ (which is $\binom{32}{6}$), then exclude the number of combinations where exactly two of the suits are missing (which is $\binom{4}{2}\binom{16}{6}$) and then by inclusion-exclusion formula add the combinations where all three suits are missing (which is $\binom{4}{3}\binom{8}{6}$). The number of combinations of all $4$ suits missing is, of course, zero.
My question is - where is my logic incorrect? I know it is, but can't seem to spot the error.
| There are four qualities that hands may have. There are hands that miss clubs, hands that have no hearts, hands without spades and hands without diamonds.
The situation is tricky because the binomial coefficients intervene in a double way. To reveal this let us use $N(x)$ and $E(x)$ the generating functions for the "hands having at least x qualities" and "hands having exactly x qualities".
Shortly, P.I.E says that $N(x)= E(x+1)$
Here
$N(x) = \binom 4 3 \binom 8 6 x^3 + \binom 4 2 \binom {16} 6 x^2 + \binom 4 1 \binom {24} 6x + \binom 4 0 \binom {32} 6$
$N(x) =112x^3 + 48048x^2 + 538384x + 906192$
$E(x) = N(x-1) = 12x^3 + 47712x^2 + 442624x + 415744$
As one may see, $442624$, the desired number, is
$442664 = \color {red}{\binom 3 1 }\binom 4 3 \binom 8 6 - \color{red}{\binom 2 1 } \binom 4 2 \binom {16} 6 + \color{red}{\binom 1 1 } \binom 4 1 \binom {24} 6$
Suppose now we have a 3-color deck of 24 cards and we are interested in all-color hands. Again,
$N(x) = \binom 3 2 \binom 8 6 x^2 + \binom 3 1 \binom {16} 6 x + \binom 3 0 \binom {24} 6 $
$N(x) =84x^2 + 24024x + 134596$
$E(x) = N(x-1) = 84x^2 + 23846x + 110656$
$110656 = \color {red}{\binom 2 0 }\binom 3 2 \binom 8 6 - \color{red}{\binom 1 0 } \binom 3 1 \binom {16} 6 + \color{red}{\binom 0 0 } \binom 3 0 \binom {24} 6$
Again, as we may see, $442624 = 4\times 110656$
=====
:) And do not be afraid of "generating functions"; they are neither generating nor functions but just mnemonical expressions that fit and help to shorten the writing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
inequality $\frac{2(x + y)^2}{2x^2 + y^2} \leq 3$ I'm looking for a "nice" way to show that the following inequality holds, i.e. without differentiating and determining the maximum:
$$
\frac{2(x + y)^2}{2x^2+ y^2} \leq 3 \quad \forall x,y \in \mathbb{R}
$$
It's rather easy to show
$$
\frac{4xy}{x^2 + y^2} \le 2
$$
and obviously
$$
\frac{2x^2 + 2y^2}{2x^2+y^2} \le 2
$$
but combining these two does not give me a sufficient low bound.
| Let $y=l.x$; we have:
$\frac{2[(l+1)x]^2}{x^2(l^2+2)}=\frac {2l^2+4l+2}{l62+2}=2+\frac{4l-2}{l^2+2}$
$\frac{4l-2}{l^2+2}<1$
because:
$(l-2)^2>0$
$l^2-4l+2+2>0$
$4l-2<l^2+2$
If $l=2$ the equality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Moment generating function of Laplace distribution step by step There are two similar posts but none of them helped me to get through the full derivation of the very simple MGF that should be according to Wikipedia:
$$
\frac {\exp(\mu t)}{1-b^{2}t^{2}}
$$
Here's my attempt:
Definition of MGF
$$
M_X(t) = \mathbb{E}\left[\exp(t X)\right]
$$
by LOTUS
$$
= \int \exp(t x) \cdot p(x) dx
$$
Plugging-in PDF
$$
= \int \exp(t x) \cdot \frac{1}{2b}\exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$
$$
= \frac{1}{2b} \int \exp(t x) \cdot \exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$
$$
= \frac{1}{2b} \int \exp \left(t x - \frac{\mid x - \mu \mid }{b} \right) dx
$$
getting rid of abs. value
$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(t x - \frac{ x - \mu }{b} \right) dx
+ \frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{- x + \mu }{b} \right) dx
$$
the left integral first
$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x - \mu }{b} \right) dx
$$
$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x }{b} \right) \exp \left(\frac{- \mu }{b} \right) dx
$$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x }{b} \right) dx
$$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{ (bt - 1) }{b} x \right) dx
$$
Now since $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \left| \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} x \right) \right|_{-\infty}^{\mu}
$$
Let's evaluate the integral first
$$
\left| \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} x \right) \right|_{-\infty}^{\mu}
$$
$$
= \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right)
-
\frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (-\infty)\right)
$$
Since $\lim_{x \rightarrow 0} e^{-x} = 0$, the second term is 0
$$
= \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right)
- 0
$$
Double check from WorlframAlpha:
$$
= \frac{b}{bt-1} \exp \left( \mu \left( t - \frac{1}{b} \right) \right)
$$
for $\frac{1}{b} < t$
Now the right-hand side integral:
$$
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{- x + \mu }{b} \right) dx
$$
$$
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(\frac{b t x + x - \mu }{b} \right) dx
$$
$$
= \frac{1}{2b} \int_{\mu}^{\infty} \exp \left(\frac{b t x + x }{b} \right) \exp \left(\frac{-\mu }{b} \right)dx
$$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{ -\mu }{b} \right) \int_{\mu}^{\infty} \exp \left(\frac{b t x + x }{b} \right) dx
$$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{ -\mu }{b} \right) \int_{\mu}^{\infty} \exp \left( \frac{b t + 1 }{b} x\right) dx
$$
Now evaluate the integral
$$
\int_{\mu}^{\infty} \exp \left( \frac{b t + 1 }{b} x\right) dx
$$
According to wolfram alpha
$$
= - \frac{b}{bt + 1}\exp\left(\mu \frac{1}{b + t}\right)
$$
for $\frac{1}{b} + t < 0$
Now sum-up these two integrals:
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right) + \\
\frac{1}{2b} \cdot \exp \left(\frac{ -\mu }{b} \right) \cdot \left( - \frac{b}{bt + 1}\exp\left(\mu \frac{1}{b + t}\right) \right)
$$
And simplify
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right) - \\
\frac{1}{2b} \cdot \exp \left(\frac{ -\mu }{b} \right) \cdot \frac{b}{bt + 1} \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$
$$
= \frac{1}{2b} \cdot \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right) - \\
\frac{1}{2b} \cdot \frac{b}{bt + 1} \cdot \exp \left(\frac{ -\mu }{b} \right) \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$
$$
= \frac{1}{b} \cdot \frac{1}{ (bt - 1) } \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right) - \\
\frac{1}{b} \cdot \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu }{b} \right) \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$
$$
= \frac{1}{b} \cdot \frac{1}{ (bt - 1) } \cdot \exp \left(\frac{- \mu }{b} +\frac{ (bt - 1) }{b} (\mu) \right) - \\
\frac{1}{b} \cdot \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu }{b} + \mu \frac{1}{b + t}\right)
$$
$$
= \frac{1}{b} \cdot \frac{1}{ (bt - 1) } \cdot \exp \left( \mu \left(t - \frac{2}{b} \right) \right) - \\
\frac{1}{b} \cdot \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu }{b} + \mu \frac{1}{b + t}\right)
$$
But I cannot move away from this ugly formula to the simple beauty. Is there just a stupid mistake above or am I doing something substantially wrong?
| There are a number of efficiencies that can be realized if we first consider the shifted exponential family $$f_Y(y) = \frac{1}{b} e^{-(y - \mu)/b}, \quad y \ge \mu$$ with location $\mu$ and scale $b > 0$. The MGF is easy to calculate: $$M_Y(t) = \int_{y=\mu}^\infty \frac{1}{b} e^{ty} e^{-y/b} e^{\mu/b} \, dy = \frac{e^{\mu/b}}{b} \int_{y=\mu}^\infty e^{-(1/b-t)y} \, dy = \frac{e^{\mu/b} e^{-(1/b - t)\mu}}{b(1/b-t)} = \frac{e^{\mu t}}{1 - bt} , \quad t < 1/b.$$ The MGF of $-Y$ is then $$M_{-Y}(t) = M_Y(-t) = \frac{e^{-\mu t}}{1 + bt}, \quad t > -1/b.$$
Now define the mixture density $$f_X(x) = \frac{1}{2} (f_{Y_1}(x) + f_{-Y_2}(x))$$ where $Y_1$ is shifted with location $\mu$, and $Y_2$ is shifted with location $-\mu$, and both share scale $b$. Then it is easy to see $X$ is Laplace with location $\mu$ and scale $b$. The MGF of $X$ is then $$M_X(t) = \frac{1}{2} \left( M_{Y_1}(t) + M_{-Y_2}(t) \right) = \frac{1}{2} \left( \frac{e^{\mu t}}{1 - bt} + \frac{e^{-(-\mu)t}}{1 + bt} \right) = \frac{e^{\mu t}}{1 - (bt)^2}, \quad |t| < 1/b.$$
In fact, this computation easily generalizes to the case where $Y_1$ and $Y_2$ do not have the same scale parameter, in which the distribution has nonzero skew.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3929288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can't find n (algebraic manipulation) I‘m trying to understand a proof on my textbook that says $\sqrt {n − 1} +\sqrt {n + 1}$ is irrational, where $n \in \mathbb{N} $. However I can't seem to show that
$$n^2 - 1 = \frac{(\frac{p^2}{q^2} - 2n)^2}{4} \to n = \frac{(p^4+4q^4)}{4p^2q^2}$$,
where $p,q \in \mathbb{Z}$，and the gcd of $p,q$ is 1. I feel really stupid, is there something I don't see?
The furthest I got is
$$n = \frac{\sqrt{2p^4+5p^4-2p^2q^2}-(p^2-q^2)}{2q^2}$$
| $$n^2 - 1 = \frac{\left(\dfrac{p^2}{q^2} - 2n\right)^2}{4}= \frac{\dfrac{p^4}{q^4} + 4n^2-4n\dfrac{p^2}{q^2} }{4}=\dfrac {p^4}{4q^4}+n^2-n\dfrac{p^2}{q^2}. $$
Subtract $n^2$ from both sides to get $$-1=\dfrac {p^4}{4q^4} -n\dfrac{p^2}{q^2} ,$$
which is $$n\dfrac{p^2}{q^2}=1+\dfrac {p^4}{4q^4}.$$
Can you take it from here (solve for $n$)?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3930958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\lim\limits_{n \to \infty}{\lfloor x \rfloor + \lfloor x^2 \rfloor + ... + \lfloor x^n \rfloor\over x^n}, 1$$\lim\limits_{n \to \infty}{\lfloor x \rfloor + \lfloor x^2 \rfloor + ... + \lfloor x^n \rfloor\over x^n}, 1<x\in \mathbb R^+$$
I have a problem solving this limit.Here is my solution. Let x be an integer, then we have:
$$\lim\limits_{n \to \infty}{ x + x^2 + ... + x^n\over x^n} = \lim\limits_{n \to \infty} {(x^n -1)x \over (x - 1) x^n } = \lim\limits_{n \to \infty} {\frac{x^{n+1}}{x^n}-\frac{x}{x^n}\over \frac{x^{n+1}}{x^n} -\frac{x^{n}}{x^n}} = \lim\limits_{n \to \infty} {x - \require{cancel} \cancelto{0}{\frac{x}{x^n}}\over x -1}= {x\over x-1}$$
But I don't know how to solve for non-integer x.
| You proved that if $x \in \mathbb{N}_{\geq 2}$ the limit is $\frac{x}{x-1}$, but we can actually prove that for any $x>1$ we have $\lim\limits_{n \to \infty} \frac{x+\ldots+x^n}{x^n} = \frac{x}{x-1}$.
$x>1$, then $(x+\ldots+x^n) -n\leq\lfloor x \rfloor + \ldots + \lfloor x^n \rfloor \leq x + \ldots +x^n$, which show us that the limit is again $\frac{x}{x-1}$ (Why?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3931879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality: $(y+z)*x^4+(x+z)*y^4+(x+y)*z^4\le((x+y+z)^5)/12$ for positive reals. The inequality is homogeneous, but I am not sure if that helps.
Also, I was thinking about considering a concave function and apply Jensen or Karamata on LHS.
| Suppose $x \geqslant y \geqslant z$ and
$$f(x,y,z) = x^4(y+z)+y^4(z+x)+z^4(x+y).$$
We have
$$\begin{aligned}f(x,y,z)-f(x,y+z,0) & = y^4(z+x)+z^4(x+y)-x(y+z)^4\\ \\& = y^4(z+x)+z^4(x+y)-x(y+z)^4 \\ \\&=-yz\left[3x(y+z)^2+y^2(x-y)+z^2(x-z)\right] \leqslant 0.\end{aligned}.$$
Therefore $f(x,y,z) \leqslant f(x,y+z,0).$ Finally, we need to prove
$$f(x,y+z,0) = x^4(y+z)+x(y+z)^4\leqslant \frac{1}{12}(x+y+z)^5.$$
Setting $a = y +z,$ the inequality become
$$12ax(a^3+x^3) \leqslant (x+a)^5,$$
or
$$(a+x)(a^2-4ax+x^2)^2 \geqslant 0.$$
Which is true. The proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3933239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}$ without L'Hôpitals Rule. The limit,
$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}=6$
is easily found using L'Hôpitals Rule. However, the exercise is to evaluate it without using this method. I've tried multiple substitutions, mainly pythagoric and reciprocal identities, but without success, the indeterminate form $\frac{0}{0}$ keeps appearing. I’ve also tried splitting the limit and doing the substitutions. I'm looking to simplify the expression in order to evaluate the limit directly.
Thanks in advance.
| I use (at the end)
$\begin{array}\\
\cos x-\sin x
&=\sqrt{2}((1/\sqrt{2})\cos x-(1/\sqrt{2})\sin x)\\
&=\sqrt{2}(\sin(\pi/4)\cos x-\cos(\pi/4)\sin x)\\
&=\sqrt{2}\sin(\pi/4-x)\\
\end{array}
$
$\begin{array}\\
v
&=\lim \limits_{x \to \frac\pi4}\dfrac{\csc^2x-2\tan^2x}{\cot x-1}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{\dfrac1{\sin^2x}-2\dfrac{\sin^2x}{\cos^2x}}{\dfrac{\cos x}{\sin x}-1}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{\cos^2x-2\sin^4x}{\sin x\cos^3 x-\sin^2x\cos^2x}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{1-\sin^2x-2\sin^4x}{\sin x\cos x(1-\sin^2x)-\sin^2x(1-\sin^2x)}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{(1-2\sin^2x)(1+\sin^2x)}{\sin x(1-\sin^2x)(\cos x-\sin x)}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{(1-2\sin^2x)(3/2)}{(\sqrt{2}/2)(\cos x-\sin x)}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{1-2\sin^2x}{(\cos x-\sin x)}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)(1+\sqrt{2}\sin x)}{\cos x-\sin x}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)(1+\sqrt{2}(\sqrt{2}/2))}{\cos x-\sin x}\\
&=\dfrac{6}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)}{\cos x-\sin x}\\
&=\dfrac{6}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{1-\sqrt{2}\sin x}{\sqrt{2}\sin(\pi/4-x)}\\
&=3\lim \limits_{x \to \frac\pi4}\dfrac{1-\sqrt{2}\sin x}{\sin(\pi/4-x)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-\sqrt{2}\sin (y+\pi/4)}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-\sqrt{2}(\sin y\cos(\pi/4)+\cos(y)\sin(\pi/4))}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-2(\sin y+\cos(y))}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{-2\sin y}{-\sin(y)}\\
&=6\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3933829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Compute $\int_0^{2 \pi} \frac{ \cos(2nx) }{ a^2 + \cos^2 x} dx$ Compute the integral
$$I_n=\int_0^{2 \pi} \frac{ \cos(2nx) }{ a^2 + \cos^2 x} dx$$
where $n$ is any integer.
Using trig identities to evaluate special cases I found answers for $n=0$ and $n=1$.
For $n=0$
$$I_0=\frac{2\pi}{ a \sqrt{a^2 +1}}$$
For $n=1$
$$I_1=2\pi\frac{ 2a( \sqrt{a^2 +1} - a ) -1 }{ a \sqrt{a^2 +1}}$$
How do I find a generic answer for any integer n?
| Assume $n\ge0$, $ a>0$ (wlog) and let $t=2x$. Then
$$I_n=\int_0^{2 \pi} \frac{ \cos(2nx) }{ a^2 + \cos^2 x } dx
=2\int_0^{2\pi} \frac{\cos(nt )}{{2a^2+1+\cos t}}dt
$$
Apply the Fourier series below and observe that only the term of $k=n$ survives the integration due to $\cos(nt ) $ in the numerator (the constant term does in the case of $n=0$)
$$\frac{1}{{2a^2+1+\cos t}}=
\frac1{2a\sqrt{1+a^2}}\left(1+2\sum_{k=1}^\infty (-1)^k (\sqrt{1+a^2}-a)^{2k}\cos (kx)\right)
$$
As a result, the integral yields
$$I_n= 2\pi (-1)^n \frac{(\sqrt{1+a^2}-a)^{2n}}{a\sqrt{1+a^2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$? Question
Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.
My attempt
Let $f(x) = 3x^3 + 7x^2 - 48x + 49$
$\text{Using Remainder Theorem,}$
\begin{align}
f(-k) &= f(k) \\
3(-k)^3 + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}
$\text{Comparing } 6k^2 + 14k - 96 = 0 \text{ with } ak^2 + bk + c = 0, a = 6, b = 14, c = -96$
\begin{align}
k = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } &= \frac{ -(14) \pm \sqrt{(14)^2 - 4(6)(-96)} }{ 2(6) } \\
k &= 3 \text{ and } -5\frac{1}{3}
\end{align}
$\therefore k = 3, -5\frac{1}{3} \text{ or } 0 $
My answer is incorrect. The correct answer is $k = 0, 4 \text{ or } -4$
| Duncan Ramage has patiently showed you where you went wrong. I would like to provide some tips on how to avoid the errors you made and simplify your proof.
There is no need to write out every term of $f(k)$ and $f(-k)$. Instead you take the difference:
$$0=f(k)-f(-k)=3(k^3-(-k)^3) + 7(k^2-(-k)^2)-48(k-(-k))+(49-49)\\
=6k^3-96k=6k(k^2-16) \implies k=0, \pm 4.$$
When you get experienced you can skip the even-order terms altogether:
$$0=f(k)-f(-k)=3(k^3-(-k)^3) -48(k-(-k))\\
=6k^3-96k=6k(k^2-16) \implies k=0, \pm 4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Prove that $\sum \frac{i+1}{i!+1}<3$
Prove that $$\sum_{i=1}^{n}\frac{i+1}{i!+1}<3$$
My approach: If $k=1$ we can see that $$S_{1}=\sum_{k=1}^{1}\frac{k+1}{k!+1}=\frac{1+1}{1!+1}=\frac{2}{2}=1<3$$ Now, suppose that the statement is true for some $n\in \mathbb{N}$, it's to say suppose that $S_{n}=\sum_{k=1}^{n}\frac{k+1}{k!+1}<3$ is true, so we need to show that is true for $n+1$, it's to say we need to show that $S_{n+1}=\sum_{k=1}^{n+1}\frac{k+1}{k!+1}<3$ is also true.
Now, we can see that
\begin{eqnarray*}
S_{n+1}&=&\sum_{k=1}^{n+1}\frac{k+1}{k!+1}\\
&=&\sum_{k=1}^{n}\frac{k+1}{k!+1}+\sum_{k=n+1}^{n+1}\frac{k+1}{k!+1}
\end{eqnarray*}
but, how can I continue from here?
| Alternative proof without induction:
Note that $\frac{i+1}{i!+1}$ looks like $\frac{i}{i!} = \frac{1}{(i-1)!}$ and $\sum_{i\ge1}\frac{1}{(i-1)!}=e < 3$. Unfortunately $\frac{i+1}{i!+1}>\frac{i}{i!} $, but we can make some adjustment:
$$\frac{i+1}{i!+1} < \frac{i+1}{i!}=\frac{i+1}{i} \cdot \frac{1}{(i-1)!}$$
Therefore
$$\sum_{i=1}^{n}\frac{i+1}{i!+1} < \frac{1+1}{1!+1}+\frac{2+1}{2!+1}+\sum_{i=3}^\infty\frac{i+1}{i!+1} < 1+1+\sum_{i=3}^\infty\frac{i+1}{i} \cdot \frac{1}{(i-1)!}\\
< 2+ \frac{4}{3}\sum_{i\ge 3}\frac{1}{(i-1)!}=2+\frac 43\cdot (e-2) \approx 2.958 < 3. \blacksquare
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?
Dr. Lizardo writes down a sequence of $n$ letters, where each letter is A, B, or C, and the letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?
What I have tried so far
The last letter is predefined in terms of "A or not" based on the first $n-1$ letters. If there are an odd number of As so far, the last one has to be A, otherwise, B or C. Therefore, there are $3^{n-1}$ ways to write the first $n-1$ letters. However, the last letter can either have 1 choice or 2 choices.
I am thinking maybe we need to do $S_n = 3^{n-1} + S_{n-1}$. However, I need a closed form, not a recurrance. How can I do this?
Thank you!
Note: $S_n = 3^{n-1} + S_{n-1}$ seems to be working for the first few values of $n$ (I tested through $n=4$), so finding a closed form of that recurrance would work as well.
| Here we show the equality of two already given answers. The number $S_n$ is given by
\begin{align*}
\color{blue}{S_n=\frac{1}{2}\left(3^n+1\right)=\sum_{k=0}^n\binom{n}{2k}2^{n-2k}\qquad\qquad n\geq 0}\tag{1}
\end{align*}
The following is valid
\begin{align*}
\sum_{k=0}^{n}\binom{n}{2k}x^{2k}+\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1}&=(1+x)^n\\
\sum_{k=0}^{n}\binom{n}{2k}x^{2k}-\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1}&=(1-x)^n\\
\sum_{k=0}^n\binom{n}{2k}x^{2k}&=\frac{1}{2}\left((1+x)^n+(1-x)^n\right)\tag{2}
\end{align*}
We obtain from (2)
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{n}{2k}2^{n-2k}}
&=2^n\sum_{k=0}^n\binom{n}{2k}\left(\frac{1}{2}\right)^{2k}\\
&=2^n\,\frac{1}{2}\left(\left(1+\frac{1}{2}\right)^n+\left(1-\frac{1}{2}\right)^n\right)\\
&=2^n\,\frac{1}{2}\left(\left(\frac{3}{2}\right)^n+\left(\frac{1}{2}\right)^n\right)\\
&\,\,\color{blue}{=\frac{1}{2}\left(3^n+1\right)}
\end{align*}
and the claim (1) follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Existence of $x_0$ s.t. $\frac{f''( x_0 )}{2}=\frac{f( a )}{( a-b ) ( a-c )}+\frac{f( b )}{( b-a ) ( b-c )}+\frac{f( c )}{( c-a ) ( c-b )}$
Let $f$ be continuous on $[a,b]$, and $f''(x)$ exists on $(a,b)$. Prove that $\forall c\in(a,b), \exists x_0\in(a,b)$, s.t.
$$
\frac{f''\left( x_0 \right)}{2}=\frac{f\left( a \right)}{\left( a-b \right) \left( a-c \right)}+\frac{f\left( b \right)}{\left( b-a \right) \left( b-c \right)}+\frac{f\left( c \right)}{\left( c-a \right) \left( c-b \right)}
$$
My Approach: Assume $$\dfrac{K}{2} = \frac{f\left( a \right)}{\left( a-b \right) \left( a-c \right)}+\frac{f\left( b \right)}{\left( b-a \right) \left( b-c \right)}+\frac{f\left( c \right)}{\left( c-a \right) \left( c-b \right)}$$
Let
$$
g\left( x \right) =\frac{K}{2}\left( \left( a-x \right) \left( x-c \right) \left( c-a \right) \right) +f\left( a \right) \left( x-c \right) +f\left( x \right) \left( c-a \right) +f\left( c \right) \left( a-x \right)
$$
Then $g(a)=g(c)=g(b)=0$.
Since $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exist $x_1\in(a,c),x_2\in(c,b)$ s.t. $g'(x_1)=g'(x_2)=0$.
And since $g'$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $x\in(x_1,x_2)\subset (a,b)$, s.t. $g''(x_0)=0$.
From
$$
g''\left( x_0 \right) =\frac{K}{2}\cdot 2\left( a-c \right) +f''\left( x_0 \right) \left( c-a \right)
$$
We have $K=f''(x_0)$.
My Question is, is the proof right? And can we generalize the original claim(that is, change $f''(x)$ to $f^{(k)}(x)$)?
Other proofs of the original claim is also welcomed.
| Your formula can be written as
$$\frac{\left | \begin{matrix} 1& 1& 1\\ a & b & c \\ f(a) & f(b) & f(c) \end{matrix} \right|}{
\left| \begin{matrix} 1& 1& 1\\ a & b & c \\ g(a) & g(b) & g(c) \end{matrix} \right|}=\frac{f''(\xi)}{g''(\xi)}
$$
(with $g(x) = x^2$).
Let $K$ be the value of the quotient. Then the function
$$F(x) =\left | \begin{matrix} 1& 1& 1\\ x & b & c \\ f(x) & f(b) & f(c) \end{matrix} \right| - K \cdot \left | \begin{matrix} 1& 1& 1\\ x & b & c \\ g(x) & g(b) & g(c) \end{matrix} \right|$$
is $0$ at $x=a$. Clearly it is also $0$ at $x=b$, and $x=c$. So it has $3$ distinct zeroes. Applying Rolle twice we find $\xi$ such that $F^{(2)}(\xi)=0$.
It is easy to generalize this to $n+1$ distinct $a_i$'s. One can also pass to the limit in these formulas, grouping several $a_i$ together. In particular, if $n$ of them are "equal", we get the quotient of Taylor series approximations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Curve adaptation with a finite number of measured values I consider a function $f(t)=\alpha\cdot 2^{-t}+\beta\cdot 2^{-2t}$ and a finite number of measured values $f_0,...,f_n$. The coefficients $ \alpha, \beta $ are calcualted by the least squares method which means
$$ \sum\limits_{i=0}^n |f(t_i)-f_i|^2$$
is minimized.
Then $$\sum\limits_{i=0}^n 2^{-t_i}\cdot f(t_i)=\sum\limits_{i=0}^n 2^{-t_i}\cdot f_i.\quad (*)$$
My idea:
I consider the expressions
$$ \begin{aligned}f_0&\stackrel{!}{=}f(t_0)=2^{-t_0}+\beta\cdot 2^{-2t_0}\\f_1&\stackrel{!}{=}f(t_1)=2^{-t_1}+\beta\cdot 2^{-2t_1}\\\vdots\\f_n&\stackrel{!}{=}f(t_n)=2^{-t_n}+\beta\cdot 2^{-2t_n}\end{aligned}$$
which can be reprensented by $$ \underbrace{\begin{pmatrix}2^{-t_0}&2^{-2t_0} \\2^{-t_1}&2^{-2t_1} \\\vdots&\vdots\\2^{-t_n}&2^{-2t_n} \end{pmatrix}}_{=:A}\cdot \underbrace{\begin{pmatrix}\alpha\\\beta \end{pmatrix}}_{=:x}=\underbrace{\begin{pmatrix}f_0\\f_1\\\vdots\\f_n \end{pmatrix}}_{=:f} $$
Then I want to minimize
$$ F((\alpha,\beta)):=\|A\cdot x-f\|_2^2=\sum\limits_{i=0}^n |f(t_i)-f_i|^2=\sum\limits_{i=0}^n \Big(\alpha\cdot 2^{-t_i}+\beta\cdot 2^{-2t_i}-f_i \Big)^2 $$
I calculate the gradiant of $ F $:
$$ \nabla F((\alpha,\beta))= 2\cdot A^T\cdot (A\cdot x-f)\stackrel{!}{=}0$$
which means I have to solve $$ A^T\cdot A\cdot x=A^T\cdot f $$
At first I have
$$ A^T\cdot A=\begin{pmatrix}\sum\limits_{i=0}^n 2^{-4t_i}&\sum\limits_{i=0}^n 2^{-3t_i}\\\sum\limits_{i=0}^n 2^{-3t_i}&\sum\limits_{i=0}^n 2^{-2t_i} \end{pmatrix}=:\begin{pmatrix}a&b\\b&c \end{pmatrix} $$
and
$$ A^T\cdot f=\begin{pmatrix}\sum\limits_{i=0}^nf_i\cdot 2^{-2t_i}\\\sum\limits_{i=0}^nf_i\cdot 2^{-t_i} \end{pmatrix}=:\begin{pmatrix}b_1\\b_2 \end{pmatrix} $$
The solution is $$ \alpha=\frac{a\cdot b_2-b\cdot b_1}{a\cdot c-b^2}\qquad \beta=\frac{c\cdot b_1-b\cdot b_2}{a\cdot c-b^2} $$
And here comes the problem. These expressions of the solution are soo complicated that I'm not able to handle them to show the upper identity (*).
Is there annother way to show or what could I do else(*)?
| Given that $$ A^T\cdot A=\begin{pmatrix}\sum\limits_{i=0}^n 2^{-2t_i}&\sum\limits_{i=0}^n 2^{-3t_i}\\\sum\limits_{i=0}^n 2^{-3t_i}&\sum\limits_{i=0}^n 2^{-4t_i} \end{pmatrix}=:\begin{pmatrix}a&b\\b&c \end{pmatrix}, $$
note that $$(A^TA)^{-1} = \frac{1}{ac - b^2}\begin{pmatrix} c & -b \\ -b & a \end{pmatrix}.$$
Let $d$ be the first column of $A$ (unfortunate nomenclature...). Then $$d^TA = \begin{pmatrix} \sum_{i=0}^n2^{-2t_i} & \sum_{i=0}^n2^{-3t_i} \end{pmatrix} = \begin{pmatrix} a & b \end{pmatrix},$$ and $$d^TA(A^TA)^{-1} = \frac{1}{ac - b^2}\begin{pmatrix} ac - b^2 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \end{pmatrix}$$ -- et voila! Since now $$d^TA(A^TA)^{-1}A'f = d^Tf,$$
which had to be shown. Note that ($\ast$) can be expressed as $d^TA\hat x = d^Tf$, where $\hat x = (A^TA)^{-1}A^Tf$ is the LS estimate of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x,y,z>0$, $n,m\in N$, $n\ge m$ prove:$\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}\ge \frac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$
Given positive numbers $x,y,z$ and $n,m$ positive integers with $n\ge m$ prove that: $$\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}\ge \frac{1}{2^m}(x^{n-m}+y^{n-m}+z^{n-m})$$
I tried doing it as follows:
If $x\ge y \ge z$, then:
$y+z\le z+x\le x+y$
Hence:
$x^n\ge y^n\ge z^n$ and $\frac{1}{(y+z)^m}+\frac{1}{(z+x)^m}\ge \frac{1}{(x+y)^m}$
From Tchebychev:
$3[\frac{x^n}{(y+z)^m}+\frac{y^n}{(z+x)^m}+\frac{z^n}{(x+y)^m}]\ge (x^n+y^n+z^n)[\frac{1}{(y+z)^m}+\frac{1}{(x+z)^m}+\frac{1}{(x+y)^m}]$
which is where I got stuck.
Could you please help me finish this question off?
| Your idea of chebyshev is right we just have to tailor it with repect to what we have to prove
WLOG $x\ge y\ge z$ then $x^{n-m}\ge y^{n-m}\ge z^{n-m}$ and ${(\frac{x}{y+z})}^m\ge{( \frac{y}{x+z})^m}\ge {(\frac{z}{x+y})}^m $
By chebyshev: $$\sum_{cyc} \frac{x^n}{{(y+z)}^m}=\sum_{cyc} x^{n-m} {(\frac{x}{y+z})}^m\ge \frac{1}{3} \sum x^{n-m}\cdot \sum {(\frac{x}{y+z})}^m \tag S$$ Now by power mean inequality and Nesbitts inequality $$\sum {(\frac{x}{y+z})}^m\ge \dfrac{{(\sum \frac{x}{y+z})}^m}{3^{m-1}}\ge \frac{3}{2^m}$$ Use this in inequality (S) directly to prove result
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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conceptual doubt in finding $\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}$ While calculating the aforementioned limits I did the following.
$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}\\ =
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -4\sqrt2 \left (\frac{1}{\sqrt 2} \cos x +\frac{1}{\sqrt 2} \sin x \right )^5}{1-\sin {2x}}\\=
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin^5 (x + \frac{\pi}{4}) \right )}{1-\sin {2x}}\\
=\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin (x + \frac{\pi}{4}) \right ) \left(1 +\sin (x + \frac{\pi}{4}) +\sin^2 (x + \frac{\pi}{4}) +\sin^3 (x + \frac{\pi}{4}) +\sin^4 (x + \frac{\pi}{4}) \right)}{1-\sin {2x}}\\$$
As $x$ approaches $\frac{\pi}{4}$, the functions $\sin (x + \frac{\pi}{4})$ and $\sin 2x$ become close enough. So, I canceled them($1-\sin {2x}$ and $ 1 - \sin (x + \frac{\pi}{4})$ ) and the limit turns out to be $20\sqrt2$.
I made the above claim after comparing the graphs. But the fact that bothers me is L'Hospital's rule yielded the answer $5\sqrt2$ (and it is the correct answer to the problem)
Can anyone please explain?
|
As $x$ approaches $\frac{\pi}{4}$ the functions $\sin (x + \frac{\pi}{4})$ and $\sin 2x$ become close enough
When we say "close enough", we are saying that $\lim_{x\to \pi/4} \dfrac{\sin (x + \frac{\pi}{4})}{\sin 2x}=1$. However, the limit
$\lim_{x\to \pi/4} \dfrac{1-\sin (x + \frac{\pi}{4})}{1-\sin 2x}=\dfrac{1}{4}\neq 1$
So we cannot cancel them simply.
We can view it from the perspective of Taylor Series:
$$
\frac{\sin \left(x+\frac{\pi }{4}\right)}{\sin (2 x)}=1+\frac{3}{2} \left(x-\frac{\pi }{4}\right)^2+\frac{19}{8} \left(x-\frac{\pi }{4}\right)^4+O\left(\left(x-\frac{\pi }{4}\right)^5\right) \\
\frac{1-\sin \left(x+\frac{\pi }{4}\right)}{1-\sin (2 x)}=\frac{1}{4}+\frac{1}{16} \left(x-\frac{\pi }{4}\right)^2+\frac{1}{96} \left(x-\frac{\pi }{4}\right)^4+O\left(\left(x-\frac{\pi }{4}\right)^5\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of triangle with incircle tangent to semicircle whose diameter is on one triangle side Triangle $\triangle ABC$ with incenter $I$ and inradius $r$ has the following property: the incircle is tangent to the semicircle with diameter $AB$ and is within the semicircle. Find the area of $\triangle ABC$ as a function of the side $AB$ and the inradius $r$.
So far I've noticed that $\angle AEF=\angle BEF=\frac{\pi}{4}$ but I don't think that that's useful in any way.
|
$\triangle ABC = \frac{1}{2} AB \times h$
Let's get onto finding $h$ and then we are done.
We know $R = \frac{AB}{2} \,( R$ being the radius of the semicircle).
$AB = h \cot A + h \cot B \implies h = \frac{AB}{\cot A + \cot B}$
Now, $\cot \frac{A}{2} = \frac{AF}{IF} = \frac{R+x}{r}$
Similarly, $\cot \frac{B}{2} = \frac{R - x}{r}$
Using the identity $\, \cot 2 \alpha = \frac{1}{2} (\cot \alpha - \tan \alpha), \,$ we have,
$2 \cot A = \frac{(R+x)^2 - r^2}{r(R+x)} \,, \, 2 \cot B = \frac{(R-x)^2 - r^2}{r(R-x)}$
$ 2(\cot A + \cot B) = \frac{(R^2-x^2)(R+x) + (R^2-x^2)(R -x) - 2 r^2R}{r(R^2-x^2)}$
Also, $R^2 - x^2 = R^2 - ((R-r)^2 - r^2) = 2Rr$
So, $ 2(\cot A + \cot B) = \frac{4R - 2r}{2r} = \frac{AB - r}{r}$
And hence $\frac{h}{2} = \frac{AB \, r}{AB - r} \, , \,$ Area $\triangle ABC = \frac{AB^2 \, r}{AB - r}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| $(x+y)^2
=x^2+2xy+y^2
=2xy+1
$
so
$x+y
=\sqrt{2xy+1}
$.
$\begin{array}\\
x^3+y^3
&=(x+y)(x^2-xy+y^2)\\
&=\sqrt{2xy+1}\dfrac{2x^2-2xy+2y^2}{2}\\
&=\sqrt{2xy+1}\dfrac{x^2+y^2+x^2-2xy+y^2}{2}\\
&=\sqrt{2xy+1}\dfrac{1+(x-y)^2}{2}\\
&\ge\dfrac12\sqrt{2xy+1}\\
\end{array}
$
so we want
$\dfrac12\sqrt{2xy+1}
\ge xy\sqrt{2}
$
or
$2xy+1
\ge 8(xy)^2
$
or,
with $z = xy$,
$8z^2-2z-1
\le 0
$.
$8z^2-2z-1
=8(z+\frac14)(z-\frac12)
$.
$0
\le (x-y)^2
=x^2-2xy+y^2
$
so
$2xy
\le x^2+y^2
=1
$
or
$0 \le z \le \frac12
$.
Therefore
$8z^2-2z-1
=8(z+\frac14)(z-\frac12)
\le 0
$
which is what we want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 10,
"answer_id": 3
} |
Let $a,b,c$ be integers such that
$\gcd(a,6)=3=\gcd(b,6)$. Show that $\gcd(a+b,6)=6$. Let $a,b,c$ be integers such that
$\gcd(a,6)=3=\gcd(b,6)$.
Show that $\gcd(a+b,6)=6$.
Attempt:
Let $d = \gcd(a+b,6)$.
Then, by definition, $d \mid (a+b)$ and $d \mid 6$. It means that $d=1,2,3$ or $6$. On the other hand, we have
$3\mid a$ and $3 \mid b$. Write
$a = 3m$ and $b=3n$ for some integers $m$ and $n$. Then, $a+b=3m+3n=3(m+n)$ and so
$3 \mid (a+b)$. But, $d \mid (a+b)$. Now, $a+b = dp$ for some integer $p$.
Thus, we have $3 \mid dp$ which means $3 \mid d$ or $3 \mid p$.
If $3 \mid d$, then $d=3k$ for some integer $k$. Hence, $d$ is either $3$ or $6$.
But, how to get $d=6$?
| Since $\gcd(a,6)=\gcd(b,6)=3$, then $a=3m$ and $b=3n$ for some odd $m$ and $n$.
Thus, $a+b=3(m+n)$ while $m+n$ is even, so $6\mid a+b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$ This integral
$$\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
is fabricated based on an interesting point which I may post later.
The question is: How would you do it?
| $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx=\int_{0}^{1} \frac{(x+2)(3(x^2-x)+1)+2x-1}{3(x^2-x)+1}dx =\int_{0}^{1} (x+2+\frac{2x-1}{3(x^2-x)+1})dx=\int_{0}^{1} (x+2+\frac{(x^2-x+\frac{1}{3})'}{3(x^2-x+\frac{1}{3})})dx=[\frac{1}{2}x^2+2x+\ln(x^2-x+\frac{1}{3})]_0^{1}=\frac{5}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Find variables from a multiplication of geometric sequences
Given $$f(x) = \sum_{i=0}^\infty a_ix^i$$
And $$f(x)(1+2x+2x^2+x^3) = \frac{1}{(1-x)^3}$$
Find the values of $a_0$, $a_1$, and $a_2$.
I try to expand them but it doesn't seem to lead me to any solution.
How can I isolate a0, a1, and a2 to find them?
| Tip:
Notice $1+2x+2x^2+x^3=\left(1+x+x^2\right)(1+x)$
Then $$\begin{split}f(x)&=\frac1{(1-x)^3(1+x)\left(1+x+x^2\right)}\\&=\frac1{1-x}\cdot\frac1{(1-x)(1+x)}\cdot\frac1{(1-x)\left(1+x+x^2\right)}\\&=\frac1{1-x}\cdot\frac1{1-x^2}\cdot\frac1{1-x^3}\\&=\left(1+x+x^2+x^3+...\right)\left(1+x^2+x^4+x^6+...\right)\left(1+x^3+x^6+x^9+...\right)\end{split}$$
Then expansion finishes it all. Try it!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How to find the series $a_{n}$ which has this generating function: $f(x)=\frac{1}{1-x^{2}}$? I'm given a generating function $f(x)$:
$$
f(x)=\frac{1}{1-x^2}
$$
Naturally, I tried to transform it into the form $f(x)=\sum_0^\infty a_n \cdot x^n$.
So far I figured out what follows:
$$
f(x)=\frac{1}{(1-x)(1+x)}=\frac{\frac{1}{2}}{1-x} + \frac{\frac{1}{2}}{1+x}=\frac{1}{2}\cdot\left( \sum_{n=0}^\infty x^n+\frac{1}{1-(-x)} \right)
$$$$
\frac{1}{1-(-x)}=1+(-x)+x^2+(-x^3)+x^4+\cdots
$$$$
\frac{1}{1-(-x)}=(1+x^2+x^4+\cdots)-(x+x^3+x^5+\cdots)=\sum_{n=0}^\infty x^{2n}-x^{2n+1}
$$
Therefore:
$$
f(x)=\frac{1}{2} \left( \sum_{n=0}^\infty x^n+x^{2n}-x^{2n+1} \right)
$$
Now I am lost. Is my train of thought good? Am I anywhere near finding the series which is generated by function $f(x)$?
| $\frac 1 {1-x^2}=1+x^2+x^4+x^6+\cdots$ for $|x| <1$ since $\frac 1 {1-r}=1+r+r^2+\cdots$ for $|r|<1$. So $a_n=1$ for $n$ even and $a_n=0$ for $n$ odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find $f(x)$ such that $f(y)-f(x)-(y-x)f'(\frac{x+y}{2})+\frac{(x-y)^2}{4}=0$
Given $f(x)$ to be differentiable such that $f(y)-f(x)-(y-x)f'(\frac{x+y}{2})+\frac{(x-y)^2}{4}=0$, then find $f(x)$
From the given equation, we can't plug in $x=y$ as it just gives $0=0$. For some reason I think $f(x)$ is a quadratic equation ( I dunno why tho ). So I set $f(x)=ax^2+bx+c$ and put it in the given equation.
\begin{equation}
ay^2+by+c-(ax^2+bx+c)-(y-x)[a(x+y)+b]+\frac{x^2-2xy+y^2}{4}=0\\
\Rightarrow x^2-2xy+y^2=0\\
\Rightarrow x=y
\end{equation}
So does that mean for all quadratic equation it satisfies this functional equation?.
| I'm assuming here the equation has to be satisfied for all $x,y$. Showing that $f$ being quadratic implies $x=y$ means that $f$ cannot be quadratic, as we could choose $x\neq y$.
As the problem is currently stated there is no such $f(x)$. First, let $x=a, y=b$.
The equation becomes
$$f(b) - f(a) - (b-a)f'\left(\frac{a+b}{2}\right)+\frac{(a-b)^2}{4} = 0$$
Then, letting $x=b, y=a$, we have
$$f(a) - f(b) - (a-b)f\left(\frac{b+a}{2}\right)+\frac{(b-a)^2}{4} = 0$$
Adding the two together, we get
\begin{align}
&\frac{(a-b)^2}{2} = 0\\
\implies& a=b
\end{align}
Which means we have to choose $x=y$ regardless of $f$. If we relax the problem so that the equation only has to be satisfied for $x=y$, it becomes trivial as you have shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.
As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.
| Hint for a more direct approach not involving limits (at least not explicitly): if $x$ is large enough, then you can conclude
\begin{align*}
3x & < (0.01) x^2; \\
1 & < (0.01) x^2; \\
x^2 & < (0.01) x^3; \mathrm{~and} \\
1 & < (0.01) x^3.
\end{align*}
For such $x$, you will have:
$$5x^2 + 3x + 1 < 5x^2 + (0.01) x^2 + (0.01) x^2 = (5.02) x^2$$
and
$$x^3 - x^2 - 1 > x^3 - 0.01 x^3 - 0.01 x^3 = (0.98) x^3.$$
Therefore, you will be able to conclude:
$$\frac{5x^2 + 3x + 1} {x^3 - x^2 - 1} < \frac{(5.02) x^2}{(0.98) x^3} = \frac{5.02}{0.98} \cdot \frac{1}{x} < \frac{10}{x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to solve $\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx$
\begin{equation}
\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx
\end{equation}
This is a MCQ question and there are 5 options to choose which are "A.$2\sqrt{3}\pi+9\ln 3$, B.$2\sqrt{7}\pi+8\ln 3$ , C.$2\sqrt{3}\pi+8\ln 3$ , D.$2\sqrt{2}\pi+2\ln 3$, E.Other solution."
How do you solve this integral?
This appears in MCQ test, so I think there should be a trick to solve this without using too much force. The test have 30 questions and 2 hours, so each question should be finished under 4mn ( I doubted it though). Here is my solution that I spend around 3h to solve it. ( Sorry for not using latex and sorry for the inconveniences )Photo
I appreciate any solution tricks. Thanks!
| Here are a few shortcuts. With $t =\sin x$
\begin{equation}
I=\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx=36\int_0^1 \frac {1}{(t^4+t^2+1)^2}dt\tag1
\end{equation}
Note
$$\left(\frac{t-t^3}{t^4+t^2+1}\right)’
= \frac{t^6+5}{(t^4+t^2+1)^2}-\frac{4}{t^4+t^2+1}
$$
Integrate both sides over $(0,\infty)$
\begin{align}
\int_0^\infty \frac{4}{t^4+t^2+1}dt &= \int_0^\infty \overset{t\to 1/t} {\frac{t^6}{(t^4+t^2+1)^2} }dt
+ \int_0^\infty \frac{5}{(t^4+t^2+1)^2}dt\\
&= \int_0^1\frac{6}{(t^4+t^2+1)^2}dt +\int_1^\infty \frac{6}{(t^4+t^2+1)^2}dt\tag2
\end{align}
Similarly, integrate over $(0,1)$
\begin{align}
\int_0^1\frac{4}{t^4+t^2+1}dt=\int_0^1\frac{5}{(t^4+t^2+1)^2}dt +\int_1^\infty \frac{1}{(t^4+t^2+1)^2}dt \tag3
\end{align}
Combine (2) and (3) to obtain
\begin{align}
\int_0^1\frac{dt}{(t^4+t^2+1)^2}= \int_0^1\frac{dt}{t^4+t^2+1}
-\frac14 \int_0^\infty \frac{dt}{t^4+t^2+1}= \frac14\ln3+\frac{\pi}{6\sqrt3}
\end{align}
Plug into (1)
$$I= 9\ln3+ 2\sqrt3{\pi}$$
P.S. The last integral is carried out as follows
$$\int \frac{1}{t^4+t^2+1}dt= \frac12 \int\frac{1-t^2}{t^4+t^2+1}dt
+ \frac12 \int\frac{1+t^2}{t^4+t^2+1}dt \\
= \frac12 \int\frac{d(t+\frac1t)}{(t+\frac1t)^2-1}
+ \frac12 \int\frac{d(t-\frac1t)}{(t-\frac1t)^2+3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Help evaluating an interesting series. I'm trying to evaluate the sum
$$
s = 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\cdots.
$$
The hint I was given was $\frac{1}{2n+1} = \int_0^1 x^{2n} \, dx$. I'm familiar with the trick of writing the $n$th term of series with a definite integral and then interchanging the order of integral and sum, but the signs on the terms are making this difficult for me. If this were a normal alternating series, we could do something like
\begin{align*}
\sum_{n=0}^{\infty}(-1)^n\frac{1}{2n+1} &= \sum_{n=0}^\infty\int_0^1 (-1)^n x^{2n} \, dx\\
&= \int_0^1 \left(\sum_{n=0}^{\infty} (-x^2)^{n}\right)dx\\
&= \int_0^1 \frac{1}{1+x^2} \, dx = \arctan(1)-\arctan(0) = \frac{\pi}{4}.
\end{align*}
I can't however seem to proceed with the strange alternating series. I know that $(-1)^{(n^2+n+2)/2}$ gives the right pattern of $++--++--\dots$, but that doesn't combine well with the $x^{2n}$ term in the integrand after exchanging sum and integral.
Any thoughts?
Edit: If I use the imaginary unit $i$, then I can write a related sum
$$
\tilde{s} = 1+\frac{1}{3}i-\frac{1}{5}-\frac{1}{7}i+\frac{1}{9}+\frac{1}{11}\cdots.
$$
Using the above trick gives
$$
\sum_{n=0}^{\infty}(i)^{n}\frac{1}{2n+1} = \sum_{n=0}^{\infty}\int_{0}^{1}(i)^nx^{2n}dx = \int_{0}^{1} \sum_{n=0}^{\infty}(ix^2)^n =\int_{0}^{1}\frac{1}{1-ix^2}dx
$$
Which I imagine I can do with some type of partial fraction decomposition.
My hope is that from this new sum I can recover my old sum by adding real and imaginary parts. My only concern is that this would essentially be a rearrangement of the series (adding the even indexed terms together and then the odd indexed terms separately) and this series doesn't converge absolutely so I'm worried that this isn't rigorously justified.
| $$ \left( 1+x^2 - x^4 - x^6 \right) \left(1 + x^8 + x^{16} + x^{24} + x^{32} + x^{40} + x^{48} \cdots \right) $$
$$ \frac{1+x^2-x^4-x^6}{1-x^8} = \frac{(1-x^4)(1+x^2)}{(1-x^4)(1 + x^4)} = \frac{1+x^2}{1 + x^4}$$
which can be integrated by partial fractions; note that
$$ x^4 + 1 = \left(x^2 - x\sqrt 2 + 1 \right) \left(x^2 + x \sqrt 2 + 1 \right) $$
We are looking for
$$ \int \frac{1}{1+(x \sqrt 2 + 1)^2} + \frac{1}{1+(x \sqrt 2 - 1)^2} dx $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integral of $1/x^2$ without power rule I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents
\begin{equation*}
\int \frac{1}{x^2} \; dx
\end{equation*}
When using integration by parts, you end up with the same integral in the rhs so it seems out of luck
\begin{equation*}
\int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}
This question is inspired by this blackpenredpen's video
Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write
\begin{equation*}
\int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}
you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.
The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that
\begin{equation*}
\int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}
and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)
| General way:
$$ x^n \cdot x^{-n} = 1$$
$$ nx^{n-1} x^{-n} + x^n \frac{d}{dx} x^{-n}=0$$
$$ \frac{n}{x^{n+1} } + \frac{d}{dx} x^{-n} = 0$$
Or,
$$ \frac{d}{dx}x^{-n} = \frac{-n}{x^{n+1} }$$
integrating both sides with $x$,
$$ x^{-n} +C = \int \frac{-n}{x^{n+1} } dx$$
Or,
$$ \frac{-n}{x^n} +C' = \int \frac{1}{x^{n+1} } dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 11,
"answer_id": 10
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In how many ways can we distribute $26$ shirts into $5$ washing machines such that at least one washing machine contains $6$ shirts or more? My attempt:
Choose a machine in $5$ ways. Then choose six shirts out of $26$ to put in the chosen machine which can be done in $\displaystyle{\binom{26}{6}}$ ways. Distribute the rest of the shirts into five machines in $5^{20}$ ways. In all there $\displaystyle{5\binom{26}{6}5^{20}}$ such possibilities.
Now, pick any two machines out of five in $\displaystyle{\binom52}$ ways. Then, first choose six shirts out of $26$ and for those choices pick another six out of $20$. That is, $\displaystyle{\binom{26}{6}\binom{20}{6}}$ choices. Next, distribute $14$ shirts into the five machines in $5^{14}$ ways. In all, there are $\displaystyle{\binom52\binom{26}{6}\binom{20}{6}5^{14}}$ such possibilities.
Continuing this way, our third and fourth cases should be $\displaystyle{\binom53\binom{26}{6}\binom{20}{6}\binom{14}{6}5^8}$ and $\displaystyle{\binom54\binom{26}{6}\binom{20}{6}\binom{14}{6}\binom{8}{6}5^2}$ possibilities, respectively.
Finally, we have
$\displaystyle{5\binom{26}{6}5^{20} + \binom52\binom{26}{6}\binom{20}{6}5^{14} + \binom53\binom{26}{6}\binom{20}{6}\binom{14}{6}5^8 + \binom54\binom{26}{6}\binom{20}{6}\binom{14}{6}\binom{8}{6}5^2}$.
Please see (and comment) if the reasoning above is correct. Thanks.
Edit:
This problem in its entirety reads like this below (if that helps). I rephrased the problem cause counting the number of favorable outcomes is the difficult part of such problems:
What is the probability that, when $26$ shirts are distributed into $5$ washing machines, that at least one washing machine contains $6$ shirts or more?
| As mentioned, the strong pigeonhole principle says that if you have more than $kn$ objects to distribute into $n$ locations, at least one location will receive more than $k$ objects. Here $k=n=5$ and $26>5\times 5$, allowing us to be certain that there are more than $5$ shirts in some washing machine.
However to show this "the hard way", let's try sharing the shirts among washing machines. I'll assume that the shirts are indistinguishable and the machines are labelled. Then we can split the shirts between machines via the sticks-and-stones principle in $\binom{26+4}{4}$ ways, total $$\binom{30}{4} = 27405$$
Then using inclusion-exclusion, we can see how many of these ways will break the constraints of only 5 shirts per machine. Pre-allocating $6$ shirts to each machine in turn gives $5\binom{24}{4}$, then pre-allocating $6$ shirts to pairs of machines gives $\binom{5}{2}\binom{18}{4}$, then triplets of machines $\binom{5}{3}\binom{12}{4}$ and finally four machines at once with pre-broken constraint is $\binom{5}{4}\binom{6}{4}$. We can't break the constraint on all five machines at once.
So using inclusion-exclusion to remove double-counting etc., the count of constraint-breaking options is \begin{align}
\binom{5}{1}\binom{24}{4}-\binom{5}{2}\binom{18}{4}+\binom{5}{3}\binom{12}{4}&-\binom{5}{4}\binom{6}{4} \\
&= 5\cdot 10626 -10\cdot 3060 + 10\cdot 495 - 5\cdot 15\\
&= 53130-30600+4950-75 \\
&= 27405
\end{align}
Certainty.
| {
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"url": "https://math.stackexchange.com/questions/3991892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving continuity of $f(x)=\sqrt{1−x^3}$ on $[0,1]$ I have to prove the continuity of the function with $\epsilon, \delta$. For this I have to prove that the function is continuous at every $x \in [0, 1]$ by proving:
$$\forall 0 < \epsilon, \exists 0 < \delta, \forall a \in [0, 1]: |x - a| < \delta \implies \left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right| < \epsilon.$$
This is where I am right now:
\begin{align}
\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|&= \frac{\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{x^3 - a^3}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{(x - a)(x^2 + ax + a^2)}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3 + 1 - a^3}\right|}
\end{align}
Any hint is welcome.
|
$$\frac{\mid x^3-a^3\mid}{\mid\sqrt{1-x^3}+\sqrt{1-a^3}\mid}$$
I'm taking it up from here...
Fix a $\delta'>0$. For simplicity, take $\delta'=1$. Then, $\forall\mid x-a\mid<1$,
$$\frac{\mid x-a\mid\mid x^2+ax+a^2\mid}{\mid\sqrt{1-x^3}+\sqrt{1-a^3}\mid}\leq\frac{\mid x-a\mid\mid x^2+ax+a^2\mid}{\mid\sqrt{1-a^3}\mid}<\frac{\mid x-a\mid\mid \left(a+1\right)^2+a\left(a+1\right)+a^2\mid}{\mid\sqrt{1-a^3}\mid}$$
Thus, for $\mid f(x)-f(a)\mid<\epsilon$, $\mid x-a\mid<\frac{\epsilon\mid\sqrt{1-a^3}\mid}{\mid \left(a+1\right)^2+a\left(a+1\right)+a^2\mid}$.
Now just take $\delta:=\min\left\{1,\frac{\epsilon\mid\sqrt{1-a^3}\mid}{\mid \left(a+1\right)^2+a\left(a+1\right)+a^2\mid}\right\}$, and you've found a $\delta>0$, for which the conditions of continuity hold for any arbitrary $\epsilon>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Substituting $x = \sec(u)$ to evaluate $\int\sqrt{x^2-1}\ dx$: but why is $\tan(u) = \sqrt{x^2 - 1}$ rather than $\tan(u) = -\sqrt{x^2 - 1}\ $? I tried evaluating $\int\sqrt{x^2-1}\ dx$ using the substitution $x = \sec(u).$
I know my method gets to the same answer as wolframalpha, namely:
$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C,\quad (*)$
but there is one step I can't justify.
When I got to $\int \tan^2(u)\sec(u)\ du = \frac{1}{2}\left(\ \tan(u)\sec(u) - \ln\left|\sec(u)+\tan(u)\right|\ \right) + C,$
I then have to substitute stuff back in in terms of $x$.
Now $x=\sec(u) \implies \tan^2(u) = x^2 - 1$. But I don't see how this implies $\tan(u) = \sqrt{x^2 - 1}$.
Comparing the graphs of $\sec(u)$ and $\tan(u)$, I don't see why not: $\ \tan(u) = -\sqrt{x^2 - 1}$, which would give:
$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ -x \sqrt{x^2-1} - \ln\left|x - \sqrt{x^2-1}\right|\ \right) + C,\quad (**)$
which is a different answer than $(*)$ ?
Now I noticed that $(*) = -(**)\ $ (ignoring the $C \to -C)$.
I can see from the graph of $\sqrt{x^2-1}$ that for $x>1$, the definite integral $\int^x_1\sqrt{t^2-1}\ dt = (*),$ and for $x<-1,\ \int^{-1}_x\sqrt{t^2-1}\ dt = (**)$
So is the indefinite integral sort of poorly defined, or would you say it is:
$\int\sqrt{x^2-1}\ dx = \pm \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C\ $ ?
| Another method to do this is, use hyperbolic functions.
Consider the change of variables $x=\cosh u$. Then for all $t]x\geq 1$
$$\int_1^x\sqrt{t^2-1}dt=\int_0^{\cosh^{-1}x}\sinh^2 u \;du$$
Since $\cosh(2u)=2\sinh^2u+1$, we have
$$\int_1^x\sqrt{t^2-1}dt=\frac12\int_0^{\cosh^{-1}x}\cosh(2u)-1 \;du=\frac14\sinh(2u)-\frac u2$$
Now,
$$\frac14\sinh(2u)-\frac u2=\frac12\left(\sinh u\cosh u-u\right)=\frac x2\sqrt{x^2-1}-\cosh^{-1}(x)$$
Hence,
$$\int_1^x\sqrt{t^2-1}dt=\frac x2\sqrt{x^2-1}-\cosh^{-1}(x)$$
Note that, for $x\geq 1$, $\cosh^{-1}x=\ln(x+\sqrt{x^2-1})$.
Similarly, for all $x\leq -1$ we have,
$$\int_{-1}^x\sqrt{t^2-1}dt=\frac x2\sqrt{x^2-1}+\cosh^{-1}(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Help in two improper integrals For a physics problem, I need to evaluate two integrals:
$$I_n=\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty} \frac{x^2 \sin^2x}{(n^2\pi^2-x^2)^2} dx,$$ and
$$J_n=\int_{0}^{\infty} \frac{x^2 \cos^2x}{((n+1/2)^2\pi^2-x^2)^2} dx, n=1,2,3..$$
At Mathematica the value of these integral turns out to be $\pi/4$ which is independent of $n$.
$I_n$ has got a pole of order 2 at $x=n\pi$ and when I calculate the residue at $x=n \pi$ is zero. But on the other hand I find that $\lim_{x \to n\pi} f(x)=0,$ so $x=n\pi$ may not be a pole of $f(x)$. I face the same problem in the case of $J_n$. Also $I_0=\pi/2$ is a standard integral. In any case, I want to evaluate both $I_n$ and $J_n$. Please help me.
| Let is use $$\int_{n\pi}^{\infty} \frac{\sin^2 t}{t^2} dt=\frac{\pi}{2}-\text{Ei}(2n \pi)$$
Then
$$I_n=\int_{0}^{\infty} \frac{x^2 \sin^2 x}{x^2-n^2\pi^2} dx$$
$$=\frac{1}{4}\int_{0}^{\infty} dx \left(\frac{\sin^2 x}{(x-n\pi)^2}+\frac{\sin^2 x}{(x+n\pi)^2}+\frac{\sin^2 x}{4n\pi(x-n\pi)}-\frac{\sin^2 x}{4n\pi(x+n\pi)}\right)$$
$$I_n=\frac{1}{4}[\pi/2-\text{Ei}(2n\pi)+\pi/2+\text{Ei}(2n\pi)]+\int_{-n\pi}^{n\pi} \frac{\sin^2 t}{t}dt=\frac{\pi}{4} $$
Similarly
$$J_n=\int_{0}^{\infty} \frac{x^2 \cos^2 x}{x^2-m^2\pi^2} dx, m=n+1/2$$
$$=\frac{1}{4}\int_{0}^{\infty} dx \left(\frac{\cos^2 x}{(x-m\pi)^2}+\frac{\cos^2 x}{(x+m\pi)^2}+\frac{\cos^2 x}{4m\pi(x-m\pi)}-\frac{\cos^2 x}{4m\pi(x+m\pi)}\right)$$
$$I_n=\frac{1}{4}[\pi/2-\text{Ei}(2m\pi)+\pi/2+\text{Ei}(2m\pi)]+\int_{-m\pi}^{m\pi} \frac{\sin^2 t}{t}dt=\frac{\pi}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to maximize a function when you cannot solve gradient = 0 I have to maximize this function : $$ f(x,y) = a\sqrt{x} + b\sqrt{y}$$
$$ a, b \in{R+*} $$
Knowing that $$ 0 ≤ x ≤ 2 − 2y $$ with $$ 0 ≤ y ≤ 1 $$
I said that f is a linear combination of 2 concave functions so it has a maximum (for 0<=x<=2-2y and 0<=y<=1). But because a,b are positive real numbers, I cannot solve $$ \frac{a}{2\sqrt{x}} = 0 $$
And it's the same for y.
How do I deal with such a situation. Thank you very much !!
| By cauchy-schwarz-inequality, $$a\sqrt{x} + b\sqrt{y} = (\sqrt2a)(\sqrt{x}/\sqrt2) + b\sqrt{y} \le \sqrt{(2a^2+b^2)(x/2 + y)} \le \sqrt{2a^2+b^2}$$
due to the constraint $x \le 2 − 2y \iff (x+2y)/2 \le 1$, with equality holds iff $$\frac{\sqrt2a}{b} = \frac{\sqrt{x}/\sqrt2}{\sqrt{y}} \iff 2a \sqrt{y} = b \sqrt{x}$$
and $x + 2y = 2$.
\begin{align}
2a \sqrt{y} &= b \sqrt{2 - 2y} \\
2a^2 y &= b^2 (1 - y) \\
y &= \frac{b^2}{2a^2 + b^2} \in (0,1). \tag{$a,b \in \mathbb{R}_+^*$} \\
x &= 2(1-y) = \frac{4a^2}{2a^2 + b^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Divisibility by n for 3- digit number We have 3 numbers:
$\; \overline{xyz}=100x+10y+z$, $\; \overline{yzx}$ and $\; \overline{zxy}$, all 3 divisible by the same number $n$.
Prove that the number $P=x^3+y^3+z^3-3xyz$ is also divisible by n.
$n \mid 100x+10y+z$
$n \mid 100y+10z+x$
$n \mid 100z+10x+y$
$n \mid 111x+111y+111z \Leftrightarrow n \mid 111(x+y+z)$ but since 111 is not a prime, we can't say for sure if $n \mid (x+y+z)$.
Also $P = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$.
So if I managed to prove that $n \mid (x+y+z)$ I think it would be sufficient, right?
Can you help me?
| COMMENT.-We have the linear system $$\begin{pmatrix}100&10&1\\1&100&10\\10&1&100\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}na\\nb\\nc\end{pmatrix}$$ whose determinant is equal to
$\Delta=998001=3^6\cdot37^2$. Calculations give
$$x=\frac{n999(10a-b)}{3^6\cdot37^2}=\frac{n(10a-b)}{3^3\cdot37}(\text{ note that }3^3\cdot37=\sqrt{\Delta})$$ and
$$y=\frac{n(10b-c)}{3^3\cdot37}\\z=\frac{n(10c-a)}{3^3\cdot37}$$
Therefore we have after calculations $$x^3+y^3+z^3-3xyz=\frac{n^3}{\Delta}(a^3+b^3+c^3-3abc)$$ This shows that if the proposition is true then $$\frac{n^2}{\Delta}(a^3+b^3+c^3-3abc)\in \mathbb Z$$ This should not be difficult to prove while holding clearly from the system the equality $n(a+b+c)=3\cdot37(x+y+z)$.
| {
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"url": "https://math.stackexchange.com/questions/4000916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)=\frac{1}{\pi}\cosh\frac{\pi\sqrt{3}}{2}$ First I rewrite $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)$ as $\prod_{n=1}^{\infty}\left(\frac{1+n^{3}}{n^{3}}\right)$, then by factor out polynomial I get $\prod_{n=1}^{\infty}\left(\frac{(1+n)(n^{2}-n+1)}{n^{3}}\right)$ which is a problem because I can't factor any further which makes me stuck on this step I would hope for any help.
| [Note] : I shall use the following identites :
*
*Identity $(1)$ : $\quad\displaystyle\left|\Gamma\left(\frac{1}{2}+\beta i\right)\right|=\frac{\pi}{\cosh(\pi\beta)}$
*Identity $(2)$ :$\quad\displaystyle\lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{\Gamma(n+\beta)}\approx n^{\alpha-\beta}$
*Identity $(3)$ : $\quad\displaystyle\prod_{k=1}^{n}(x+k)=\frac{\Gamma(n+x+1)}{\Gamma(x+1)}$
We proceed as follow :
\begin{align*}
\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)&=\prod_{n=1}^{\infty}\left(\frac{n^{3}+1}{n^{3}}\right)\\ \\
&=\prod_{n=1}^{\infty}\left[\frac{(1+n)(n^{2}-n+1)}{n^{3}}\right] \\ \\
&=\lim_{\gamma\to\infty}\prod_{n=1}^{\gamma}\left[\frac{(n+1)\left(n+\frac{-1+i\sqrt{3}}{2}\right)\left(n+\frac{-1-i\sqrt{3}}{2}\right)}{n^{3}}\right] \\ \\
&=\lim_{\gamma\to\infty}\frac{\Gamma(\gamma+2)\cdot\displaystyle\frac{\Gamma\left(\gamma+\displaystyle\frac{1+i\sqrt{3}}{2}\right)}{\Gamma\left(\frac{1+i\sqrt{3}}{2}\right)}\displaystyle\cdot\frac{\Gamma\left(\gamma+\frac{1-i\sqrt{3}}{2}\right)}{\Gamma\left(\frac{1-i\sqrt{3}}{2}\right)}}{\Gamma(\gamma+1)\Gamma(\gamma+1)\Gamma(\gamma+1)} \\ \\
&=\frac{1}{\displaystyle\underbrace{\Gamma\left(\frac{1+i\sqrt{3}}{2}\right)\Gamma\left(\frac{1-i\sqrt{3}}{2}\right)}_{\text{Reflection}}} \\ \\
&=\frac{1}{\displaystyle\frac{\pi}{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}} \\ \\
&=\frac{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}{\pi}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Is the matrix $A^4+A^3-3A^2-3A$ invertible?
Let $A$ be a real $5 \times 5$ matrix satisfying $A^3-4A^2+5A-2I=O$. Is the matrix $A^4+A^3-3A^2-3A$ invertible?
Consider the polynomial $t^3-4t^2+5t-2=(t-1)^2(t-2)$. The minimal polynomial of $A$ divides $(t-1)^2(t-2)$. Let $p(t)$ be the minimal polynomial of $A$.
If $p(x)=t-1$ or $t-2$, then $A$ is a scalar multiple of the identity and $A^4+A^3-3A^2-3A=-4I \text{ or }6I$, so $A^4+A^3-3A^2-3A$ is invertible.
If $p(t)=(t-1)^2(t-2)$ or $(t-1)(t-2)$, then $A$ is similar to a upper triangular matrix $J$ with diagonal entries $1,2$. Then $A^4+A^3-3A^2-3A=P(J^4+J^3-3J^2-3J)P^{-1}$ and $\det(A^4+A^3-3A^2-3A)=\det(J^4+J^3-3J^2-3J)$. Since $J^4+J^3-3J^2-3J$ is upper triangular and the diagonal entries do not vanish, $\det(J^4+J^3-3J^2-3J) \ne 0$.
If $p(t)=(t-1)^2$, then by the same reasoning, $A^4+A^3-3A^2-3A$ is invertible.
Is there more efficient way to solve this kind of problems or I have to discuss any possible situations every time?
| In general, the eigenvalues (with algebraic multiplicity) of $P(A)$ for any polynomial $P$ are the values $P(\lambda)$ for every eigenvalue $\lambda$ of $A$. In particular, the first condition gives you that all eigenvalues of $A^3-4A^2+5A-2I$ are $0$, so
$$(\lambda-1)^2(\lambda-2)=\lambda^3-4\lambda^2+5\lambda-2=0$$
for every eigenvalue $\lambda$ of $A$, so $A$ can only have the eigenvalues $1$ and $2$. As a result, every eigenvalue of $A^4+A^3-3A^2-3A$ will be
$$\lambda^4+\lambda^3-3\lambda^2-3\lambda$$
for $\lambda\in\{1,2\}$, and you can manually check each $\lambda$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the coefficient of $x^{16}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4$ We are supposed to find the coefficients of it, I wanted to know if my approach is right here. The final answers seems a bit iffy.
$$(1+x+\dots+x^5)^4=\left(\frac{1-x^6}{1-x}\right)^4=(1-x^6)^4(1-x)^{-4}$$
Using the binomial theoerem, I got the following:
$$
(1-x^6)^4=\sum_{k\ge0}(-1)^k\binom{4}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-4}=\sum_{k\ge0}(-1)^k\binom{-4}{k}x^k=\sum_{k\ge0}\binom{4+k-1}{k}x^k
$$
So the $x^{16}$ coefficient is
$$
\binom{4}{0}\binom{8+16-1}{16}-\binom{4}{1}\binom{8+10-1}{10}+\binom{4}{2}\binom{8+4-1}{4}\
$$
$$
= 1(245157)-4(19448)+6(330)\
$$
$$
= 245157-77792+1980\
= 169345
$$
Not too sure if I did this right. I think my math may be off somewhere but can't figure out where.
| $$1+x+x^2+\cdots+x^5=\frac{1-x^{6}}{1-x}=(1-x^{6})(1-x)^{-1}$$
so
$$(1+x+x^2+\cdots+x^5)^4=\left(\frac{1-x^{6}}{1-x}\right)^4=(1-x^{6})^4(1-x)^{-4}=(1-4x^{6}+\cdots)(1+4x+10x^2+20x^3+35x^4+56x^5)$$
There is only one $x^4$ term, and you can see immediately that its coefficient is $35$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Global maximum and minimum on a sphere I have a sphere $x^2+y^2+z^2=R^2$ and i have $T=C(x^2+2y^2+3z^2+2xy+2xz)$ temperature function. I need to find maximum and minimum of this temperature function.
First i tried to calculate $\partial_x f=2x$,
$ \partial_y f=2x$,
$ \partial_z f=2z $
$\partial_x T=2x+2y+2z $, $\partial_y T=4y+2x $, $\partial_z T=6z+2x $
Then I equate them $2x=\lambda(2x+2y+2z)$, $2y=\lambda(4y+2x)$, $2x=\lambda(6z+2x)$
But I cannot find lambda from here. What am I doing wrong?
| $T=C(x^2+2y^2+3z^2+2xy+2xz)$
As $C$ is a constant, we take $f(x,y,z) = x^2+2y^2+3z^2+2xy+2xz$
where $g(x,y,z) = x^2+y^2+z^2 - R^2 = 0$
$f(x,y,z) = \lambda g(x,y,z)$
Taking derivatives wrt $x,y,z$,
$(1-\lambda)x+y+z = 0$ ...(i)
$x + (2-\lambda)y = 0$ ...(ii)
$x + (3-\lambda)z = 0$ ...(iii)
From $(ii)$ and $(iii)$,
$y = -\frac{x}{2-\lambda}, z = -\frac{x}{3-\lambda}$ (for $\lambda \ne 2,3$)
Plugging into $(i)$,
$((1-\lambda)(2-\lambda)(3-\lambda) + 2\lambda - 5)x=0$
So either $x=0$ or
$-\lambda^3+6\lambda^2-9\lambda+1=0$ ($x\ne 0$)
$\lambda \approx 0.12061, 2.3473, 3.5321$.
We already know $y, z$ in terms of $x$ and $\lambda$. Plugging these values and using the constraint $x^2+y^2+z^2=R^2$ will give us possible values of $x$ and from there onx, values of $y, z$. We then need to check these critical points for maxima and minima.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it permitted to use $\tan 90^\circ$ to prove an equation/identity even if it is not defined? defined? Domain of tangent function limits to all real numbers except odd integral multiples of $\frac{\pi}{2}$.
I noticed when proving an equality that using $\tan 90^\circ$ yielded the exact same result. Here what I did:
For a triangle $\mathrm{ABC}$, prove that $$\tan \frac{A}{2}\tan\frac{ B}{2 }+\tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2}\tan \frac{A}{2}=1$$
Since $\mathrm{A+B+C = \pi}$
$$\mathrm{\frac{A+B+C}{2} = \frac{\pi}{2}}$$
I applied identity for sum of angles
$$\tan{(A/2 + B/2 + C/2)}=\frac{\tan \frac{A}{2} +\tan \frac{B}{2} +\tan \frac{C}{2}-\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}{1-\tan \frac{A}{2}\tan\frac{ B}{2 }-\tan \frac{B}{2} \tan \frac{C}{2} - \tan \frac{C}{2}\tan \frac{A}{2}}$$
Then I reciprocated both side to get L.H.S. to be equal to 0 and I was able to prove the given equation.
Why did I arrive at the right conclusion by using the value out of domain for tangent function?
| Just like there is an identity for the tangent of a sum of three angles, there is an identity for the cotangent of a sum of three angles.
Not surprisingly, you can write a version of it just by taking the reciprocal of the formula for the tangent of the sum:
$$
\cot\left(\alpha + \beta + \gamma\right)
= \frac{1 -\tan \alpha \tan\beta - \tan\beta \tan\gamma - \tan\gamma \tan\alpha}
{\tan\alpha + \tan\beta + \tan\gamma - \tan\alpha \tan\beta \tan\gamma}.
$$
Plug in $\alpha = \frac A2,$ $\beta = \frac B2,$ $\gamma = \frac C2$:
$$
\cot\left(\frac A2 + \frac B2 + \frac C2\right)
= \frac{1 - \tan\frac A2 \tan\frac B2 - \tan\frac B2 \tan\frac C2 - \tan\frac C2 \tan\frac A2}
{\tan\frac A2 + \tan\frac B2 + \tan\frac C2 - \tan\frac A2 \tan\frac B2 \tan\frac C2}.
$$
Now just use the fact that $\cot\left(\frac\pi2\right) = 0.$
Intuitively, this seems to be what you were trying to do.
The main difference in the methods is that $\cot\left(\frac\pi2\right) = 0$
is a true equation with well-defined quantities on each side, wherease
$1/\tan\left(\frac\pi2\right) \stackrel?= 0$ is not.
| {
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The segment joining the vertices of altitudes A triangle $ABC$ with sides $AB=15,BC=14,AC=13$ is given. $AA_1=12$ and $BB_1$ are heights. Find $A_1B_1$.
I was thinking about the Ptolemy's theorem: $$A_1B_1\cdot CH=B_1C\cdot HA_1+B_1H\cdot CA_1,$$ where $H$ is the intersection of the heights (the orthocenter). We can use it because $HA_1CB_1$ is cyclic: $$\measuredangle HB_1C+\measuredangle HA_1C=180^\circ.$$ I was able to find that $CA_1=5$. I am not sure this is the easiest solution, because we have to find $5$ more segments. Any thoughts on the problem will be appreciated. Thank you!
| Equate the area of the triangle to find $BB_1 = \frac{12 \times 14}{13}$.
$A_1C = 5$ as you mentioned.
Applying Pythagoras, $B_1C^2 = BC^2 - BB_1^2 \implies B_1C = \frac{70}{13}$
Now find $\cos C = \frac{AC^2 + BC^2 - AB^2}{2 AC \cdot BC} = \frac{5}{13}$
$A_1B_1^2 = A_1C^2 + B_1C^2 - 2 A_1C \ B_1C \cos C = 25 + (\frac{70}{13})^2 - \frac{3500}{13^2} = (\frac{75}{13})^2$
Hence $A_1B_1 = \frac{75}{13}$
| {
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Show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$. For $n\ge 0$, show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$.
Over a well ordered set of non-negative integers, induction can be used.
Base case: $n=0$
$\binom{x}{0} \binom{y}{0} = 1.\frac{(y)!}{(y)! \,\,(0)!}=1$.
Taking the rhs, get :
$\binom{x+y}{0} = 1$.
Induction hypothesis: it is true for natural number $n$.
So, $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$
Inductive step: need show for next natural $n+1$
$\sum_{k=0}^{n+1} \binom{x}{k} \binom{y}{n-k+1} = \binom{x+y}{n+1}$
This should lead to :
$\binom{x+y}{n} + \binom{x}{n+1}\binom{y}{0}= \binom{x+y}{n+1}$
get on lhs:
$\binom{x+y}{n} + 1.\binom{x}{n+1}$
$\implies \frac{(x+y)!}{n!(x+y-n)!} + \frac{x!}{(n+1)!(x-n-1)!}$
Please help to proceed further, or suggest alternative approach.
| There are much better proof strategies in my opinion, but here's an inductive proof of the Vandermonde convolution identity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$X \sim Beta(m,2)$ and $P \left(X\le\dfrac{1}{2}\right)=\dfrac{1}{2}$ find Variance
Let $X$ follow a beta distribution with parameters $m(<0)$ and $2$. If
$P \left(X\le\dfrac{1}{2}\right)=\dfrac{1}{2},$ then $Var(X)$ ?
$(A)\ \ \dfrac{1}{10}\space\ \ \ (B)\ \ \dfrac{1}{20}\ \ \ (C)\ \
\dfrac{1}{25} \ \ \ (D)\ \ \dfrac{1}{40}$
$V(X)=\dfrac{mn}{(m+n)^2(m+n+1)}=\dfrac{2m}{(m+2)^2(m+3)} \ \ \ \ \ \ ---->(1)$
$\int_{0}^{\frac{1}{2}}x^{m-1}(1-x)dx=\left(\frac{x^{m}}{m}\right)_{0}^{\frac{1}{2}}-\left(\frac{x^{m+1}}{m+1}\right)_{0}^{\frac{1}{2}}=\left(\frac{1}{2^{m}m}\right)-\left(\frac{1}{2^{m+1}(m+1)}\right)=\frac{B(m,2)}{2}\ \ \ \ \ \ ---->(2)$
Now I plugged in values by hit and trial
For $m=2$
$\dfrac{2m}{(m+2)^2(m+3)}=\dfrac{1}{20}$
also satisfies $(2)$
$\dfrac{1}{2^2\cdot2}-\dfrac{1}{24}=\dfrac{1}{12}=\dfrac{B(2,2)}{2}$
For $m=3$
$\dfrac{2m}{(m+2)^2(m+3)}=\dfrac{1}{25}$
but
$\dfrac{1}{2^3\cdot3}-\dfrac{1}{2^4 \cdot 4}=\dfrac{1}{12}\ne\dfrac{B(3,2)}{2}$
This question came $2$ marks so there must be an easy way out which I am not able to figure out. Please suggest me alternate method.
| The simple approach is to say that if a $\mathrm{Beta}(a,b)$ distribution has $a>b$ then $\mathbb P(X \le \frac12) \lt \frac12$ and its median is above $\frac12$, while if $a<b$ then then $\mathbb P(X \le \frac12) \gt \frac12$ and its median is below $\frac12$.
This can be proved by comparing the densities at $x$ and $1-x$ and then integrating over the half intervals. When $a=b$ you get $\mathbb P(X \le \frac12) = \frac12$ and a median of $\frac12$ by symmetry.
So here you have $m=2$ and $\mathrm{Beta}(2,2)$ is symmetric with a median of $\frac12$ and a variance of $\frac1{20}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$ Question:
$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $
Answer key:
$\implies \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $
I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:
$\implies (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $
$ \implies (\sin(\theta)-\cos(\theta)-1) \leq0 \because (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $
Doubt:
I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.
I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too
| I have re-written this answer because, with the help of WolframAlpha here, I was surprised to discover
$$0\le \big(3-\sin(2\pi x)\big)\sin(\pi x - \frac{\pi}{4})- \sin(3\pi x +\frac{\pi}{4})\le 2\sqrt{2}\\
\implies (2 n - 1)<x\le \frac{2 \big(\pi n - \frac{3 π}{8}\big)}{\pi} \land n \in\mathbb{ Z}\\
\land \implies \frac{2 \big(\pi n + \frac{\pi}{8}\big)}{\pi}\le x\le\frac{(4 n + 1)}{2} \land n\in\mathbb{Z} $$
What the first "implies" means for $\space 0\le x\le 100\space$ is that there are discreet ranges of values that make this true as in these samples for $1\le n\le 6$
$$(1\le x \le \frac{5}{4}=1.25)\quad
(3\le x \le \frac{13}{4}=3.25)\quad
(5\le x \le \frac{21}{4}=5.25)\quad \\
(7\le x \le \frac{29}{4}=7.25)\quad
(9\le x \le \frac{37}{4}=9.25)\quad
(11\le x \le \frac{45}{4}=11.25)\quad $$
What the second "implies" means is that the following ranges apply for $\space x$ for $1\le n\le 6$.
$$\frac{9}{4}=2.25\le x \le \frac{5}{2}=2.5\qquad
\frac{17}{4}=4.25\le x \le \frac{9}{2}=4.5\\
\frac{25}{4}=6.25\le x \le \frac{13}{2}=6.5\qquad
\frac{33}{4}=8.25\le x \le \frac{17}{2}=8.5\\
\frac{41}{4}=10.25\le x \le \frac{21}{2}=10.5\qquad
\frac{49}{4}=12.25\le x \le \frac{25}{2}=12.5$$
We can see that each value of $n$ generates $\space x\space$ with a total "coverage of $\space 0.5\qquad $ Note that $\quad 1\le n \le 24\implies 1\le x \le 97\quad$ so there are $24\times0.5=12$ "full-numbers-worth" of $0\le x \le 100$ that satisfy $I.\quad$ However, the first "implies" above permits
$n=25\implies 99\le x \le 99.25\space $ and the second "implies" peermits $n=0.0\implies 0.25\le x \le 0.5\quad $. There are a total of $100$ intervals of $0.25$ each so the members run $\frac14$ of the time.
This means that there members swim $\frac34$ of the time and the ratio of $\frac{swim}{run}$ is $\quad 3:1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring $a^4+b^4+(a-b)^4$ I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.
I rewrite it as:
$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.
| We begin with the special case that $a=x$ and $b=1$ to obtain $f(x)=x^4 +(x-1)^4 =2g(x),$ where $$g(x) =x^4-2x^3+3x^2-2x+1.$$ Since $g(x)$ has no real roots, it equals the product of two irreducible quadratic polynomials. Let us guess that these two quadratic polynomials are equal, so that $g(x)=(x^2+cx+1)^2.$ Happily, $c=-1$ works. Therefore,
$$f(x)=2(x^2-x+1)^2.$$
In the above factorization replace $x$ by $a\over b$ and multiply both sides by $b^4$ to get the factorization of the original expression.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using $\sec(x)$ for integral.
Find the undefined integral $\int \frac{\sqrt{x^2+4x}}{x^2}\mathrm{dx}$
$$\displaystyle\int \dfrac{\displaystyle\sqrt{x^2+4x}}{x^2}\mathrm{dx}$$
I tried to rearrange the square root and I got:
$$\sqrt{(x+2)^2-4}$$ and I substitute the $x+2$ with $2\sec(u)$ so indeed I got these two:
$$x+2=2\sec(u) \\
\sqrt{x^2+4x}=2\tan(u) \\
\mathrm{dx}=2\tan(u)\sec(u)\mathrm{du}$$
And the integral turns out like this:
$$\displaystyle \int \dfrac{\tan^2(u)\sec(u)}{(\sec(u)-1)^2}\mathrm{du}$$
And I continued to rearrange the integral:
$$\displaystyle\int \dfrac{1-\cos^2(u)}{\cos(u)(1-\cos(u))^2}\mathrm{du}$$
I apply partial fractions method, saying $\cos(u)=u$ without integral sign and I ended up with:
$$\displaystyle\int \dfrac{1}{\cos(u)}\mathrm{du}+2\displaystyle\int \dfrac{1}{1-\cos(u)}\mathrm{du}$$
One can easily integrate those integrals se I skip the calculating part. In the end I get:
$$\ln\left|\sec(u)+\tan(u)\right|+2(-\cot(u)-\csc(u))+\mathrm{C}$$
I drew a triangle:
Knowing $\sec(u)=\dfrac{x+2}{2}$ says us $\cos(u)=\dfrac{2}{x+2}$
As I calculate I got:
$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{2x-8}{\sqrt{x^2+4x}}+\mathrm{C}$$
Bu the answer key is:
$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{8}{x+\sqrt{x^2+4x}}+\mathrm{C}$$
I have been thinking where my wrong is for five hours but I couldn't find anything. Please if you see any gap tell me. Thanks.
| Hint:
Another way:
For $x>0,$
$$I=\int\dfrac{\sqrt{x^2+4x}}{x^2}\ dx=\int\dfrac{\sqrt{x+4}}{x^{3/2}}\ dx$$
Integrating by parts,
$$I=\sqrt{x+4}\int\dfrac{dx}{x^{3/2}}-\int\left(\dfrac{d\sqrt{x+4}}{dx}\int\dfrac{dx}{x^{3/2}}\right)dx$$
$$=-\sqrt{x+4}\cdot\dfrac{2}{\sqrt x}+\int\dfrac{dx}{\sqrt{(x+2)^2-2^2}}$$
Now observe that $\dfrac8{\sqrt{x^2+4x}+x}=\dfrac{8(\sqrt{x^2+4x}-x)}{4x}=2\sqrt{\dfrac{x+4}x}-2$
| {
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"url": "https://math.stackexchange.com/questions/4020561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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how to continue after this substitution?
Given $abc=1$, prove that $${\frac{a}{a^2+2}}+ {\frac{b}{b^2+2}}+{\frac{c}{c^2+2}} \leqslant 1 $$
I tried substitution $a={\frac{1}{x}},b={\frac{1}{y}},c={\frac{1}{z}}$ but can't finish.
| Let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ we have to prove
$$\sum_{cyc}\frac{xy}{x^2+2y^2}\le 1$$ Now as $x^2+y^2\ge 2xy$ it suffices to prove
$$\sum_{cyc}\frac{x}{2x+y}\le 1 $$ $$\iff \sum_{cyc} \frac{y}{2x+y}\ge 1$$ which is true as
$$\sum_{cyc} \frac{y}{2x+y}= \sum_{cyc} \frac{y^2}{2xy+y^2}\ge \frac{{(x+y+z)}^2}{2xy+y^2+2zx+x^2+2yz+z^2}=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$
Prove that (where $a,b,c>0$) $$\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$$
I have found a proof for this problem but it is very lengthy and is not nice.(Its not using computer though).I shall post it later.
Background:
Here however is a very similar problem
If $a,b,c>0$ then prove that $$\sum_{cyc} \frac{a(a^2+bc)}{b+c}\ge a^2+b^2+c^2$$
There is a beautiful proof of this by Michael Rozenberg (arqady)
WLOG $a\ge b\ge c$ rewrite the inequality as $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\ge 0$$ but $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\geq\frac{a(a-b)(a-c)}{b+c}+\frac{b(b-a)(b-c)}{a+c}=$$
$$=(a-b)(\frac{a(a-c)}{b+c}-\frac{b(b-c)}{a+c})\geq0$$
However this method doesnt work...
I am looking for a clean and smooth proof (without using BW,uvw or complete expanding)
I am not planning to disclose how I proved the inequality as it may spoil the fun!
| We want to show
$$
\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}
$$
The following steps mainly show how this inequality can be maximally simplified.
The final steps then have been proved by the OP Albus Dumbledore himself, and others.
Due to homogeneity, let $a+b+c =1$, then equivalently
$$
\sum_{cyc} \frac{(a-1+1)(a^2+2bc)}{1-a} \ge \frac{1}{2} $$
$$-\sum_{cyc} (a^2+2bc) + \sum_{cyc} \frac{a^2+2bc}{1-a} \ge \frac{1}{2}$$
$$-(a+b+c)^2+ \sum_{cyc} \frac{a^2+2bc}{1-a} \ge \frac{1}{2}$$
$$\sum_{cyc} \frac{a^2-1+1+2bc}{1-a} \ge \frac{3}{2}
$$
$$-\sum_{cyc} (1+a)+\sum_{cyc} \frac{1+2bc}{1-a} \ge \frac{3}{2}
$$
$$\sum_{cyc} \frac{1+2bc}{b+c} \ge \frac{11}{2}
$$
Now we have
$$\frac{2bc}{1-a} =\frac{2bc}{b+c} = b+c -a-1 + \frac{1 -(a^2+b^2+c^2)}{b+c} $$
which leads to
$$(2 -(a^2+b^2+c^2))\sum_{cyc} \frac{1}{b+c} \ge \frac{15}{2}
$$
Now we have isolated a single sum, which is the main benefit of this answer.
This sum can be evaluated:
$$\sum_{cyc} \frac{1}{1-a} = \frac{3 - (a^2+b^2+c^2)}{1 - 2abc - (a^2+b^2+c^2)}
$$
which leaves to show
$$(2 -(a^2+b^2+c^2))(3 -(a^2+b^2+c^2)) \ge \frac{15}{2} (1 - 2abc - (a^2+b^2+c^2))
$$
Let $x = a^2+b^2+c^2$ then we have to show
$$2x^2 + 5x + 30 abc \ge 3
$$
Note that by now, no single change has been made to the original question, since all transformations are equivalences.
The last inequality can be proved (amongst other methods) by Schur, which has been done by the OP Albus Dumbledore himself, and others, see here. $\qquad \Box$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine whether the given points are interior point of the given set Let $B = \{(x,y) \in \Bbb{R}^2 \mid -1 \le x \lt 2, 0 \lt y \le 2 \}
\cup \{(x,y) \in \Bbb{R}^2 \mid 5 \lt x \le 7, y = 1 \}$. Determine whether a point $(0,1)$
is an interior point of $B$.
I got a little bit confuse here, since a point $(0,1)$ exactly lies on vertical line $y=1$, in which it does mean that $(0,1)$ be an interior point of not.
Here's my attempt:
Fix $0 < r = \frac{1}{2}$. Note that
\begin{equation*}
B((0,1), \frac{1}{2}) = \{(x,y) \in \Bbb{R} \mid x^2 + (y-1)^2 < \frac{1}{4} \}.
\end{equation*}
Let $(x,y) \in B((0,1), \frac{1}{4})$. Then,
\begin{align*}
x^2 + (y-1)^2 < \frac{1}{4} \\
x^2 < \frac{1}{4} \wedge (y-1)^2 < \frac{1}{4} \\
-\frac{1}{2} < x < \frac{1}{2} \wedge -\frac{1}{2} < y-1 < \frac{1}{2} \\
-\frac{1}{2} < x < \frac{1}{2} \wedge \frac{1}{2} < y < \frac{3}{2}.
\end{align*}
Hence, $x \in (-\frac{1}{2}, \frac{1}{2}) \subseteq [-1,2)$ and $y \in (\frac{1}{2}, \frac{3}{2})
\subseteq (0,2]$. Thus, forall $(x,y) \in B((0,1), \frac{1}{2})$, we have $(x,y) \in B$.
In another words, $B((0,1), \frac{1}{2}) \subseteq B$. Therefore, there exists $r>0$
such that $B((0,1), \frac{1}{2}) \subseteq B$. Hence, $(0,1)$ is an interior point of $B$.
Am I true? If not, any explain how to show it?
Thanks in advanced.
| Your proof is fine, but verbose. You could also note that $U=(-1,2) \times (0,2)$ is open (open rectangle) and a subset of $B$ so that any point of $U$ is an interior point of $B$, including $(0,1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all $x,y,z \in \mathbb R$ Such that $\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y=3 $ Find $x,y,z \in \mathbb R$ Such that :
$$\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y =3$$
My Attempt:
I’ve turned this into a system of equations :
$$\cases{\frac{1}{x}+y+z =3 \\ \frac{1}{y}+x+z =3 \\ \frac{1}{z}+x+y=3 } \iff \cases{1+xy+zx=3x \\1+xy+zy=3y \\1+xz+yz=3z}$$
Multiplying some equations by $-1$ and adding them together we get:
$$z(y-x)=3(y-x)\iff z=3$$
Notice that $y\ne x$.
You can play a little bit with $x,y$ ’s values, you will end up with :
$$(x,y,z) \in \{(3,\frac{-1}{3},3), (-3, \frac{1}{3},3)\}$$
Edit: user pointed out in comments that the first triple of my solution doesn’t work in all cases, but i don’t know why.
And the values of $x,y,z$ Can swap places because of the symmetry in the equations.
My question is what would happen if $x=y$.
And there is one more thing to notice is that one of the obvious solutions is $$x=y=z=1$$
Thank you.
| Given such $x,y,z\in\Bbb{R}$, in particular you have
$$\frac1x+y+z=\frac1y+x+z,$$
and hence also $x-\tfrac1x=y-\tfrac1y$. By symmetry we see that
$$x-\frac1x=y-\frac1y=z-\frac1z=c,$$
for some constant $c$, and so $x$, $y$ and $z$ are all roots of
$$T^2-cT-1=0.\tag{1}$$
A quadratic polynomial has at most two real roots, so without loss of generality we have $x=y$. We distinguish two cases:
*
*If $x=y=z$, then we get the identity
$$\frac1x+x+x=3,$$
and hence $2x^2-3x+1=0$. This quadratic factors as
$$2x^2-3x+1=(2x-1)(x-1),$$
yielding the two solutions $(x,y,z)=(1,1,1)$ and $(x,y,z)=(\tfrac12,\tfrac12,\tfrac12)$.
*If $z\neq x$, then also $z\neq y$ and so $xz=yz=-1$ because $x$ and $z$ are distinct roots of the quadratic $(1)$. Then from
$$\frac1z+x+y=3,$$
we find that also
$$3z=1+xz+yz=1+(-1)+(-1)=-1,$$
which shows that $z=-\tfrac13$ and hence $x=y=3$. This yields the three solutions
$$(x,y,z)=(3,3,-\tfrac13),\qquad (x,y,z)=(3,-\tfrac13,3),\qquad(x,y,z)=(-\tfrac13,3,3).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the chance to get two pairs from a hand of five cards if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards What is the chance to get two pairs if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards, thus I have 55 cards? Can anybody help me !!
Is my solution right?
$$\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1} + \binom{13}{2}\binom{5}{2}\binom{5}{2}\binom{11}{1}\binom{5}{1} + \binom{13}{2}\binom{6}{2}\binom{6}{2}\binom{11}{1}\frac{\binom{6}{1}}{\binom{55}{5}}$$
First, I chose $2$ ranks of $13$ and then $2$ suits of $4$ (this is for hearts and clovers). Then I chose $2$ ranks of $13$ and then $2$ suits of $5$ suits (this is for diamonds, but I am not sure). Then I chose $2$ ranks of $13$ and then $2$ suits of $6$ suits (this is for spades, but I am not sure).
Lastly, I divided all on $\binom{55}{5}$
| Your approach seems fine but there are a few mistakes.
Number of ways to choose two pairs -
i) A pair of queen and a pair of king
$\displaystyle \binom{6}{2}\binom{5}{2}\binom{44}{1}$
ii) A pair of queen and a pair from other $11$ ranks (not king)
$\displaystyle \binom{6}{2} \binom{11}{1} \binom{4}{2} \binom{45}{1}$
iii) A pair of king and a pair from other $11$ ranks (not queen)
$\displaystyle \binom{5}{2} \binom{11}{1} \binom{4}{2} \binom{46}{1}$
iv) Two pairs from other $11$ ranks (other than queen and king)
$\displaystyle \binom{11}{2} \binom{4}{2} \binom{4}{2} \binom{47}{1}$
Now the probability is sum of $(i), (ii), (iii), (iv)$ divided by $\displaystyle {55 \choose 5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$ for $1 \le x \le n-1$.
Let $n \ge 3$, $x \in \Bbb R$ such that $1 \le x \le n-1$. Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$.
For $n=3$, $1 \le x \le 2$ we want to show $|x(x-1)(x-2)(x-3)|=x(x-1)(2-x)(3-x) \le 1$.
We see that some large bounds are easy to guess:
$$
x(x-1)(2-x)(3-x) \le 2 \times 1 \times 1 \times 2=4
$$
but not precise enough. Another try:
$$
x(x-1)(2-x)(3-x) \le x(3-x) \le 9/4
$$
the last inequality is from the study of $f(x)=x(3-x)$.
Same problem.
For $n=3$ and $1 \le x \le 2$ let $0 \le y=x-1 \le 1$.
We get using AM-GM
\begin{align}
|x(x-1)(x-2)(x-3)| &= |(y+1)(y)(y-1)(y-2)|\\
&=y(1-y^2)(2-y)\\
&\le (\frac{3-y^2}{3})^3\\
&\le 1
\end{align}
For the general case I tried to use the same approach:
let $1 \le x \le n-1$ ie. $0 \le y = (x-1)/(n-2) \le 1$. We have
\begin{align}
\prod_{k=0}^n |x-k|
&=
\prod_{k=0}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l |(n-1)y +1-k| \prod_{k=l+1}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l ((n-1)y +1-k) \prod_{k=l+1}^n (k-1-(n-1)y)\\
\end{align}
where $l = \lfloor (n-1)y+1\rfloor$. Applying AM-GM doesn't really help here. Suppose $y<1$ ie. $l<n$, we see that
$$
\prod_{k=0}^l ((n-1)y +1-k) \le \prod_{k=0}^l (n-k)=\frac{n!}{(n-l-1)!}
$$
and
$$
\prod_{k=l+1}^n (k-1-(n-1)y) \le \prod_{k=l}^{n-1} (k) = \frac{(n-1)!}{(l-1)!}
$$
So
$$
\prod_{k=0}^n |x-k|\le
\frac{(n-1)!}{(l-1)!}\frac{n!}{(n-l-1)!}
$$
The bound is too big...
I also tried this approach: let $f(x)= \prod_{k=0}^n (x-k)$, $f'(x)=f(x)(\sum_{k=0}^n \frac{1}{x-k})$.
I'm trying to find informations about $x$ such that $f'(x)=0$. Let $x \in \Bbb R - [|0,n|]$. So $\sum_{k=0}^n \frac{1}{x-k}=0$. But I'm stuck here.
My questions are:
*
*Does someone have a hint or a proof?
*Can someone give me some advice on how to handle such problems during an exam?
Thanks :)
| It's easy to notice that the function $f(x) =\prod_{k=0}^n |x-k|$ is a symmetric function at $x = \frac{n}{2}$ (in other words, $f(x) = f(n-x)$). So, it suffice to prove $f(x) \le \frac{(n-1)!}{2}$ in the interval $x \in [1, \frac{n}{2}]$.
For $x \in [p,p+1)$ with $p\in \{0,...,[\frac{n}{2}]-1\}$, we have
$$ f(x+1) = \prod_{0 \le k \le n}|(x+1)-k| = \frac{x+1}{|x-n|}\prod_{0 \le k \le n}|x-k| = \frac{x+1}{|x-n|} f(x)$$
As $x \in [p,p+1)$ with $p\in [0,...,([\frac{n}{2}]-1)]$, we have then $x < \frac{n-1}{2}$, so $\frac{x+1}{|x-n|} <1$ or $f(x+1) < f(x)$.
We deduce that for all $p\in \{0,...,[\frac{n}{2}]-1\}$
$$\max_{x \in [p+1,p+2)}f(x) = \max_{x \in [p,p+1)}f(x+1) < \max_{x \in [p,p+1)}f(x)$$
or
$$\max_{x \in [1,\frac{n}{2}]}f(x) = \max_{x \in [1,2)}f(x)$$
So, it suffices to prove $f(x) \le \frac{(n-1)!}{2}$ in the interval $x \in [1, 2)$.
In this interval, we have
$$f(x) = \prod_{0 \le k \le n}|x-k| = x(x-1)(2-x) \prod_{3 \le k \le n}(k-x) = x(x-1)(4-2x) \frac{1}{2}\prod_{3 \le k \le n}(k-x) \tag{1}$$
By AM-GM inequality,
$$x(x-1)(4-2x) \le \left(\frac{x + (x-1) + 2(2-x)}{3} \right)^3 = 1 \tag{2}$$
and because $x \in [1, 2)$, then
$$\prod_{3 \le k \le n}(k-x) \le \prod_{3 \le k \le n}(k-1) = (n-1)! \tag{3}$$
From (2) and (3), we deduce that (1) holds true. Hence, for all $x\in [1,n-1]$
$$f(x) \le \frac{(n-1)!}{2}$$
Note: The equality can't occur because the equality in (2) occurs if $x = x-1 = 4-2x$, which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$.
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.
I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?
| It's easy to see if $a= 2^km; m$ odd and $b = 2^jn;n $odd that $k=j$ (we can factor out $2^{\min(3j,3k)}$ from $a^3 + b^3$) so wolog we can assume $a,b$ are both odd.
If $a \ne b$ and wolog $a < b$ we have $a= 2^x -d;b =2^x + d$ where $d$ is odd and so
$(a^2 -ab+b^2) = (2^x-d)(2^x-d) - (2^x+d)(x^2 - d) +(2^x+d)(2^x+d)=$
$4\cdot 2^{2x} +3d^2$ is .... odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
If positive reals $a$ and $b$ satisfy $a\sqrt{a}+b\sqrt{b}=183, a\sqrt{b}+b\sqrt{a}=182$, find $\frac{9}{5}(a+b)$. Question from Math Olympiad:
Suppose $a, b$ are positive real numbers such that $a\sqrt{a} + b\sqrt {b} = 183$ and $a\sqrt{b} + b\sqrt {a} = 182$. Find $\frac{9}{5}(a+b)$.
My approach:
$a\sqrt{a} + b\sqrt {b} = 183$
$a\sqrt{b} + b\sqrt {a} = 182$
Therefore $(\sqrt{a} + \sqrt{b})(a+b) = 365$
I only see two integral possibilities:
$1 \times 365 $ and $5 \times 73 $
And on putting either of them, I am getting $657$ and $131.4$, whereas the answer is an integral $73$.
Please help me with the general method to go about the same.
This is the picture from the source.
| Hint: Let $x=\sqrt a$ and $y=\sqrt b$. You know that
$$x^3+y^3=183$$
and
$$xy(x+y)=182.$$
Since
$$x^3+3xy(x+y)+y^3=(x+y)^3,$$
you can use this to find $x+y$; see if you can use this to find $\frac 95(x^2+y^2).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$a + b + c = 0$ Prove that $\frac{a^2 + b^2 + c^2}{2} * \frac{a^3 + b^3 + c^3}{3} = \frac{a^5 + b^5 + c^5}{5}$ The question says: $a + b + c = 0$
Prove that $$\frac{a^2 + b^2 + c^2}{2} * \frac{a^3 + b^3 + c^3}{3} =
\frac{a^5 + b^5 + c^5}{5}$$
So I started with simplifying each part of expression
*
*$$ \frac{a^2 + b^2 + c^2}{2} = \frac{(a+b+c)^2 -2ab-2ac-2bc}{2} =
-(ab + ac + bc)$$
Because $a+b+c=0$
*$$ \frac{a^3+b^3+c^3}{3} = \frac{(a+b+c)^3 - 3a^2b-3ab^2-3a^2c-3ac^2-3b^2c-3bc^2-6abc}{3}
= -(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2 -2abc) $$
So, my question is, what can I do with the second part? Like $\frac{(a+b+c)^2(a+b+c)^3 - ...}{5}$
| From $a+b+c=0$, we can replace every occurrence of $a$ with $-b-c$, turning this into a two variable problem. Then, after simplifying the various expressions, we find
\begin{align*}
\frac{(-b-c)^2+b^2+c^2}{2} &= b^2+bc+c^2 \\
\frac{(-b-c)^3+b^3+c^3}{3} &= -b^2c-bc^2 \\
\frac{(-b-c)^5+b^5+c^5}{5} &= -b^4c -2b^3c^2 -2b^2c^3 -bc^4
\end{align*}
It is straightforward to show the product of the first two is the third.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find $K$ in $2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$? The problem is as follows:
Which value should have $K$ in the expression from below so it makes an identity?
$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&11\\
2.&12\\
3.&14\\
4.&15\\
\end{array}$
I'm confused exactly how to get that particular value for $K$.
The only thing which I was able to spot is that: (Using manipulations on the left side of the equation)
$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$
$2\csc 30 ^\circ (2\cos 7x \cos x)(2\cos 4x \cos 2x)-1$
$4 (2\cos 7x \cdot \cos 2x)(2 \cos x \cdot 4x)-1$
$4(\cos 9x + \cos 5x)(\cos 5x+\cos 3x)-1$
$4(\cos 9x \cos 5x +\cos 9x \cos 3x + \cos 5x \cos 5x + \cos 5x \cos 3x)-1$
$2(\cos 14x + \cos 4x + \cos 11x + \cos 6x + 1+\cos 10x + \cos 8x + \cos 2 x) - 1$
However at the point I reached here then I got stuck it seems that this might not be the intended strategy. What could be done here to solve this problem?. Is there any trick or what?.
Could someone help me with an orderly method to solve this without much fuss?.
| If it is an identity, it will be true for any small angle $x$.
Using Taylor series, we have
$$\cos(nx)=1-\frac{n^2 }{2}x^2+O\left(x^4\right)$$ This makes the lhs to be
$$4 (1-32 x^2+1-18 x^2)(1 - 18 x^2+1-2 x^2)-1=15-560 x^2+O\left(x^4\right) \tag 1$$
Now, for the rhs
$$\sin(Kx)=K x-\frac{K^3 }{6}x^3+O\left(x^5\right)$$
$$\csc(x)=\frac{1}{x}+\frac{x}{6}+O\left(x^3\right)$$
$$\sin(Kx)\csc(x)=K+\frac{1}{6} \left(K-K^3\right) x^2+O\left(x^3\right)\tag 2$$
Compare $(1)$ and $(2)$. This gives $K=15$ and you can check that, for $K=15$
$$\frac{1}{6} \left(K-K^3\right)=-560$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Finding the particular solution of a differential equation using at least three different methods. Find the particular solution $(x^2+6y^2)dx-4xydy=0$; when $x=1$, $y=1$ using at least three different methods.
I have done the first two. Can somebody help me with the third method.
Method 1: Homogenous Equation
Let $y=vx; dy=vdx+xdv$
$(x^2+6x^2v^2)dx-4x(vx)(vdx+xdv)=0$
$(x^2+2x^2v^2)dx-4x^3vdv=0$
$(1+2v^2)dx-4xvdv=0$
$\int\frac{dx}{4x}-\int\frac{v}{1+2v^2}dv=0$
$\ln{x}-\ln{(2v^2+1)}=C$
$\ln{(\frac{x}{2v^2+1})}=\ln{C}$
$\frac{x}{2v^2+1}=\frac{1}{C}$
$C=3$
$3x=2v^2+1$
$3x^3=2y^2+x^2$
$2y^2=x^2(3x-1)$
The particular solution by method 1 is $2y^2=x^2(3x-1)$.
Method 2: Bernoulli Equation
$2y\frac{dy}{dx}-\frac{x^2+6y^2}{2x}=0$
$2y\frac{dy}{dx}-\frac{3y^2}{x}=\frac{x}{2}$
Let $v=y^2; dv=2ydy$
$\frac{dv}{dx}-\frac{3v}{x}=\frac{x}{2}$
$P(x)=-3x^{-1}$; I.F.$=e^{-3\int x^{-1}dx}=x^{-3}$
$vx^-3=\frac{1}{2}\int\frac{dx}{x^2}$
$2vx^{-3}=-x^{-1}+C^{-1}$
$2y^2x^{-3}+x^{-1}=C^{-1}$
$C=\frac{1}{3}$
$2y^2+x^2=3x^3$
$2y^2=x^2(3x-1)$
The particular solution by method 2 is also $2y^2=x^2(3x-1)$.
| We can check if the ODE is exact:
$$\Delta=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=\frac{\partial}{\partial y}(x^2+6y^2)-\frac{\partial}{\partial x}(-4xy)=16y\ne0.$$
So it's not exact, but as
$$\frac{\Delta}{N}=\frac{16y}{-4xy}=-\frac{4}{x}=f(x),$$
we can get an integrating factor with the form
$$\mu(x)=\exp\left(\int f(x)\mathrm{d}x\right)=\exp\left(\int-\frac{4}{x}\mathrm{d} x\right)=e^{-4\log x}=x^{-4}.$$
Now, let's apply our IF, multiplying both sides by it:
$$x^{-4}(x^2+6y^2)\mathrm{d}x-4x^{-3}y\,\mathrm{d}y=0,$$
Let's check if the new ODE is exact (it must be, but just in case),
$$\Delta'=\frac{\partial M'}{\partial y}-\frac{\partial N'}{\partial x}=\frac{\partial}{\partial y}\bigl[x^{-4}(x^2+6y^2)\bigr]-\frac{\partial}{\partial x}(-4x^{-3}y)=0.$$
So, we are looking for a function $F(x,y)$ that satisfy
\begin{align}
\frac{\partial F}{\partial x} & =x^{-4}(x^2+6y^2);\\
\frac{\partial F}{\partial y} & =-4x^{-3}y.
\end{align}
From the second condition,
$$F(x,y)=\int-4x^{-3}y\,\mathrm{d}y+\varphi(x)=-2x^{-3}y^2+\varphi(x).$$
Now, we can plug $F$ in the first condition:
\begin{align*}
\frac{\partial}{\partial x}\bigl[-2x^{-3}y^2+\varphi(x)\bigr]=x^{-4}(x^2+6y^2) & \Longrightarrow6x^{-4}y^2+\varphi'(x)=x^{-4}(x^2+6y^2)\Longrightarrow\\
& \Longrightarrow\varphi'(x)=x^{-2}\Longrightarrow\varphi(x)=-\frac{1}{x}.
\end{align*}
So, the solution takes the form $F(x,y)=C$, and is
$$-2x^{-3}y^2-\frac{1}{x}=C\Longrightarrow2y^2=-Cx^3-x^2.$$
If $y=1$ when $x=1$,
$$2=-C-1\Longrightarrow C=-3,$$
and we we have
$$2y^2=3x^3-x^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Given $2\sin A + \sin B = 2\sin C$ in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$ Given $2\sin A + \sin B = 2\sin C $ where $A,B,C$ are angles in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$
so $\sin C - \sin A = \frac{ \sin B}{2}$ so $2\cos \frac{C+A}{2} \sin \frac{C-A}{2} = \frac{ \sin B}{2} = 2\sin \frac{B}{2} \sin \frac{C-A}{2} $ so $\sin \frac{C-A}{2} = \frac{1}{4}$
I feel like I am awfully close here, but how do I finish it?
| In terms of the sides and circumradius:
$$2\sin A + \sin B = 2\sin C\iff \dfrac{a}{R} + \dfrac{b}{2R} = \dfrac{c}{R}\iff 2c=b+2a.$$
Then, you can turn your expression into a function of two variable, say $a,b$, using the same Law of Sines and take derivatives, using the formula:
$$R = \dfrac{abc}{4S},$$
where $S$ is the area.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4047955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24]=(z+x-7)[y(z+x)+24]$, find $x^2+y^2+z^2$
Let x, y, z be pairwise distinct real numbers, if $$(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24] $$ $$=(z+x-7)[y(z+x)+24]$$, find $x^2+y^2+z^2$
I've tried many ways but couldn't find a working way to solve it. I tried letting $(x+y-7)[z(x+y)+24] = k$ and $t = x + y + z$ but none of these gives useful transformation as far as I can see. Could somebody shed some lights on this? Thanks in advance.
| Let $S=x+y+z$. Given is
$$P(x)=(S-x-7)\{x(S-x)+24\}$$
$$=x^3-x^2(2S-7)+x(S^2-7S-24)+24S-168$$
has same value as $P(y), P(z)$.
This means $x,y,z$ are roots of the polynomial, $P(x)+\text{constant}$.
By Vieta's, sum of roots is $S=2S-7 \Rightarrow S=7$. Also
$$xy+yz+zx=-(7^2-7\cdot 7 -24)=24$$
Hence $x^2+y^2+z^2=7^2-2\cdot 24=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4048713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Decomposition of $\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}$
Determine $\alpha,\beta,\gamma,\delta$ in
$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{\alpha s+\beta}{(s+3)^2+25}+\frac{\gamma s+\delta}{s^2+25}$$
I have come across the partial fraction and trying to decompose it. I have tried the methods used by multiplying out and making one variable to 0 to find the others etc but It does not work here. Any ideas how I could approach this?
| Apply your method to proceed as follows
$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}
=\frac{A}{s+5i} + \frac{B}{s-5i} +\frac{C}{s+3+i5}+ \frac{D}{s+3-i5}$$
Then
\begin{align}
&A = \lim_{s\to-5i} \frac{10(s+6)(s+5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{-45+68i}{327}\\
&B = \lim_{s\to5i} \frac{10(s+6)(s-5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{-45-68i}{327}\\
&C = \lim_{s\to-3-5i} \frac{10(s+6)(s+3+5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{45-41i}{327}\\
&D= \lim_{s\to-3+5i} \frac{10(s+6)(s+3-5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{45+41i}{327}\\
\end{align}
and
\begin{align}
& \lambda = A+B = -\frac{90}{327}\\
& \delta =(B -A)\>5i= \frac{680}{327}\\
& \alpha = C +D = \frac{90}{327}\\
& \beta =(C+D) \>3+(D -C ) \>5i= -\frac{140}{327}\\
\end{align}
Thus
$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}
=\frac{10}{327}\left( \frac{68-9s}{s^2+25} + \frac{9s-14}{(s+3)^3+25}\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}≥\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Question : Prove $$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
$(a, b, c \in \mathbb{R}^+)$
I tried to solve it like this :
$$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \; (\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2})$$
Am I doing this right? How can I finish this problem?
| Problems of this type can often be solved by showing, using AM-GM, that each term on the right hand side is not greater than some linear combination of the terms on the left hand side.
In this particular situation,
$$
4\cdot \frac{a^2}{b^3} + 6\cdot \frac{b^2}{c^3} + 9\cdot\frac{c^2}{a^3} \ge 19 \cdot \sqrt[19]{\left(\frac{a^2}{b^3}\right)^4 \left(\frac{b^2}{c^3}\right)^6 \left(\frac{c^2}{a^3}\right)^9} = 19 \cdot \sqrt[19]{\frac{a^8}{a^{27}}} = 19 \cdot \frac{1}{a}.
$$
Adding this inequality to its cyclic variants yields the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $1+\cos(\theta)+\dots+\cos (n\theta)=\frac12+\frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$
Suppose that $n \geq 1$ is a natural number.
(i) Show that for $z$ in $\mathbb{C}$, provided $z \neq 1$,
$$ \frac{1-z^{n+1}}{1-z} = 1+z+...+z^n.$$
(ii) Using de Moivre’s Theorem, part (b)(i) and standard trigonometric identities, show that for $\theta$ in $(0,2\pi)$,
$$ 1+\cos(\theta) + \dots + \cos (n\theta) = \frac{1}{2} + \frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$$
I have set $z=a+bi$ and subbed this into the top equation but dont seem to be getting anwhere with this problem. Any guidance will be great!
| Part (i) should be straightforward if you are familiar with finding the sum of a geometric series:
\begin{align}
S &= 1+z+z^2+\ldots+z^{n-1}+z^n \\
zS &= z+z^2+z^3+\ldots + z^n+z^{n+1} \\
zS - S &= z^{n+1}-1 \\
S(z-1) &= z^{n+1}-1 \\
S &= \frac{z^{n+1}-1}{z-1} = \frac{1-z^{n+1}}{1-z} \, . \\
\end{align}
For part (ii), you should first consider the geometric series
$$
1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta} \, .
$$
Since $e^{ik\theta}=\cos k\theta+i\sin k\theta$ for $k \in \mathbb{Z}$, we know that
\begin{align}
&1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta}\\=&1+(\cos\theta+i\sin\theta)+(\cos 2\theta+i\sin 2\theta)+\ldots+(\cos n\theta+i\sin n\theta)
\end{align}
Hence,
$$
1+\cos\theta+\cos 2\theta+\ldots+\cos n \theta=\Re(1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta}) \, .
$$
| {
"language": "en",
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Question on The multiplicative magic square Consider this magic square:
$$\begin{array} {|r|r|}\hline a & b & c \\ \hline d & e & f \\ \hline j & h & i \\ \hline \end{array} $$
Where $a,b,c,d,e,f,j,h,i\in \mathbb N^*$ and the $\gcd$ of all nine elements is $1$.
This magic square is A multiplicative one such that the product in each row, column and diagonal is equal to $P$
The first question: Prove that $e^3=P$
The second question: construct a multiplicative magic square with the divisors of $100$
My attempt:
It’s not actually an attempt, it is just an observation, since the $\gcd(a,b,c,d,e,f,j,h,i)=1$ And $$aei=cej=beh=def=P$$
$e\in\mathbb N^* \implies e\geq1$, so we can divde by it:
$$ai=cj=bh=df$$
Wich mean that $a\mid cj,bh,df$ And $i\mid cj,bh,df$ and so on..., but what’s next?
| You can start with any additive magic square. The following are the simplest
\begin{array}{c}
\begin{pmatrix}
0 & 2 & 1 \\
2 & 1 & 0 \\
1 & 0 & 2 \\
\end{pmatrix} &
\begin{pmatrix}
1 & 0 & 2 \\
2 & 1 & 0 \\
0 & 2 & 1 \\
\end{pmatrix} &
\begin{pmatrix}
1 & 2 & 0 \\
0 & 1 & 2 \\
2 & 0 & 1 \\
\end{pmatrix} &
\begin{pmatrix}
2 & 0 & 1 \\
0 & 1 & 2 \\
1 & 2 & 0 \\
\end{pmatrix} \\
\end{array}
Pick one or more and associate each to an array of some chosen base raised to those powers.
For example.
\begin{array}{c}
\begin{pmatrix}
2^0 & 2^2 & 2^1 \\
2^2 & 2^1 & 2^0 \\
2^1 & 2^0 & 2^2 \\
\end{pmatrix} &
\begin{pmatrix}
3^1 & 3^0 & 3^2 \\
3^2 & 3^1 & 3^0 \\
3^0 & 3^2 & 3^1 \\
\end{pmatrix} &
\begin{pmatrix}
5^1 & 5^2 & 5^0 \\
5^0 & 5^1 & 5^2 \\
5^2 & 5^0 & 5^1 \\
\end{pmatrix}
\end{array}
Simplify.
\begin{array}{c}
\begin{pmatrix}
1 & 4 & 2 \\
4 & 2 & 1 \\
2 & 1 & 4 \\
\end{pmatrix} &
\begin{pmatrix}
3 & 1 & 9 \\
9 & 3 & 1 \\
1 & 9 & 3 \\
\end{pmatrix} &
\begin{pmatrix}
5 & 25 & 1 \\
1 & 5 & 25 \\
25 & 1 & 5 \\
\end{pmatrix}
\end{array}
Create one array by multiplying pointwise.
\begin{pmatrix}
15 & 100 & 18 \\
36 & 30 & 25 \\
50 & 9 & 60 \\
\end{pmatrix}
| {
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} |
prove this ODE solution is genuine? Prove that the solution to:
$y'-2y\tan x=2x\sec x$ is $$y=\frac{2\left(x\sin\left(x\right)+\cos \left(x\right)\right)+C}{\cos^2\left(x\right)}$$
I have found the derivative: $$\frac{2x\cos^3\left(x\right)+\sin \left(2x\right)\left(2\left(x\sin \left(x\right)+\cos \left(x\right)\right)+C\right)}{\cos ^4\left(x\right)}$$
But cannot substitute derivative into original equation to prove
| Treating it as a linear first order ODE, then proceed to find the integrating factor: $$\mu = e^{\int -2\tan xdx}= e^{2\ln (\cos x)}= \cos^2 x $$
$$\implies (y\cos^2x)' = 2x\sec x\cdot \cos^2x=2x\cos x$$
$$\implies y\cos^2x= \displaystyle \int 2x\cos xdx$$
$$ = \int 2xd(\sin x)$$
$$= 2x\sin x- \int 2\sin xdx$$
$$= 2x\sin x+2\cos x+C$$
$$\implies y = \dfrac{2x\sin x+2\cos x +C}{\cos^2x}$$ as claimed above.
| {
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Prove that $3^{n + 1} \ge n^4 + n^2 + 1$ for any $n \ge 5$ Prove that $3^{n + 1} \ge n^4 + n^2 + 1$ for any $n \ge 5$, using mathematical induction.
I've thought about writing the equation like that: $3^{n + 1} \ge (n^2 + 1)^2 - n^2$, but all for nothing.
| You can check the base case of $n=5$ - add that into your question.
For the inductive step, let's re-cast the inequality as $3^{n+1} - n^4-n^2-1 \geq 0$.
$$\begin{align}
3^{n+2} - (n+1)^4-(n+1)^2-1 &= \color{#0B2}{3^{n+1}} + 2\cdot3^{n+1} \color{#0B2}{- n^4} - 4n^3 - 6n^2 - 4n -1\\
&\quad\quad\quad \color{#0B2}{-n^2}-2n-1 \color{#0B2}{-1}\\
&\geq 2\cdot3^{n+1} - 4n^3 - 6n^2 - 6n - 2\\
&= 2(3^{n+1} \color{#03B}{ - 2n^3 - 3n^2 - 3n - 1})\\
\end{align}$$
Now note that since $n>5$, $n^4+n^2+1 > 5n^3+5n+1 > \color{#03B}{2n^3+3n^2+3n+1}$ & finish
| {
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Decomposition of a sum of cubes I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can't understand how rewriting the red term as $x^2+12y^2$, without trivially executing the squares. Can you give me a hint?
| Use $(a+b)+(a-b)=2a$
and $(a+b)^2+(a-b)^2=2a^2+2b^2$
and $(a-b)(a+b)=a^2-b^2$
Then
$$
\underbrace{(\color{#C00}{(x+2y)}+\color{#090}{(x-2y)})}_{2x}\left(\!\vphantom{2^2}\right.\underbrace{\color{#C00}{(x+2y)}^2+\color{#090}{(x-2y)}^2}_{2x^2+8y^2}-\underbrace{\color{#C00}{(x+2y)}\color{#090}{(x-2y)}}_{x^2-4y^2}\left.\!\vphantom{2^2}\right)
$$
| {
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} |
Shared area of $2$ circles (polar integral) Given $x^2+y^2=4y$ and $x^2+y^2=4x$ find the shared area of the 2 circles
What I tried was :
so first I transformed them to polar coordinates
I got $r=4cos\theta$ for $x^2+y^2=4x$ and $r=4sin\theta$ for $x^2+y^2=4y$
After that I did $4cos\theta=4sin\theta$ then $\theta=\frac{\pi}{4}$ I am not sure but according to this I believe that $\frac{\pi}{4}≤\theta≤\frac{\pi}{2}$ lastly I did the integral
$\int_\frac{\pi}{4}^\frac{\pi}{2}\frac{1}{2}((4cos\theta)^2-(4sin\theta)^2) d\theta $
after integrating I got $-4$ which is a wrong answer.
what am I doing wrong? thanks for any help and tips!
Edit: I forgot to mention that I havent studied double integrals yet , I learned polar system and this formula $\int_\alpha^\beta\frac{1}{2}(r(\theta))^2 d\theta $
|
Please note that for $0 \leq \theta \leq \frac{\pi}{4}$, you are bound by circle $4 \sin\theta $ and for $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$, you are bound by circle $4 \cos\theta$.
So the area is given by, $A = \displaystyle 2 \int_0^{\pi/4} \int_0^{4\sin\theta} r \ dr \ d\theta$
or $ \ \displaystyle 2 \int_0^{\pi/4} \frac{1}{2}(4\sin\theta)^2 \ d\theta$
You can see that the area between $0 \leq \theta \leq \frac{\pi}{4}$ and between $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$ are same. So we multiply by $2$ or use the fact that $\sin\theta = \cos (\frac{\pi}{2} - \theta)$.
| {
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Proving $(A+B)^{-1} = A^{-1} + B^{-1}$ when there exists $J$ so that $J^2 = -I$ I'm having trouble proving this biconditional statement:
Prove that there exist $A, B \in \mathcal{M}_{n\times n}(\mathbb{R}^n)$ such that $(A+B)^{-1} = A^{-1} + B^{-1}$ if and only if there exists $J \in \mathcal{M}_{n \times n}(\mathbb{R}^n)$ such that $J^2 = -I_n$.
Here, $I_n$ is the $n \times n$ identity matrix. So far, none of my ideas have panned out and I don't feel like I have enough information to solve this problem. Any hints would be appreciated!
| $(\Longrightarrow)$
Since
$(A+B)(A^{-1}+B^{-1})= I + AB^{-1}+BA^{-1} + I=(A+B)(A+B)^{-1}=I$
$AB^{-1}+BA^{-1}=AB^{-1}+(AB^{-1})^{-1}=-I$ and $ (AB^{-1})^2+I=-AB^{-1}$
and if we let $X=AB^{-1}$, then we have $X^2+X+I=O$.
We can do a change of variables $X=Y+C$
$(Y+C)^2+(Y+C)+I=Y^2 +(2C+I)Y+C^2+C+I$
and by letting $C=-\frac{1}{2}I$, then we get $Y^2+0Y+\frac{3}{4}I=Y^2+\frac{3}{4}I=O$
now $\left(\frac{2}{\sqrt{3}}Y\right)^2+I=Z^2+I=0$ and $Z=\frac{2}{\sqrt{3}}\left(X+\frac{1}{2}I\right)$ satisfies the properties of $J$.
$(\Longleftarrow)$
Given $J$ s.t. $J^2=-I$, then let $Z=-\frac{1}{2}I+\frac{\sqrt{3}}{2}J$
then $\left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)\left(-\frac{1}{2}I-\frac{\sqrt{3}}{2}J\right)= \frac{1}{4}I^2-\frac{3}{4}J^2= \frac{1}{4}I+\frac{3}{4}I=I$ and $Z$ is invertible.
$Z$ also satisfies $Z^2 + Z +I =O$
$\left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)^2 + \left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)+ I=\frac{1}{4}I^2-\frac{\sqrt{3}}{2}J+\frac{3}{4}J^2-\frac{1}{2}I+\frac{\sqrt{3}}{2}J+I$
$=\frac{1}{4}I-\frac{3}{4}I-\frac{1}{2}I+I=O$
Given an invertible matrix $B$, let $ZB=A$, and then $Z=AB^{-1}$ where $A$ is invertible because $Z$ and $B$ are inverible.
$(A^{-1}+B^{-1})=(A+B)^{-1}$ because $Z+Z^{-1}=AB^{-1}+BA^{-1} = -\frac{1}{2}I + \frac{\sqrt{3}}{2}J -\frac{1}{2}I -\frac{\sqrt{3}}{2}J =-I$
thus $I+AB^{-1}+BA^{-1}+I=(A+B)(A^{-1}+B^{-1})=I$.
| {
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Can you solve $\int 1/(x^2+1)^2\, dx$ I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$
to solve the anti-derivative : $$\int \frac{1}{(x^2+1)^2}\, dx$$
so :$$\frac{1}{(x^2+1)^2}=\frac{A}{x^2+1}+\frac{B}{x^2+1}\iff \frac{1}{x^2+1}=A+B$$
$$1+0x^2=(A+B)x^2+(A+B)1$$
Thus I got this system of equations:
$$\cases{A+B=1 \\ A+B=0}$$ Which is impossible. I also tried to use substitution but it didn't help me.
| If we use the substitution $x = \tan \theta$, $dx = \sec^2 \theta \,d\theta$, the integral becomes \begin{align*}
\int \frac{1}{(x^2+1)^2}\, dx &= \int \frac{\sec^2 \theta}{(\tan^2\theta + 1)^2}\,d\theta\\
&= \int \frac{\sec^2\theta}{\sec^4 \theta}\,d\theta\\
&= \int\cos^2\theta \,d\theta\\
&= \frac{1}{2}(\theta + \sin\theta\cos\theta) + C.
\end{align*} I will leave that last integral to you, but we can use the double angle formula for cosine. So, we just need to undo the substitution. Given that $x = \tan\theta$, we can conclude that $\theta = \arctan x$ and $$\sin\theta = \frac{x}{\sqrt{1+x^2}}\quad\text{and}\quad \cos\theta = \frac{1}{\sqrt{1+x^2}}.$$
This follows from the fact that $$x^2 + 1 = \tan^2\theta + 1 = \sec^2\theta \implies \cos^2\theta = \frac{1}{1+x^2}.$$
This gives us
$$\int \frac{1}{(x^2+1)^2}\, dx = \frac{1}{2}(\theta + \sin\theta\cos\theta) +C= \frac{1}{2}\left(\arctan x + \frac{x}{1+x^2}\right) +C.$$
| {
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Expressing $ (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ in terms of powers of $5$ I was doing my preview while some problems turned out to make me feel sucky.
It's said that every element $ x $ of $ (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ could be expressed uniquely as
$$ x \equiv\pm 5^{\alpha} \bmod 2^k $$
where $ 1\le\alpha\le 2^{k-2} $, and the exponent $$ e=2^{k-2} $$
which I understood as the $ l.c.m $ of the orders of all $ x \in (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ equals to $ 2^{k-2} $
All these above were written in my teacher's PPT for the number theory class. Since I'm doing a preview, I don't quite get that. Why is it and how to understand it?
Any explanation would be greatly appreciated!
| The following is an excerpt from a book whose draft I wrote last year. It answers your question in essentially the same way as in the comments by reuns, but in explicit detail.
Problem. Take the following steps for all integers $n\ge 3$:
*
*Prove by induction on $n$ that $\nu_2 \left(5^{2^{n-2}}-1\right) =
n.$
*Show that the $\text{ord}_{2^n}(5) = 2^{n-2}.$
*Deduce that
$S=\left\{\pm 5^k : k\in \left[2^{n-2}\right] \right\}$ is a reduced
residue system modulo $2^n.$
Solution. Each step leads to the next one.
*
*We will show by induction on $n\ge 3$ that, for each integer $n\ge 3,$ there exists an odd integer $x_n$ such that $5^{2^{n-2}} = 1 + x_n \cdot 2^n.$ Since $x_n$ will be shown to be odd, no power of $2$ higher than $2^n$ can divide $5^{2^{n-2}}-1,$ while it is true that $2^n$ does divide $5^{2^{n-2}}-1.$ Thus, we will have proven that $\nu_2 \left(5^{2^{n-2}}\right) = n.$
The base case $n=3$ holds because $$5^{2^{3-2}} - 1 = 24 = 3\cdot 2^3.$$ Now suppose $5^{2^{n-2}} = 1 + x_n \cdot 2^n$ for some integer $n\ge 3$ and odd integer $x_n.$ Squaring the equation yields
$$5^{2^{n-1}} = 1 + x_n \cdot 2^{n+1} + x_n^2 \cdot 2^{2n} = 1 + x_n(1+x_n 2^{n-1})\cdot 2^{n+1}\equiv 1\pmod{2^{n+1}}.$$ Since $x_n$ is odd and $n\ge 3,$ so is $$x_{n+1}=x_n(1+x_n 2^{n-1}).$$
*We know from the first part that $$5^{2^{n-2}} \equiv 1\pmod{2^n}.$$ So the order of $5$ modulo $2^n$ must divide $2^{n-2}.$ Suppose, for contradiction, that there exists an integer $i$ such that $3\le i\le n$ and $5^{2^{n-i}}\equiv 1\pmod{2^n}.$ Squaring this sufficiently many times (taking it to an exponent of $2^{i-3},$ to be precise) we get $5^{2^{n-3}} \equiv 1\pmod{2^n}.$ So there exists an integer $y_n$ such that $$5^{2^{n-3}} = 1 + y_n 2^n.$$ Squaring this yields $$5^{2^{n-2}} = 1 + y_n 2^{n+1} + y_n^2 2^{2n} = 1+(2y_n + y_n^2 2^n)2^n,$$ which implies that the $x_n$ from the last part is the even number $2y_n + y_n^2 2^n$. This is a contradiction.
*Note that the order of $5$ modulo $2^n$ is $2^{n-2}=\frac{\varphi(2^n)}{2}$ and that the elements of $S$ are all in fact units. We could show that $S$ consists of precisely all of the units (that is, invertible elements, which are the odd integers in this case) by showing that none of the $5^i$ coincide with the $-5^j$ modulo $2^n.$ If $5^i\equiv -5^j\pmod {2^n}$ then cancelling the smaller of the two powers from both sides (or either, if they are equal) would yield a power of $5$ congruent to $-1$ modulo $2^n.$ It suffices to show that $$5^k \not\equiv -1\pmod{2^n}$$ for all $k.$ If the congruence held, then we could reduce it to $$5^k \equiv -1 \equiv 3\pmod{4}$$ using $4\mid 2^n.$ This is contradictory because $$5^k \equiv 1^k \equiv 1\pmod{4}.$$ Therefore, there are no overlaps between the $5^i$ and $-5^j.$ This produces $$2\cdot 2^{n-2}=2^{n-1}=\varphi(2^n)$$ distinct units modulo $2^n,$ which is the maximal number and so this is the set of all units modulo $2^n.$
| {
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Probability, dependent Events I am trying to understand some of the tasks and would be happy if someone could check my solution and possibly point out any mistakes.
In advance, I am not asking for a solution, because I would like to work it out by myself.
The purpose of this exercise is to check whether events A and B are independent of each other.
Three dices are thrown.
The events are defined as follows: | All dice have the same number of dots.
B: The total sum of the dices is less than 5.
Therefore is
$A = \left \{ \left ( 1,1,1 \right ), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6) \right \}$
$B = \left \{ \left ( 1,1,1 \right ), (1,1,2), (1,2,1), (2,1,1) \right \}$
Some intermediate results:
$P\left ( A \right ) = \frac{6}{216} = \frac{3}{108}$
$P\left ( not A \right ) = 1 - P(A) = 1 - \frac{3}{108} = \frac{35}{36}$
$P(B) = \frac{4}{216} = \frac{1}{54}$
$P(A\cap B) = \frac{1}{216}$
$P(notA\cap B) = \frac{3}{216}$
A and B are independent if
$P(B|notA) = P(B|A)$
$
P(B|A) = \frac{P(A\cap B)}{P(A)} = \frac{\frac{1}{216}}{\frac{6}{216}} = \frac{1}{6}
$
$
P(B|notA) = \frac{P(notA\cap B)}{P(notA)} = \frac{\frac{3}{216}}{\frac{35}{36}} = \frac{1}{70}
$
Thus, the events A and B are dependent.
Thank you in advance for your time.
| {
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Finding stationary points of $f(x,y) = x^2+y^2+\beta xy+x+2y$ I am working on a problem which can be seen below. Part i) was fine but part ii) is causing me some issues. Now the First Order Necessary Conditions state that for the interior case, then $$\nabla f(x^*) = 0$$ which I have attempted to use to find $x^*$. Taking the derivatives of the objective function I get $$\begin{bmatrix}2x + \beta y +1 \\
2y + \beta x +2 \end{bmatrix} = \begin{bmatrix}0\\ 0 \end{bmatrix}$$ which I have solved to get $(y-x) = \frac{1}{\beta - 2}$. Now I am convinced that I am wrong here because this does not seem plausible.
I would really appreciate if someone could give me some hints.
Thanks
| You got:
$$\begin{bmatrix}2 & \beta \\ \beta & 2 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-1\\ -2 \end{bmatrix}$$
Let's perform one step of row reduction:
$$\begin{bmatrix}1 & \frac{1}{2}\beta \\ 0 & 2-\frac{1}{2}\beta^2 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-\frac{1}{2}\\ \frac{1}{2}\beta-2 \end{bmatrix}$$
The system has no solution if $2-\frac{1}{2}\beta^2=0$ but $\frac{1}{2}\beta-2 \neq 0$, so when $\beta=2$ or $\beta=-2$. The system has infinitely many solutions if $2-\frac{1}{2}\beta^2= 0$ and $\frac{1}{2}\beta-2 = 0$, which cannot happen.
When $\beta\neq 2$ and $\beta\neq -2$ we do one more step of row reduction:
$$\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-\frac{1}{2}-\frac{1}{2}\frac{\beta^2-4\beta}{4-\beta^2}\\ \frac{\beta-4}{4-\beta^2} \end{bmatrix}$$
You can now readily read the solution.
| {
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Are both values of cos15° that I obtained equal and if yes, why do I get such different-looking values? I found the value of $\cos{15}^\circ$ using 2 methods.
Method 1: using $\cos{(a-b)} = \cos{a}\cos{b}+\sin{a}\sin{b}$
$\cos{(45-30)^\circ} = \cos{45^\circ}\cos{30^\circ}+\sin{45^\circ}\sin{30^\circ}$
$\cos{15^\circ} = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}$
$\cos{15^\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$
Method 2: using $\cos{2\alpha} = 2\cos^2{\alpha}-1$
$\cos{(2_\dot{}15)^\circ} = 2\cos^2{15^\circ}-1$
$\cos{30^\circ} = 2\cos^2{15^\circ}-1$
substituting $\cos{15^\circ}$ with $x$ and $\cos{30^\circ}$ with $\frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{2} = 2x^2-1$
$2x^2-1-\frac{\sqrt{3}}{2} = 0$
solving for $x$, we get $x = \frac{\pm\sqrt{\sqrt{3}+2}}{2}$ and therefore $\cos{15^\circ} = \frac{\pm\sqrt{\sqrt{3}+2}}{2}$
Note: Only the positive value out of the two values we have got can be the correct value of $\cos{15\circ}$ since $\cos{15\circ}$ cannot be a negative value. So going forward I shall only refer to the positive value obtained in Method 2 as the solution from Method 2.
Now my question is: Are both answers equal? If yes, why do we even get these two answers?
The reason I ask equal and not correct is because the universal answer everywhere on the internet and in books is that $\cos{15^\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$
So I am confused as to how you can get completely different-looking answers. Now a quick use of the calculator reveals that the decimal approximation of both answers is the same i.e. 0.96592... . So if that is the case then why do we get such different-looking solutions?
| To see that $$\sqrt {\sqrt 3 +2}=\frac {\sqrt 3 +1}{\sqrt 2}$$
simply square both sides.
Since both terms are positive we see that our desired equality is equivalent to $$\sqrt 3+ 2 =\frac {3+1+2\sqrt 3}{2}=2+\sqrt 3$$
and we are done.
As a general remark, algebraic relations can be very complicated and unintuitive. It is certainly a good idea to compare numerical expressions, as you did, as sanity checks.
| {
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If $a^3=20a^2+b^2+c^2-a-340$, $b^3=20b^2+c^2+a^2-b-340$, $a^3=20a^2+a^2+b^2-c-340$, what is value of $abc$? Question : If $a^3=20a^2+b^2+c^2-a-340$, $b^3=20b^2+c^2+a^2-b-340$, $a^3=20a^2+a^2+b^2-c-340$, what is value of $abc$?
I think I should add this and get close to $abc$, but I can't think about this. I know that answer is 19. Please help me
| $$20a^2+b^2+c^2-a-340-a^3=0$$
$$20b^2+c^2+a^2-b-340-b^3=0$$
$$20c^2+a^2+b^2-c-340-c^3=0$$
Taking note of @Ishraaqparvez comments
Maybe before you find the product of $abc$, you should first know what their corresponding values are
$$(a-19)^4(a^2+1)^4(a^3-22a^2+a+340)(a^4+5a^2-38a-356)(a^4-41a^3+422a^2+283a-6459)^2=0$$
$$(b-19)^4(b^2+1)^4(b^3-22b^2+b+340)(b^4+5b^2-38b-356)(b^4-41b^3+422b^2+283b-6459)^2=0$$
$$(c-19)^4(c^2+1)^4(c^3-22c^2+c+340)(c^4+5c^2-38c-356)(c^4-41c^3+422c^2+283c-6459)^2=0$$
Note that the equations are symmetric, so $a,b,c$ are different solutions of the same equation
$$(x-19)(x^2+1)$$
$$x^3-22x^2+x+340$$
Then their real product is $-340$
| {
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Show that $\frac{\arctan(x^2)}{\ln(1+x^2)}$ is bounded by $\frac{1+\sqrt{2}}{2}$ Given $F(x) = \frac{\arctan(x^2)}{\ln(1+x^2)},$ which is defined for $x>0.$ I would like to show that it is bounded above by $\frac{1+\sqrt{2}}{2}$.
I tried to differentiate the function in order to find maximal points however I've reached an equation which I don't know how to solve :
$$F'(x) = \frac{\frac{2x}{1+x^4}\ln(1+x^2)-\arctan(x^2)\frac{2x}{1+x^2}}{\ln^2(1+x^2)}\overset{{\rm ?}}{=}0$$
Any help would be appreciated.
| We can simplify the problem by treating the function in terms of $x^2$. That is, the maximum value of $\arctan(x^2)/\log(1+x^2)$ is the same as that of $\arctan x/\log(1+x)$ which has derivative $$\frac1{\log^2(1+x)}\left(\frac{\log(1+x)}{1+x^2}-\frac{\arctan x}{1+x}\right).$$ Setting this to zero yields an expression for the stationary point $x_s$ $$\frac{\log(1+x_s)}{1+x_s^2}=\frac{\arctan x_s}{1+x_s}\implies\frac{\arctan x_s}{\log(1+x_s)}=\frac{1+x_s}{1+x_s^2}.$$ To find the maximum value of $(1+x)/(1+x^2)$, we set its derivative $(1-2x-x^2)/(1+x^2)^2$ to zero which yields $x=-1\pm\sqrt2$. We know that $x_s>0$ (since the original function is a function of $x^2$) so it follows that $$\frac{\arctan x^2}{\log(1+x^2)}\le\frac{1+(-1+\sqrt2)}{1+(-1+\sqrt2)^2}=\frac{1+\sqrt2}2$$ as desired.
| {
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Is the series $ 1-1+1-1+1+\cdots = -1+1-1+1-1+1-\cdots $? We have the series
$$ S = 1-1+1-1+1-1+1 \cdots $$
If we manipulate S, we get that
$$ 1 - S = S $$
$$ 1 = 2S $$
$$ \frac{1}{2} = S $$
Also, if we re-order S, we get that
$$ S = (1-1)+(1-1)+(1-1)\cdots $$
$$ = -1+1-1+1-1+1-1+1\cdots $$
So $$ -S = S + 1 $$
Solving for S we get that
$$ S = -\frac{1}{2} $$
My question is, is the new S re-ordered considered the same series as S? When we can say that two series are the same in that context? Thanks.
| One of the simplest way to get this series will be with the help of Binomial Expansion of $\frac{1}{2}$ as follows
$\frac{1}{2} = \frac{1}{1+1 }= (1+1)^{-1}$
Expansion will be as follows
$1^{-1} + \frac{(-1)}{1!}(1)^{-2} \cdot(1)^1+\frac{(-1)\cdot(-2)}{2!}1^{-3}\cdot1^2+... = 1-1+1-1+…$
Clearly now in your question
LHS = $\frac{1}{2} $
RHS = $-1+\frac{1}{2} = \frac{-1}{2} $
In this context RHS $\neq$ LHS
Some more interesting series with Binomial Expansion will be awesome, may be sometimes divergent series also
For example
Expansion of $\frac{1}{3}$ can be done in 2 ways
Conventional and logic way of expansion will be $$\frac{1}{3} = \frac{1}{2+1 }= (2+1)^{-1}$$ which will expand into
$$ 2^{-1}+ \frac{-1}{1!}\cdot 2^{-2}+\frac{{(-1)}\cdot{(-2)}}{2!}\cdot2^{-3}+\frac{{(-1)}\cdot{(-2)}\cdot(-3)}{3!}\cdot2^{-4}...$$
$$=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...$$
Another way to put this Binomial Expansion will be $$\frac{1}{3} = \frac{1}{1+2}= (1+2)^{-1}$$
Expansion now will give different result
$$ 1^{-1}+ \frac{-1}{1!}\cdot 1^{-2}\cdot2^1+\frac{{(-1)}\cdot{(-2)}}{2!}\cdot2^{2}+\frac{{(-1)}\cdot{(-2)}\cdot(-3)}{3!}\cdot2^{3}...$$
And this will be $$1-2^1+2^2-2^3+...$$ a divergent series alternating from $-\infty$ to $+\infty$
| {
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Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers
If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .
Below is my solution approach :-
As $x$ is a negative integer, hence $|x-1|$ in the $2 \le |x-1||y+3| \le 5$ will be come $-(x-1)$ or $(1-x)$ and the main equation will transform into $2 \le (1-x)|y+3| \le 5$.
$1st$ case when $y+3 \ge 0 \Rightarrow y \ge -3 \Rightarrow y \in \{{-3,-2,-1}\} $ as $y$ is a negative integer :
For $y=-3$ we can see that there is no valid solution for $x$ as $|y+3|$ part will become $0$, hence this case is invalid.
For $y=-2$ we get the solution set for $x$ to be $x \in \{{-4,-3,-2,-1}\} $ and total number of solutions possible in this case is $4$.
For $y=-1$ we get the solution set for $x$ to be $x \in \{{-1}\} $ and total number of solutions possible in this case is $1$.
So for this $1st$ case when $y+3 \ge 0$ we have in total 5 solutions.
$2nd$ case when $y+3 \lt 0 \Rightarrow y \lt -3$ and in this case $y$ will have infinite values and I am not able to proceed from here.
The answer for the total number of solutions provided is $10$ and you can see that I've been able to find out the $5$ solutions in my $1st$ case. Can someone please guide or help me about how to proceed in the 2nd case?
Thanks in advance !
| If $x,y$ are integers then $|x-1|$ and $|y+3|$ are non negative integers.
So just rewrite them as $m,n$ and we have $1 \le mn \le 5$.
Those are small and simple numbers.
We can have $nm = 1,2,3,4,5$
But you said $x,y$ are both *negative integers. So $x \le -1$ and so $x -1 \le -2$ so $|x- 1| \ge 2$.
so we can have:
$nm = 2$ and $|x-1| =2$ and $|y + 3| = 1$ so $x-1= -2$ and $x =-1$ while $|y+3| = 1$ so $y+3 = \pm 1$ so $y = 1-3=-2$ or $y = -1 -3 = -4$.
or
$nm =3$ and $|x-1| = 3$ and $|y+3| = 1$ so $x-1 =-3$ and $x = -2$ while, as above, y=-2, -4$.
or $nm = 4$ and either $|x-1|=2$ and $|y+3|=2$. Or $|x-1| =4$ and $|y+3| =1$.
If $|x-1|=|y+3|= 2$ then $x = -1$ and $y+3 =\pm 2$ so $y = -1; -5$.
And if $|x-1|=4$ and $|y+3| =1$ then $x-1 = -4$ and $x =-3$ and $y= -2,4$ (as above).
And if $nm = 5$ then $|x-1| = 5$ so $x-1 = -5$ and $x =-4$ and $|y+3| = 1$ and $y=-2,-4$.
So all solutions are.
$(x,y) = \{$
$(-1, -2),(-1, -4), (-2,-2),(-2,-4), (-1,-1),(-1,-5), (-3,-2)(-3,-4), (-4,-2),(-4,-4)$
$\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\prod_{k=1}^m \tan \frac{k\pi}{2m+1}$
How to evaluate this?
$$\prod_{k=1}^m \tan \frac{k\pi}{2m+1}$$
My work
I couldn't figure out a method to solve this product.
I thought that this identity could help.
$$\frac{e^{i\theta}-1}{e^{i\theta}+1}=i\tan \frac{\theta}{2}$$
By supposing $z=e^{\frac{2k\pi}{2m+1}i}$, then,
$$i\tan \frac{k\pi}{2m+1}=\frac{z-1}{z+1}$$
So, the product
$$\displaystyle\prod_{k=1}^m \tan\frac{k\pi}{2m+1}=\displaystyle\prod_{k=1}^m \frac{z-1}{i(z+1)}=\displaystyle\prod_{k=1}^m \frac{e^{\frac{2k\pi}{2m+1}i}-1}{i(e^{\frac{2k\pi}{2m+1}i}+1)}$$
which is getting more complicated.
Answer is $\sqrt{2m+1}$. Any help is appreciated.
| So you can start by breaking the product to get:
$$\prod_{k=1}^{2m}\tan{(\frac{k\pi}{2m+1})}=\prod_{k=1}^{2m}\sin{(\frac{k\pi}{2m+1})}\prod_{k=1}^{2m}\frac{1}{\cos{(\frac{k\pi}{2m+1}})}$$
Now :
$$\prod_{k=1}^{2m}\cos{(\frac{k\pi}{2m+1})}=\frac{(-1)^{m}}{2^{2m}}$$
Proof:
$$\cos{(\frac{k\pi}{2m+1})}=\exp{(\frac{ki\pi}{2m+1})}\frac{(1+\exp{(\frac{-2ik\pi}{2m+1}}))}{2}$$
If we have polynomial $z^{2m+1}-1$ then its $2m+1$ roots of unity are $\exp{(\frac{-2ik\pi}{2m+1}}):1\le k \le 2m+1$
Thus :
$$z^{2m+1}-1=-\prod_{k=1}^{2m+1}(-z+\exp{(\frac{-2ik\pi}{2m+1}}))$$
$$\therefore \frac{(-1)^{2m+1}-1}{2}=-1=-\prod_{k=1}^{2m}(1+\exp{(\frac{-2ik\pi}{2m+1}}))$$
$$\therefore \prod_{k=1}^{2m}\cos{(\frac{k\pi}{2m+1})}=\frac{1}{2^{2m}}\exp(\frac{i(2m)(2m+1)\pi}{2(2m+1)})=\frac{(-1)^{m}}{2^{2m}}$$
Now we have to evaluate : $$\prod_{k=1}^{2m}\sin{(\frac{k\pi}{2m+1})}$$
Using similar process like one above we get:
$$\prod_{k=1}^{2m}\sin{(\frac{k\pi}{2m+1})}=(-1)^{m}\lim_{z\to1}\frac{z^{2m+1}-1}{(z-1)(2i)^{2m+1}}=\frac{(2m+1)(-1)^{m}}{(2i)^{2m}}$$
Okay we have these two products but how do they relate to the original problem. Well
$$\tan{(\frac{(2m+1-k)\pi}{2m+1})}=\tan{(\pi-\frac{k\pi}{2m+1})}=-\tan{(\frac{k\pi}{2m+1})}$$
thus:
$$\prod_{k=1}^{2m}\tan{(\frac{k\pi}{2m+1})}=\prod_{k=1}^{m}\tan{(\frac{k\pi}{2m+1})}\prod_{k=m+1}^{2m}\tan{(\frac{k\pi}{2m+1})}=(-1)^{m}(\prod_{k=1}^{m}\tan{(\frac{k\pi}{2m+1})})^{2}$$
$$\prod_{k=1}^{2m}\tan{(\frac{k\pi}{2m+1})}=\frac{(2m+1)}{(2i)^{2m}}2^{2m}=(-1)^{m}(2m+1)$$
$$\therefore (-1)^{m}(\prod_{k=1}^{m}\tan{(\frac{k\pi}{2m+1})})^{2}=(-1)^{m}(2m+1)$$
$$\therefore \prod_{k=1}^{m}\tan{(\frac{k\pi}{2m+1})}=\sqrt{2m+1} $$
| {
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Has the equation $x^2-21 = 17y$ integer solutions? Has the equation $x^2-21 = 17y$ integer solutions?
Attempt:
I saw this: The equation $x ^ 2 + py + a = 0$ can be solved as an integer precisely, if $-a$ is a quadratic remainder modulo p.
I get: $x^2-17y-21=0$
Now i have to show $21$ is quadratic remainder modulo $-17$? I dont know if this is correct...
$(\frac{21}{-17}) = (\frac{3}{-17}) * (\frac{7}{-17})$
for $(\frac{3}{-17}) = (-1) (\frac{-17}{3})(\text{Quadratic reciprocity})= (\frac{17}{3}) = (\frac{2}{3}) = -1$
for $(\frac{7}{-17}) = (-1) (\frac{-17}{7})(\text{Quadratic reciprocity})= (\frac{17}{7}) = (\frac{3}{7}) = (-1)(\frac{7}{3})(\text{Quadratic reciprocity}) = (\frac{2}{3}) = -1$
insert, we get:
$(-1) * (-1) =1$ and we have integer solutions?
| $$x^2-21=17y$$
Let $x=3k+2$, $k\in\mathbb Z^{+}∪{0}$, we have
$$\begin{align}x^2-21&=(3k+2)^2-21\\
&=9k^2+12k-17\end{align}$$
Then let, $k=17m, m\in\mathbb Z^{+}$, we get
$$\begin{align}9k^2+12k-17=&9\times 17^2m^2+12\times 17m-17&\end{align}$$
Finally,
$$\begin{align}y=\frac{x^2-21}{17}
&=\frac{9\times 17^2m^2+12\times 17m-17}{17}\\
&=153m^2+12m-1\end{align}$$
One of the solution sets:
\begin{align}\color {gold}{\boxed {\color{black}{x=51m+2, y=153m^2+12m-1.}}}\end{align}
Therefore, we have infinitely many integer solutions.
| {
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Differentiate $\arcsin(\frac{1}{\sqrt{1 + x^2}})$ This is from Calculus by Michael Spivak, 3rd Edition, Chapter 15, problem 1 (iv):
Differentiate $f(x) = \arcsin\left(\frac{1}{\sqrt{1 + x^2}}\right).$
Here's my attempt:
\begin{align}f'(x) &= \left[\arcsin\left(\frac{1}{\sqrt{1 + x^2}}\right)\right]' \\
&= \arcsin'\left(\frac{1}{\sqrt{1 + x^2}}\right)\cdot \left(\frac{1}{\sqrt{1 + x^2}}\right)' \\
&= \frac{1}{\sqrt{1 - \left(\frac{1}{\sqrt{1 + x^2}}\right)^2}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{1}{\sqrt{1 - \frac{1}{1+x^2}}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{1}{\sqrt{\frac{1 + x^2 - 1}{1+x^2}}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{-x}{\frac{\sqrt{x^2}}{\sqrt{1+x^2}} \cdot \sqrt{1 + x^2}(1+x^2)} \\
&=\frac{-x}{|x|} \cdot \frac{1}{1+x^2}.\end{align}
Therefore
$$f'(x) =
\begin{cases}
\dfrac{-1}{1+x^2}, & \text{if $x > 0$} \\[1ex]
\dfrac{1}{1+x^2}, & \text{if $x < 0$}
\end{cases}
$$
In the Answer Book the published solution is $\frac{-1}{1+x^2}$.
It seems Spivak is discounting the possibility that $x < 0$. Why might this be? Is there a reason $x$ cannot be negative here? Did I make a mistake? Perhaps with this part of simplifying the denominator?
$$\sqrt{\frac{x^2}{1+x^2}} \cdot \sqrt{1 + x^2} = \sqrt {x^2} = |x|$$
I suspect there's something simple I'm not seeing. I'm just now encountering these inverse trigonometric functions, so my understanding isn't very solid yet.
Edit: This error could have become a misleading, possibly demoralizing waste of time, but thanks your insightful answers it's instead proven to be more instructive than the first 3 (omitted, correct) parts of the problem.
| Let us verify in another way:
Let $y=\arcsin\dfrac1{\sqrt{1+x^2}}\implies\dfrac\pi2\ge y\ge0$
and $x^2=\csc^2y-1$
If $x\ge0, y=\text{arccot}x=\dfrac\pi2-\arctan x$
If $x<0, x=-\cot y, $
Consequently, $y=\cdots=\dfrac\pi2-\arctan(-x)=\dfrac\pi2+\arctan x$
| {
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Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$ Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$
My attempt: For all $\epsilon > 0$ there is a $\delta > 0$ such that $\left | \frac{xy(y^2-x^2)}{x^2+y^2} \right| < \epsilon \rightarrow \sqrt{x^2+y^2} < \delta$.
Note important facts: $ (1) \ \sqrt{x^2} < \sqrt{x^2 + y^2}$ and $ (2) \ x^2 < x^2 + y^2$
$$\left | \frac{xy(y^2-x^2)}{x^2+y^2}\right | = \left | \frac{\sqrt{x^2}\sqrt{y^2}(y^2-x^2)}{x^2+y^2} \right| < \left| \frac{\sqrt{x^2 + y^2}\sqrt{x^2 + y^2}(y^2-x^2)}{x^2+y^2}\right | = |y^2-x^2| $$
From $(2)$ we can show that :
$$|y^2-x^2| < |x^2+y^2-x^2| = |y^2| < |x^2 + y^2| < \delta^2 $$
Hence let $\epsilon = \delta^2$. Does this relationship work?
| Your last line of inequalities should be changed to $|y^{2}-x^{2} |= |y^{2}|+|x^{2}|\leq y^{2}+x^{2}$ without the intermediate steps (which are wrong).
If $x=\sqrt 3$ and $y=1$ the inequality $|y^{2}-x^{2} |\leq y^{2}$ is false.
| {
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Determining limit of a sequence If $$\lim_{n\to \infty}\left(\frac{1}{1^2} + \frac{1}{2^2} +\cdots+ \frac{1}{n^2} \right)=\frac{\pi^2}{6}$$
If $$\lim_{n\to \infty}\left(\frac{1}{1^3\cdot 2^3} + \frac{1}{3^3\cdot 4^3} +\cdots+ \frac{1}{n^3\cdot(n+1)^3} \right)=S$$
How to determine $S$?
I tried making a general pattern
~ $ \frac{n+1-n}{n^3\cdot(n+1)^3} $ = $ \frac{1}{n^3\cdot(n+1)^2} - \frac{1}{n^2\cdot(n+1)^3} $ this is not giving me any cancellation of terms neither any involvement of above series
| HINT: $$\dfrac{1}{n^3(n+1)^3}=6\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)-3\left(\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}\right)+\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right).$$ Now, telescope.
| {
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$(\frac1a+\frac12\frac{x}{a+2}+\frac{1\cdot3}{2\cdot4}\frac{x^2}{a+4}+...)(1+\frac12x+\frac{1\cdot3}{2\cdot4}x^2+...)=\frac1a(1+\frac{a+1}{a+2}x+...)$ For $a>0$, prove $$\left(\frac{1}{a}+\frac{1}{2}\cdot\frac{x}{a+2}+\frac{1\cdot 3}{2\cdot 4}\cdot \frac{x^2}{a+4}+\cdots\right) \cdot
\left( 1+\frac{1}{2}\cdot x+\frac{1\cdot 3}{2\cdot 4}\cdot x^2 +\cdots \right)
= \frac{1}{a} \left[
1+\frac{a+1}{a+2}x+\frac{(a+1)(a+3)}{(a+2)(a+4)} x^2 +\cdots
\right]$$
I'm a bit lost on how to prove it.
First I tried to prove it by induction but the expansion of $x^n$'s coefficient on the left side is complex.
Then I tried to simplify it. Call it $S_1 S_2 = S_3$,
apparently $S_2 = \sum_{n\ge 0} 4^{-n}{2n \choose n} x^n = \frac{1}{\sqrt{1-x}}$, but I got problem in $S_1$.
What I have done is:
Denote $c_n = 4^{-n} {2n \choose n}$, then
$$\left( \sum_{n\ge 1} \frac{c_n}{n} x^n \right)' = \sum_{n\ge 1} c_n x^{n-1} = \frac1x \left( \sum_{n\ge 0} c_n x^n - 1 \right) = \frac1x\left(\frac{1}{\sqrt{1-x}}-1\right) $$
Let $f(x) = -2 \ln (1+\sqrt{1-x})$, then $f' = \frac1x\left(\frac{1}{\sqrt{1-x}}-1\right)$.
Integral from $0$ to $x$ leads to
$$\sum_{n\ge 1} \frac{c_n}{n} x^n = \int_0^x \frac1x\left(\frac{1}{\sqrt{1-x}}-1\right) dx = f(x) |_0^x =\ln 4 - 2 \ln (1+\sqrt{1-x})$$
To work out $S_1=\sum_{n\ge 1} \frac{c_n}{2n+a} x^n = \frac12 \sum_{n\ge 1} \frac{c_n}{n+b} x^n$, where $b = a/2$, define $g(x) = 2S_1 = \sum_{n\ge 1} \frac{c_n}{n+b} x^n $. Then we have
$$ \frac{(x^b g)'}{x^b} = g' + \frac{b}{x}g = \frac1x\left(\frac{1}{\sqrt{1-x}}-1\right) $$
So
$$g = \frac{1}{x^b} \left( \int_0^x \frac{x^{b-1}}{\sqrt{1-x}} dx - \frac{x^b}{b} + c \right) $$
, where $c$ is a constant.
But then I'm stuck at the $ \int_0^x \frac{x^{b-1}}{\sqrt{1-x}} dx$
integral.
| After forming the Cauchy product for the LHS, we see that it suffices to prove that $$\sum_{k=0}^n\frac{c_k c_{n-k}}{a+2k}=\frac1a\prod_{k=1}^n\frac{a+2k-1}{a+2k}\tag{*}\label{coeffs}$$ where $c_n=\displaystyle\prod_{k=1}^n\frac{2k-1}{2k}$ as in the OP.
But \eqref{coeffs} is just the partial fraction decomposition of its RHS: $$f(a):=\frac1a\prod_{k=1}^n\frac{a+2k-1}{a+2k}=\sum_{k=0}^n\frac{b_k}{a+2k},$$ where $b_k$ are constants (that don't depend on $a$, and) determined by $$b_j=\lim_{a\to-2j}(a+2j)f(a)=\frac{\prod_{1\leqslant k\leqslant n}(2k-2j-1)}{\prod_{0\leqslant k\leqslant n,\ k\neq j}(2k-2j)}=c_j c_{n-j}$$ after writing the numerator as $\prod_{k=1}^j\cdot\prod_{k=j+1}^n$ and the denominator as $\prod_{k=0}^{j-1}\cdot\prod_{k=j+1}^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Why is the definite integral of $\int_{1}^{\infty} \frac {\ln (1+x^2)}{x^2}$ equal $\frac{\pi}{2} + \ln(2)$? I currently have a question on why when you partially integrate the definite integral.
$$\int_{1}^{\infty} \frac {\ln (1+x^2)}{x^2} = \frac{\pi}{2} + \ln(2) $$
I obtained this result via plugging it into a WolframAlpha right after partially integrating the function by hand. Though what confuses me and makes me curious is why $\frac{\pi}{2}$ is one of the results of it, rather than anything that has nothing to do with trigonometrical equations on the surface. Is there any identity i was not aware of ?
Anyone that can provide an answer is thanked in advance.
|
For fun :)
Let
$$I(y) = \int_{1}^{\infty} \frac {\ln (1+y x^2)}{x^2}\mathrm dx$$
Then, we need to find $I(1)$.
Differentiate with respect to $y$:
$$\begin{align}
I'(y) &= \int_1^\infty \frac{x^2}{(1+yx^2)x^2}\mathrm dx \\
&= \int_1^\infty \frac{\mathrm dx}{1+(x\sqrt y)^2} \\
&= \frac{1}{\sqrt y}\arctan(x\sqrt y)\bigg|_1^\infty\\
&=\frac{\pi}{2\sqrt y}-\frac{\arctan(\sqrt y)}{\sqrt y}\end{align}$$
Integrate to find $I(y)$:
$$\begin{align}
I(y) &= \int \frac{\pi}{2\sqrt y}-\frac{\arctan(\sqrt y)}{\sqrt y}\mathrm dy \\
&= \pi\sqrt y - 2\sqrt y\arctan(\sqrt y) + \ln(1+y) + C\end{align}$$
Since $I(0) = \int 0 = 0$, we must have $C \equiv 0$. Therefore $$I(y) = \pi\sqrt y - 2\sqrt y\arctan(\sqrt y) + \ln(1+y)$$ Thus
$$\boxed{I(1) = \frac{\pi}{2} + \ln 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4099571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integral $ \int_a^b ((b-x)(x-a))^{n-1}x^{\frac12-n}dx $ How can I prove the integral equality below
$$ \int_a^b ((b-x)(x-a))^{n-1}x^{\frac{1}{2}-n}dx = \frac{2^{2n-1}( (n-1)!)^2}{(2n-1)!}\left(\sqrt b-\sqrt a\right)^{2n-1} $$
I have tried to find a recurrence integrating by parts:
$$I_n:= \int_a^b ((b-x)(x-a))^{n-1}x^{\frac{1}{2}-n}dx\\=\int_a^b \frac{2}{3-2n} ((b-x)(x-a))^{n-1}(x^{\frac{3}{2}-n})'dx\\=\int_a^b \frac{2(n-1)}{2n-3} ((b-x)(x-a))^{n-2}(a+b-2x)x^{\frac{3}{2}-n} dx\\=\frac{2(a+b)(n-1)}{2n-3}I_{n-1}-\frac{4(n-1)}{2n-3} \int_a^b ((b-x)(x-a))^{n-2}x^{\frac{5}{2}-n}dx$$
But, I don't know how to continue.
Maybe it is related to Beta function :
$$B(n,1/2)=\frac{2^{2n-1}}{(2n-1)!}((n-1)!)^2$$
| Indeed, the integral can be recast into the beta function $B(n,1/2)$ as shown below. Substitute $x=\sqrt{ab}\>t^2$ to reexpress the integral as
\begin{align}
I= & \int_a^b ((b-x)(x-a))^{n-1}x^{\frac12-n}dx \\
=& 2 \sqrt[4]{ab}\int_{\sqrt[4]\frac ab }^{\sqrt[4]\frac ba } \left(b+a-\sqrt{ab}(t^2+\frac1{t^2})\right)^{n-1}dt \\
\overset{t\to\frac1t}=& \sqrt[4]{ab}\int_{\sqrt[4]\frac ab }^{\sqrt[4]\frac ba } \left(b+a-\sqrt{ab}(t^2+\frac1{t^2})\right)^{n-1}(1+\frac1{t^2})dt \\
=& \sqrt[4]{ab}\int_{\sqrt[4]\frac ab }^{\sqrt[4]\frac ba } \left((\sqrt b-\sqrt a)^2-\sqrt{ab}(t-\frac1{t})^2\right)^{n-1}d(t-\frac1{t})\\
=& \>(\sqrt b-\sqrt a)^{2n-1}\int_{-1}^1 \left(1-y^2\right)^{n-1}dy\\
\end{align}
where the substitute $y= \frac{\sqrt[4]{ab}\>(t-\frac1t)}{\sqrt b-\sqrt a}$ is used in the last step. Note that $B(1,1/2)=2$
\begin{align}
\int_{-1}^1 \left(1-y^2\right)^{n-1}dy
& = \int_{0}^1 \left(1-t\right)^{n-1}\>t^{-\frac12}\>dt\\
&=B\left(n, \frac12\right)
= \frac{2(n-1)}{2n-1} B\left(n-1, \frac12\right)=\cdots\\
&= \frac{2^{2n-1}\> ((n-1)!)^2 }{(2n-1)!}
\end{align}
Thus
\begin{align}
I=\frac{2^{2n-1}\> ((n-1)!)^2 }{(2n-1)!}\left(\sqrt b-\sqrt a\right)^{2n-1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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Fractions in Questions and Answers
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