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Finding a limit to negative infinity with square roots: $\lim\limits_{x\to -\infty}(x+\sqrt{x^2+2x})$ Find the limit of the equation
$$\lim_{x\to-\infty} (x+\sqrt{x^2 + 2x})$$
I start by multiplying with the conjugate:
$$\lim_{x\to-\infty} \left[(x+\sqrt{x^2 + 2x})\left({x - \sqrt{x^2 + 2x}\over x - \sqrt{x^2+2x}}\right)\right]$$
$$\lim_{x\to-\infty} {x^2 - (x^2 + 2x)\over x - \sqrt{x^2+2x}}$$
$$\lim_{x\to-\infty} {-2x\over x - \sqrt{x^2+2x}}$$
divide by highest power of denominator
$$\lim_{x\to-\infty} {(\frac1x)(-2x)\over (\frac1x) x - ({1\over \sqrt{x^2}})\sqrt{x^2+2x}}$$
$$\lim_{x\to-\infty} {-2\over 1 - \sqrt{1+\frac2x}} = {-2 \over 1-\sqrt{1 + 0}} = {-2 \over 0}$$
but I know this is wrong as the answer is $-1$. Where did I mess up? Thanks.
|
Notice that since $x$ is tending to $-\infty$ then $x=-\sqrt{x^{2}}$. So when you let $x$ enter the square root in the denominator the minus that was originally there becomes a plus due to cancellation. That is:
$$\frac{-2}{\frac{x-\sqrt{x^{2}+2x}}{x}}=\frac{-2}{1-\frac{1}{x}\sqrt{x^{2}+2x}}=\frac{-2}{1+\sqrt{1+\frac{2}{x}}}$$
which tends to $-1$ as $x$ tends to $-\infty$.
|
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|
Solve the equation: $\frac{z}{z-5}+\frac{1}{3}=-\frac{5}{5-z}$ Solve the equation: $\frac{z}{z-5}+\frac{1}{3}=-\frac{5}{5-z}$
First $z$ cannot be equal to $5$.
First, I multiplied $z$ with $3$, $1$ with $z-5$ and $-5$ with both. Eliminating the denominators gives me:$3z + z-5 = - 15$. Simplified, I'm left with: $4z-5=-15$.
Somehow I don't see how this can be solved. If it were $15$, I could at least say that $4\cdot5-5=15$. But that there aren't any solutions because $z$ cannot be $5$.
What am I not seeing?
|
This equation has no solution. Change sides to see that
$$\frac{z}{z-5}+(1/3)+\frac{5}{5-z}=0 \\
Or,\frac{z}{z-5}+(1/3)+\frac{-5}{z-5}=0 \\
Or, \frac{z-5}{z-5}+(1/3)=0$$
Which is impossible.
|
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|
Decompose a fraction in a sum of two Let's say that I have this fraction:
$$ \frac{2x}{x^2+4x+3}$$
I would like to decompose in two fraction:
$$ \frac{A}{x+3} + \frac{B}{x+1}$$
Which is the procedure for that? :)
|
$$ \frac{2x}{x^2+4x+3}=\frac{2x}{x^2+x+3x+3}=\frac{2x}{x(x+1)+3(x+1)}=$$
$$=\frac{2x}{(x+3)(x+1)}=\frac{A}{x+3}+\frac{B}{x+1}$$
$$2x=Ax+A+Bx+3B=(A+B)x+A+3B$$
You needs to solve the system
$$A+B=2,A+3B=0$$
|
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Evaluate the limit of $\ln(\cos 2x)/\ln (\cos 3x)$ as $x\to 0$ Evaluate Limits
$$\lim_{x\to 0}\frac{\ln(\cos(2x))}{\ln(\cos(3x))}$$
Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by $\stackrel{LHR}{=}$. LHR stands for L'Hôpital Rule)
\begin{align*}
\lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\right)\right)}{\ln \left(\cos \left(3x\right)\right)}\right)&\stackrel{LHR}{=}\\
&= \lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\right)\right)}{\ln \left(\cos \left(3x\right)\right)}\right)&\\
&=\lim _{x\to \:0}\frac{\left(\ln \left(\cos \left(2x\right)\right)\right)'}{\left(\ln \left(\cos \left(3x\right)\right)\right)'}\\
&=\lim _{x\to \:0}\left(\frac{-\frac{2\sin \left(2x\right)}{\cos \left(2x\right)}}{-\frac{3\sin \left(3x\right)}{\cos \left(3x\right)}}\right)\\
&=\lim _{x\to \:0}\left(\frac{2\sin \left(2x\right)\cos \left(3x\right)}{3\cos \left(2x\right)\sin \left(3x\right)}\right)\\
&\stackrel{LHR}{=}\lim _{x\to \:0}\frac{\left(2\sin \left(2x\right)\cos \left(3x\right)\right)'}{\left(3\cos \left(2x\right)\sin \left(3x\right)\right)'}\\
&=\lim _{x\to \:0}\left(\frac{4\cos \left(2x\right)\cos \left(3x\right)-6\sin \left(3x\right)\sin \left(2x\right)}{9\cos \left(3x\right)\cos \left(2x\right)-6\sin \left(2x\right)\sin \left(3x\right)}\right)\\
&=\lim _{x\to \:0}\left(\frac{2\left(2\cos \left(2x\right)\cos \left(3x\right)-3\sin \left(3x\right)\sin \left(2x\right)\right)}{3\left(3\cos \left(2x\right)\cos \left(3x\right)-2\sin \left(3x\right)\sin \left(2x\right)\right)}\right)\\
&=\frac{\lim _{x\to \:0}\left(2\left(2\cos \left(2x\right)\cos \left(3x\right)-3\sin \left(3x\right)\sin \left(2x\right)\right)\right)}{\lim _{x\to \:0}\left(3\left(3\cos \left(2x\right)\cos \left(3x\right)-2\sin \left(3x\right)\sin \left(2x\right)\right)\right)}=\dfrac{4}{9}
\end{align*}
*
*Could we do it in others ways?
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If you know $\lim_{x\to0}\frac{\ln(1+ x)}{x} = 1$, then
\begin{align}
\lim_{x\to 0} \frac{\ln(\cos 2x)}{\ln(\cos 3x)} = \lim_{x\to 0} \frac{\ln(1 + \cos 2x - 1)}{\cos 2x -1} \frac{\cos 3x -1} {\ln(1 + \cos 3x - 1)} \frac{\cos 2x -1}{\cos 3x -1}
\end{align}
We have $\lim_{x\to 0} \frac{\ln(1 + \cos 2x - 1)}{\cos 2x -1} = 1$, $\lim_{x\to 0} \frac{\ln(1 + \cos 3x - 1)}{\cos 3x -1} = 1$
and $\lim_{x\to 0}\frac{\cos 2x -1}{\cos 3x -1} = \frac{4}{9}$, since $\cos x = 1 - \frac{x^2}{2} + o(x^3)$ when $x \to 0$. We can also use L'Hopital's rule here
|
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|
Evaluate indefinite integral. $\int\frac{dy}{\sqrt{169 + y^2}}$
Evaluate $$\int\frac{dy}{\sqrt{169 + y^2}}$$
I have solved the problem, but don't seem to be getting the right answer. I get $\ln|\sqrt{13+y}+y|+C$ as the answer, and it is not the answer, how can I solve this?
|
Use the substitution $x=13\tan{u}$, then $dx=13\sec^2{u} \ du$
\begin{align}
\int\frac{dx}{\sqrt{169+x^2}}
&=\int\frac{13\sec^2{u} \ du}{\sqrt{169+169\tan^2{u}}}\\
&=\int\sec{u} \ du\\
&=\ln|\sec{u}+\tan{u}|+c\\
&=\ln\left|\frac{x}{13}+\sqrt{1+\left(\frac{x}{13}\right)^2}\right|+c
\end{align}
Alternatively, we can use the hyperbolic substitution $x=13\sinh{u}$, then $dx=13\cosh{u} \ du$. Thus
\begin{align}
\int\frac{dx}{\sqrt{169+x^2}}
&=\int\frac{13\cosh{u} \ du}{\sqrt{169+169\sinh^2{u}}}\\
&=\int\frac{13\cosh{u} \ du}{13\cosh{u}}\\
&=\operatorname{arcsinh}\left(\frac{x}{13}\right)+c
\end{align}
These 2 answers are equal.
|
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|
Evaluate $\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$
Evaluate $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$
I tried by taking $x^2$ out of the root by taking it common.
i.e: $$\lim_{x \to -\infty} \left(\frac{x\sqrt{\frac{1}{x^2}+1}-x}{x} \right)$$
and then cancelling the x in numerator and denominator
$$\lim_{x \to -\infty} \left(\frac{\sqrt{\frac{1}{x^2}+1}-1}{1} \right)$$
then substituting $x= -\infty$ in the equation, we get,
$$\lim_{x \to -\infty} \left(\frac{\sqrt{0+1}-1}{1} \right)$$
which equals to $0$. But it is not the correct answer.
What have I done wrong.
|
Hint: Multiply numerator and denominator by $\sqrt{1+x^2} + x$.
$$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)\cdot \frac{\sqrt{1+x^2} +x}{\sqrt{1 + x^2}+x} = \lim_{x\to -\infty}\frac{1 + x^2 - x^2}{x(\sqrt{1 + x^2} + x)}$$
$$= \lim_{x\to -\infty}\frac 1{x\sqrt{1 + x^2} + x^2}$$
Can take it from here?
|
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|
If $4x^2+y^2=40$ and $xy=-6$, find the value of $2x+y$. If $4x^2+y^2=40$ and $xy=-6$, find the value of $2x+y$.
I tried the following,
As we know, $(a+b)^2=a^2+b^2+2ab$
Therefore, $(4x^2+y^2)^2=(4x^2)^2+(y^2)+2(4x^2*y^2)$
$=16^4+y^4+8x^2y^2$
What should I do know? Please help. I am stuck.
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$$(2x+y)^2=4x^2+4xy+y^2=(4x^2+y^2)+4xy=40+4(-6)=16 \Rightarrow 2x+y=\pm 4$$
|
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|
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form EDIT: Please see EDIT(2) below, thanks very much.
I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written in the same form: $49$ is already shown, and $7$ is $2^2+3\cdot 1^2$.
I'm looking for a general proof of this using infinite descent. I tried to let $a^2+3b^2=xy$, where $x,y$ are positive integer factors of the LHS, but couldn't continue from here I'm afraid. However, I have managed to prove that integers of the form $a^2+3b^2$ are closed under multiplication, i.e. the product of two of them is of the same form. I was hoping this would help.
I also had another question if you have time: Is the representation in the form $a^2+3b^2$ unique? For example, is it possible for an integer, say $N$, to be $N=a^2+3b^2=x^2+3y^2$ with $a\ne x$ and $b\ne y$?
EDIT: Sorry, here is the "closed under multiplication" proof I found:
Using Diophantus' Identity we have $(a^2+3b^2)(x^2+3y^2)=(a^2+(\sqrt{3}b)^2)(x^2+(\sqrt{3}y)^2=(ax+\sqrt{3}\cdot \sqrt{3} y)^2)+(a\cdot \sqrt{3} y - \sqrt{3} b\cdot x)^2=(ax+3by)^2+3(ay-bx)^2$ as required.
EDIT (2): I am very sorry. The question should be:
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form.
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This is false as stated. $4$ has an expression as $a^2 + 3 b^2,$ with $a=b=1,$ but $2$ does not. $25$ has such an expression, with $a=5,b=0,$ but $5$ itself does not. The same applies to any $q^2,$ where $q \equiv 2 \pmod 3$ is prime.
|
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|
Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Inequality
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality
$$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$
I stumbled upon this question some days ago and been trying AM-GM to find the solution but so far have been unsuccessful.
|
here is a elementary method:
WLOG, let $a=$Min{$a,b,c$} $\implies a \le1 ,x=bc$
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=3((a+b+c)^2-3ab-3ac-3bc)=3(9-3a(3-a)-3bc)$
LHS$=\dfrac{a(3-a)+bc}{abc}=\dfrac{a(3-a)+x}{ax},$
RHS$= 9-3a(3-a)-3bc+abc+\dfrac{7}{4}=\dfrac{43}{4}-3a(3-a)+(a-3)x$
LHS-RHS$=\dfrac{(3-a)}{x}\left(x^2-\dfrac{4a^3-36a^2+43a-4}{4a(3-a)}x+1 \right)$
$f(x)=x^2+\dfrac{g_1(a)}{4a(3-a)}x+1 ,g_1(a)=-4a^3+36a^2-43a+4$
it is trivial that $g_1(a)>0 \implies f(x)>0 $
when $g_1(a)<0 ,f_{min}=1-\left(\dfrac{g_1(a)}{2\times4a(3-a)}\right)^2=\dfrac{(2a-1)^2(4-3a)(12a^3-44a^2+67a-4)}{64a^2(3-a)^2}$
$12a^3-44a^2+67a-4+g_1(a)=8a(3-a)>0 \implies 12a^3-44a^2+67a-4>0 \implies \\f_{min} \ge 0$
when $a=\dfrac{1}{2}, f_{min}=0 \implies x= -\dfrac{g_1(a)}{2\times4a(3-a)}=1$
$b+c=\dfrac{5}{2},bc=1 \implies (\dfrac{1}{2},2)$
|
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How prove $\angle ABC=60^o$ for triangle ABC if $BD = BF = AB-AC$ and $\frac{1}{{AD}}+\frac{1}{{CF}}=\frac{1}{{BD}}$? In acute triangle $ABC$ with $AB > AC$,we consider points $D$, $F$ (internal) of $AB$, $BC$ respectively, such that $BD = BF = AB-AC$.
If: $\frac{1}{{AD}}+\frac{1}{{CF}}=\frac{1}{{BD}}$, to how prove that $\angle ABC ={60^\circ}$?
|
From $BD = BF = AB-AC$ you can deduce that $AD = AC$.
Let $r = BD = BF$ and $R = AC = AD$.
From the other equation we deduce that $CF = \displaystyle\frac{Rr}{R-r}$.
Using the law of cosines:
$$(R+r)^2 + \left(r + \frac{Rr}{R-r}\right)^2 - R^2 = 2(R+r)\left(r + \frac{Rr}{R-r}\right)\cos B$$
Or equivalently:
$$(R+r)^2(R-r)^2 + (2Rr-r^2)^2 - R^2(R-r)^2 = 2(R+r)(R-r)(2Rr-r^2)\cos B $$
So:
$$\frac{(R+r)^2(R-r)^2 + (2Rr-r^2)^2 - R^2(R-r)^2}{2(R+r)(R-r)(2Rr-r^2)} = \cos B \\ \Rightarrow \frac{(2Rr+r^2)(R-r)^2 + r^2(2R-r)^2 }{2r(R+r)(R-r)(2R-r)} = \cos B\\ \Rightarrow \frac{(2R+r)(R-r)^2 + r(2R-r)^2 }{2(R+r)(R-r)(2R-r)} = \cos B \\ \Rightarrow \frac{1}{2}·\frac{2r^3 - 4Rr^2 + R^2r +2R^2}{r^3-Rr^2-R^2r+2R^2} = \cos B$$
We want $\cos B = 1/2$. So we want the biq quotient to be $1$. That only can be if: $$r^3 - 3r^2R+2R^2r = 0 \Rightarrow r(r-2R)(r-R)=0$$
Now $r \neq 0$ because $r = AB - AC$ and $AB > AC$.
So we have to prove that either $R = r$ or $2R = r$
Note that $r = R$ is not possible because of the second equation:
$$\frac{1}{R} + \frac{1}{CF} = \frac{1}{r}$$
So if $r = R$ then $\displaystyle\frac{1}{CF} = 0$ wich cannot be.
The same reasoning goes for $2R = r$:
$$\frac{1}{R} + \frac{1}{CF} = \frac{1}{2R}$$
What means that $CF = -R$ Wich is impossible. Note that since $CF \displaystyle\frac{Rr}{R-r}$ then we would have $Rr = (r-R)R$ what would mean that $R = 0$ so we wouldn't have a triangle.
Therefore, $\angle ABC \neq 60^\circ$
|
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Proving these trigonometric sums $\sum\limits_{k=0}^{n-1}\sin\frac{2k^2\pi}{n}=\frac{\sqrt{n}}{2}\left(\cos\frac{n\pi}{2}-\sin\frac{n\pi}{2}+1\right)$ Can someone help me to prove that:
$$ \sum_{k=0}^{n-1}\sin\frac{2k^2\pi}{n}=\frac{\sqrt{n}}{2}\left(\cos\frac{n\pi}{2}-\sin\frac{n\pi}{2}+1\right)$$
$$\sum_{k=0}^{n-1}\cos\frac{2k^2\pi}{n}=\frac{\sqrt{n}}{2}\left(\cos\frac{n\pi}{2}+\sin\frac{n\pi}{2}+1\right)$$
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Let $A = \sum_{k=0}^{n-1}\cos\frac{2k^2\pi}{n}$ and $B = \sum_{k=0}^{n-1}\sin\frac{2k^2\pi}{n}$ then we are looking to show:
$$ \sum_{k=0}^{n-1} e^{2\pi i k^2/n} = A + iB = e^{\tfrac{\pi i n}{2}}\sqrt{n}\tfrac{1+i}{2} $$
I believe this is the Gauss sum
I have seen the Gauss sum for primes $p$ but not necessarily for $n$. However,
You know the absolute value because you can find the norm, i.e. multiply by the complex conjugate, or use Parseval Theorem
$$ \bigg|\bigg|\sum_{k=0}^{n-1} e^{2\pi i k^2/n}\bigg|\bigg|^2 =
\sum_{k=0}^{n-1} e^{2\pi i k^2/n} \times
\overline{\sum_{k=0}^{n-1} e^{2\pi i k^2/n}}
= 1\cdot 1 + \dots 1 + \cdot 1 = n$$
The norm of our Gauss sum is $\sqrt{n}$ and we need to compute the sign.
|
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Trigonometric formula simplifies to $\sin x\cos x[\tan x+\cot x]$ Again, I have a little trouble figuring out how we got from the first step to the next one. It would be really appreciated if someone could help me out.
$$
\begin{split}LHS &= \cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi +x)\left[\cot\left(\frac{3\pi}{2}-x\right) +\cot(2\pi+x) \right] \\ &= \sin x\cos x[\tan x+\cot x]\end{split}
$$
I've tried the following: $\cot(3\pi/2)$ is $0$, so it would leave out $-\cot x$, that's where I lose the thread. $\cos(3\pi/2)$ is 0,
so that leaves out $\cos x$ and $\cos2\pi$ is one and that's where I lose the thread.
Note: that this is not an assignment question, these are all solved examples, only for my practice. Source image
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Replace the terms on the left side of your left side of the identity with these, and you will end up with the right hand side of your identity.
$$\require{cancel}\cos \left( \frac{3\pi}2+x \right)=\cancelto{0}{\cos \frac{3\pi}2}\cos x - \cancelto{1}{\sin \frac{3\pi}2}\sin x=\sin x$$
$$\require{cancel}\cos \left(2\pi+x \right)=\cancelto{1}{\cos 2\pi}\cos x - \cancelto{0}{\sin 2 \pi}\sin x=\cos x$$
$$\require{cancel}
\cot\left(\frac{3\pi}2 - x\right) = \frac{\cos(\frac{3\pi}2 - x)}{\sin(\frac{3\pi}2 - x)} = \frac{\cancelto{0}{\cos\frac{3\pi}2} \cos x + \cancelto{1}{\sin\frac{3\pi}2} \sin x }{\cancelto{1}{\sin\frac{3\pi}2} \cos x - \cancelto{0}{\cos\frac{3\pi}2} \sin x} = \frac{\sin x}{\cos x} = \tan x$$
$$\require{cancel}
\cot\left(2\pi + x\right) = \frac{\cos(2\pi + x)}{\sin(2\pi + x)} = \frac{\cancelto{1}{\cos 2\pi} \cos x - \cancelto{0}{\sin 2\pi} \sin x }{\cancelto{0}{\sin 2\pi} \cos x + \cancelto{1}{\cos 2\pi} \sin x} = \frac{\cos x}{\sin x} = \cot x$$
|
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|
Proving that a number is non-negative? The numbers $a$,$b$ and $c$ are real.
Prove that at least one of the three numbers
$$(a+b+c)^2 -9bc \hspace{1cm} (a+b+c)^2 -9ca \hspace{1cm} (a+b+c)^2-9ab$$
is non-negative.
Any hints would be appreciated too.
|
Hint:
If all three numbers are negative, then:
$$ab > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} ac > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} bc > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm}$$
Therefore, if we multiply the three inequalities:
$$a^2b^2c^2 > \left(\frac{a+b+c}{3}\right)^6$$
Or equivalently:
$$\left(\sqrt[3]{abc}\right)^6 > \left(\frac{a+b+c}{3}\right)^6$$
Do you know any inequality you can use here do disprove this?
|
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Prove $ x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$
I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ we have on the left side of the equation $x^2-1$ and on the right side: $(x-1)(x+1)$ which when distributed is $x^2-1$. So my base case holds.
Now I assume that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ for some $n$. However, this is where I am stuck. Am I trying to show $x^{n+1}-1=(x-1)(x^n + x^{n-1}+x^{n-2}+...+x+1)$? I am still a novice when it comes to these induction proofs. Thanks
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You may just open hooks in right half of expression:
$$(x - 1)(x^{n - 1} + x^{n - 2} + \dots + x + 1) = x * (x^{n - 1} + x^{n - 2} + \dots + x + 1) - 1 * (x^{n - 1} + x^{n - 2} + \dots + x + 1) = (x^n + x^{n - 1} + \dots + x^2 + x) - (x^{n - 1} + x^{n - 2} + \dots + x + 1) = x^ n - 1 $$
No problems)
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|
greatest common divisor and solution in integers The greatest common divisor of 203 and 147; $gcd(203,147)=7$.
Thus how can we find all the solution in integers $x,y$ of the equation $203x + 147y=7$?
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First off we can use the Extended Euclidean Algorithm to find some "first" values of x and y.
$$\begin{array}{rclcrcllr}
1 & \cdot & 203 & + & 0 & \cdot & 147 & = & 203 \\
0 & \cdot & 203 & + & 1 & \cdot & 147 & = & 147 \\
1 & \cdot & 203 & - & 1 & \cdot & 147 & = & 56 \\
-2 & \cdot & 203 & + & 3 & \cdot & 147 & = & 35 \\
3 & \cdot & 203 & - & 4 & \cdot & 147 & = & 21 \\
-5 & \cdot & 203 & + & 7 & \cdot & 147 & = & 14 \\
8 & \cdot & 203 & - & 11 & \cdot & 147 & = & 7 \\
-21 & \cdot & 203 & + & 29 & \cdot & 147 & = & 0 \\
\end{array}$$
These last two are important! The first tells us one solution of the equation; the second tells us a thing that, when I add it in to the first, doesn't change the right side.
So, our result is: $203x+147y=7 \rightarrow (x,y)=(8-21k,-11+29k)$
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Prove that $f$ is differentiable in $(0,0)$ if and only if $\lim_{t\to0+} g(t)$ exists Let $g:[0,\infty)\to\mathbb{R}$ be a mapping and $f(x,y)=xg(\sqrt{x^2+y^2})$ for all $(x,y)\in\mathbb{R^2}$.
Prove that $f$ is differentiable in $(0,0)$ $\iff$ $\lim_{t\to0+} g(t)$ exists.
My attempt:
$\implies:$ We suppose that $f$ is differentiable in $(0,0)$, that is, there is a linear mapping $A=(a_1 \,\, a_2)$ such that $$f(x,y)-f(0,0)-A(x,y) = R(x,y)$$ where R is a remainder term satisfying $\lim_{(x,y)\to(0,0)}\frac{R(x,y)}{|(x,y)|} = 0$.
Given the definition of $f$, we see that $f(0,0)=0$, so that $$f(x,y)-A(x,y) = R(x,y) \\ \iff xg\big(\sqrt{x^2+y^2}\big)-A(x,y) = R(x,y) \\ \iff g\big(\sqrt{x^2+y^2}\big) = \frac{R(x,y)}{x}+\frac{1}{x}A(x,y) = \frac{R(x,y)}{x}+\frac{1}{x}(a_1x + a_2y).$$
Now we use polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$, which gives $$g(r) = \frac{R(r\cos\theta, r\sin\theta)}{r\cos\theta}+\frac{1}{r\cos\theta}(a_1r\cos\theta+a_2r\sin\theta) \\= \frac{R(r\cos\theta, r\sin\theta)}{r\cos\theta}+a_1+a_2\frac{\sin\theta}{\cos\theta}$$
Now taking the limit as $r\to0$ the first term vanishes and we see that the limit is $$a_1+a_2\frac{\sin\theta}{\cos\theta}.$$
$\impliedby:$ Now we suppose that $\lim_{t\to0+} g(t)$ exists. I will calculate the two partial derivatives and then show that the matrix $$A = \big(\partial_1 f(0,0) \,\,\, \partial_2 f(0,0)\big)$$ is indeed the derivative of $f$ in $(0,0)$.
We have that $$\partial_1 f(0,0) = \lim_{t\to0} \frac{f(t,0)}{t} = \lim_{t\to0}\frac{tg(t)}{t}=\lim_{t\to0}g(t) = M$$ for some finite $M$; and $$\partial_2 f(0,0)=\lim_{t\to0}\frac{f(0,t)}{t} = 0.$$
We shall now investigate $$\lim_{(x,y)\to(0,0)} \frac{f(x,y) -f(0,0) - A(x,y)}{\sqrt{x^2+y^2}}.$$ Using polar coordinates this becomes $$\lim_{r\to0} \frac{r\cos\theta \,g(r) - Mr\cos\theta}{r} = \lim_{r\to0} \cos\theta \,g(r) - M\cos\theta = 0,$$ as desired.
This post turned out much longer than I had anticipated -- I'm sure I made it far more complicated than necessary :)
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For the first implication, if $f$ is differentiable at $(0,0)$, then in particular it admits a partial derivative with respect to $x$ at $0$, so that the following limit exists :
$$
\frac{\partial f}{\partial x}(0,0) = \lim_{x \to 0^+} \frac{f(x,0) - f(0,0)}{\sqrt{x^2+0^2}} = \lim_{x \to 0^+} \frac{x g\left( \sqrt{x^2+0^2} \right) - 0}{x} = \lim_{x \to 0^+} g(x).
$$
Conversely, define $M \overset{def}= \lim_{x \to 0^+} g(x)$. Then
$$
\lim_{\sqrt{x^2+y^2} \to 0} \frac{f(x,y) - Mx}{\sqrt{x^2+y^2}} = \lim_{\sqrt{x^2+y^2}} \frac{x g \left( \sqrt{x^2+y^2} \right) - Mx}{\sqrt{x^2+y^2}} = \lim_{\sqrt{x^2+y^2} \to 0} \left( \frac{x}{\sqrt{x^2+y^2}} \right) \left( g \! \left( \sqrt{x^2+y^2} \right) - M \right) = 0.
$$
The last limit is equal to zero because
$$
\left| \left( \frac{x}{\sqrt{x^2+y^2}} \right) \left( g \!\left( \sqrt{x^2+y^2} \right) - M \right) \right| \le \left| \, g \! \left( \sqrt{x^2+y^2} \right) - M \right| \to 0.
$$
This proves $f$ is differentiable (with derivative $A(x,y) = Mx$, as expected).
You were almost there! Honestly, I traced on your steps. You did a good job =)
Hope that helps,
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Evaluating the limits $\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$ and $\lim_{(x,y)\to(\infty,8)}(1+\frac{1}{3x})^\frac{x^2}{x+y}$ I got the following problem:
Evaluate the following limits or show that it does not exist:
$$\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$$
and
$$\lim_{(x,y)\to(\infty,8)}\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}$$
I tried for an hour and half evaluating each of those limits but I failed and I got nothing useful to share.
Some hints will be appreciated.
Thanks.
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By setting:
$$x:=r\cos\theta \\ y:=r\sin\theta$$
You can turn: $$\lim_{\substack{
x\rightarrow\infty\\
y\rightarrow\infty}}
{f(x,y)}$$ into: $$\lim_{r\rightarrow\infty}{g(r,\theta)}.$$
So that:
$$ \frac{2x-y}{x^2-xy+y^2}=\frac{1}{r}\cdot\frac{2\cos\theta-\sin\theta}{(1-\cos\theta\sin\theta)}\,\overset{r\rightarrow\infty}{\longrightarrow}\,0$$
The second one is easier. By noting that:
$$
\lim_{\substack{
x\rightarrow\infty\\
y\rightarrow 8}}{\frac{y}{x}} = 0
$$
We get:
$$ \left(1+\frac{1}{3x}\right)^{\frac{x^2}{x+y}}=\left(\left(1+\frac{1}{3x}\right)^{3x}\right)^{\frac{1}{3+3\frac{y}{x}}} \longrightarrow e^{\frac{1}{3}}$$
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|
How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with degree 3 such that $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots
Note that $\dfrac{\pi}{12}$, $\dfrac{9\pi}{12}$, $\dfrac{17\pi}{12}$ are solution of equation $\cos3\theta=\dfrac{1}{\sqrt{2}}$ and $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are distinct number.
We have $\cos3\theta=4\cos^3\theta-3\cos\theta$. Let $x=\cos\theta$, therefore
$\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots of $4x^3-3x=\dfrac{1}{\sqrt{2}}.$
I want method similar to this to find $P(x)$.
Thank you.
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HINT: If $r$ is a root of a polynomial then $(x-r)$ is a factor. You have four roots.
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|
Using sum/product of roots of a cubic equation to solve given expression $(a+b)^3+ (b+c)^3 + (c+a)^3$ If $a, b, c$ are the roots of the equation $7x^3- 25x +42 =0$, then the value of the expression $(a+b)^3+ (b+c)^3 + (c+a)^3$ is?
I tried to solve this but wasn't able to simplify the term to be able to apply sum and product of roots of equations. Could someone help me through simplification of the expression to make this solvable?
Since $a+b+c = \frac{-25}{7}$, $ab + bc + ca = 0$ and $abc = -42$, if i factorize the expression i get
$(2[(a+b+c)^3 - 3a^2(b+c) - 3 c^2 (a+b) -3b^2(a+c) -6abc]) - 3(a+b)^2(b+2c+a) - 3(c+a)^2(c+2b+a) - 3(b+c)^2(c+2a+b))$
This is what i get, after my best attempts at simplification. Since this is an mcq type question, i believe this does not require such mammoth efforts. Could someone provide me a hint and possibly an easier way of simplification/application of the concept?
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Using Vieta's formula $\displaystyle a+b+c=0,(-1)^3abc=42$
$$\implies \sum (a+b)^3=-\sum c^3$$
Method $\#1:$
As $a,b,c$ individually satisfy the given equation, we can write $$7a^3=25a-42\text{ etc.}$$
Again using Newton's
Power Sum formula $$7\sum a^3=25\sum a-3\cdot42=\cdots$$
Method $\#2:$
Like If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.,
$$a^3+b^3+c^3=3abc$$
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How prove this inequality $\sum\limits_{cyc}\frac{1}{a+3}-\sum\limits_{cyc}\frac{1}{a+b+c+1}\ge 0$ show that:
$$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$
where $abcd=1,a,b,c,d>0$
I have show three variable inequality
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$.
Prove that
$$\frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$$
also see:can see
：http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=243
from this equality,I have see a nice methods:
I think this Four varible inequality is also true
First,Thank you Aditya answer,But I read it your solution,it's not true
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Partial Proof:
For general case of n variables, the inequality converts to:
$$\sum_i^n \frac1{1+a_1+a_2+\cdots+a_n-a_i}\le \sum_i^n\frac1{n-1+a_i}$$
Similiar to the given proof we can convert $\frac {a_1}{a_1+(n-1)}$ like this:
$$\frac{a_1}{a_1+(n-1)}=\frac{a_1}{a_1+(n-1)(a_1a_2\cdots a_n)^{1/n}}$$
Dividing numerator and denominator by $a_1^{1/n}$:
$$=\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+(n-1)\left(\frac{a_1a_2\cdots a_n}{a_1}\right)^{1/n}}
$$
Finally using AM-GM gives:
$$\frac{a_2+a_3+\cdots+a_n}{(n-1)}\ge\left(a_2a_3\cdots a_n\right)^{1/(n-1)}$$
Or:
$$(n-1)\left(a_2a_3\cdots a_n\right)\le (a_2+a_3+\cdots+a_n)^{n-1}$$
$$=\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+(n-1)\left(\frac{a_1a_2\cdots a_n}{a_1}\right)^{1/n}}
\ge\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+a_2^{(n-1)/n}+\cdots+a_n^{(n-1)/n}}$$
So, $$\sum_i^n\frac{a_i}{a_i+(n-1)}\ge\sum_i^n\frac{a_1^{(n-1)/n}}{a_1^{(n-1)/n}+a_2^{(n-1)/n}+\cdots+a_n^{(n-1)/n}}=1\tag{i}$$
We have proved what we need for the general case of n-variables, try putting $n=3$.
Since product of all numbers is 1, we can define new fractions as: Let $$\displaystyle a_1:=\frac{x_1}{x_2},a_2:=\frac{x_2}{x_3},\cdots,a_n:=\frac{x_n}{x_1}$$
Notice that in the given proof:
$$\frac{b}{ab+b+1}=\frac{x_2/x_3}{x_1/x_2.x_2/x_3+x_2/x_3+1}=\frac{x_2}{x_1+x_2+x_3 }$$
Now similiar to the given proof we can show that(step unproven):
$$\frac{2}{a_1+(n-1)}-\frac{1}{1+a_2+a_3+\cdots+a_n}-\frac{x_2}{x_!+x_2+\cdots+x_n }\ge0$$
$$\frac{2}{a_1+(n-1)}\ge \frac{1}{1+a_2+a_3+\cdots+a_n}+\frac{x_2}{x_1+x_2+\cdots+x_n }$$
$$\frac1{a_1+(n-1)}+\frac1{a_1+(n-1)}\ge\frac{1}{1+a_2+a_3+\cdots+a_n}+\frac{x_2}{x_1+x_2+\cdots+x_n }$$
Since $\displaystyle \sum_i^n\frac{x_2}{x_1+x_2+\cdots+x_n }=1$
$$\frac1{a_1+(n-1)}+\frac1{a_1+(n-1)}\ge\frac{1}{1+a_2+a_3+\cdots+a_n}+1$$
$$\sum_{cyc}\frac{1}{a_1+(n-1)}-\sum_{cyc,i}\frac1{1+a_2+a_3+\cdots+a_n }\ge1-\sum_{cyc}\frac{1}{a_1+(n-1)}\ge0 $$
The$\ge0$ part, we have proved in (i).
So,
$${\large \sum_{cyc}\frac{1}{a_1+(n-1)}\ge\sum_{cyc,i}\frac1{1+\sum_{cyc,j\ne i}a_j}}\Box$$
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|
To find maximum value If $A>0,B>0$ and $C>0$ and further it is known that $A+B+C=\frac{5\pi}{4}$,then find the maximum value of $\sin A+\sin B+\sin C$
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From a purely algebraic point of view, elimating $C$ from the constraint leads to maximizing $$F=\sin \left(A+B-\frac{\pi }{4}\right)+\sin (A)+\sin (B)$$ Taking derivatives $$F'_A=\cos \left(A+B-\frac{\pi }{4}\right)+\cos (A)=0$$ $$F'_B=\cos \left(A+B-\frac{\pi }{4}\right)+\cos (B)=0$$ This implies $A=B$ and then $$\cos(2A-\frac{\pi }{4})+\cos(A)=2 \cos \left(\frac{A}{2} -\frac{\pi }{8}\right) \cos \left(\frac{3A}{2}
-\frac{\pi }{8}\right)=0$$ and the only acceptable solution is $A=\frac{5 \pi }{12}$ with $B=C=A$ and then a maximum value equal to $$\frac{3 \left(1+\sqrt{3}\right)}{2 \sqrt{2}}$$
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Is it true that $\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1$? This question is inspired by the formula $$\displaystyle\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots = \zeta(2)-1,$$
see for instance this question.
It appears that for any integer $n\geq 0$, we have
$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1.$$
I believe that Yiorgos' answer could be generalized to prove the above formula. Can anyone pull it off?
Side question: If $n$ is replaced by a complex number $s$, for which values of $s$ does the integral converge? (We interpret $x^s$ as $\exp(x\log s)$, where $\log s$ is the principal branch of the complex log.) If the above formula is true, then for any integer $n\geq 0$, have $$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((n+1)x^n-nx^{n-1}) dx = \zeta(n+2) - 1.$$
Is it true that for complex value of $s$ for which the integral converges, we have
$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((s+1)x^s-sx^{s-1}) dx = \zeta(s+2)-1?$$
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Letting $t=x^{-1}$ so $dx=-t^{-2}dt$ this integral can be rewritten as:
$$\begin{align}\int_{1}^\infty \lfloor t\rfloor^{-1} t^{-(n+2)}dt &= \sum_{k=1}^\infty \frac{1}k\int_{k}^{k+1}t^{-(n+2)}dt \\&= \frac{1}{n+1}\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(n+1)}-(k+1)^{-(n+1)}\right)
\end{align}$$
That's pretty much the exact same as the proof for $n=0$.
Now,
$$\begin{align}
\frac{1}{k}(k+1)^{-(n+1)}&= \frac{1}{k(k+1)}(k+1)^{-n}\\
&=\left(\frac{1}{k}-\frac{1}{k+1}\right)(k+1)^{-n} \\
&=\frac{1}{k}(k+1)^{-n}- (k+1)^{-(n+1)}\\
&=\dots\text{ induction applied here on }\frac1k(k+1)^{-n}\dots\\
&=\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}-\cdots-\frac{1}{(k+1)^{n+1}}
\end{align}$$
So $$\frac{1}{k}\left(k^{-(n+1)}-(k+1)^{-(n+1)}\right) = \left(\sum_{j=2}^{n+1} \frac{1}{(k+1)^j}\right) + \frac{1}{k^{n+2}}+\frac{1}{k+1}-\frac1k$$
This yields the result you want.
The side question (at least when $s\neq -1$) essentially asks about when:
$$\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(s+1)}-(k+1)^{-(s+1)}\right)$$ converges. When $s$ is real and $s>-1$, the terms are positive, and bounded above by the sequence $k^{-(s+1)}-(k+1)^{-(s+1)}$, which is a telescoping sequence, so it converges.
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Simplify the derivative of $y=\frac{3x}{\sqrt{x-3}}$ First of I'm still not totally clued up on how to format all this math properly so some of it may look a bit strange, any help in fixing it would be greatly appreciated. Anyhow onto the math...
So I have to differentiate
$$y= \frac{3x}{\sqrt(x-3)}$$
And I used the quotient rule and now I'm at
$$\frac{3(x-3)^{1/2} - 3/2 x (x-3)^{-1/2}}{x-3}$$
The expected answer is
$$\frac{3(x-6)}{ 2(x-3)^{3/2}}$$
So I'm slightly off, any tips?
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$$\frac{dy}{dx}=\frac{d}{dx} \left ( \frac{3x}{(x-3)^{\frac{1}{2}}} \right)=\frac{(3x)'(x-3)^{\frac{1}{2}}-3x \cdot ((x-3)^{\frac{1}{2}})'}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{\frac{1}{2}-1}}{x-3}=\frac{3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}}{x-3}=\frac{2(x-3)^{\frac{1}{2}} \cdot (3(x-3)^{\frac{1}{2}}-\frac{3}{2}x(x-3)^{-\frac{1}{2}}) }{2(x-3)^{\frac{1}{2}} \cdot (x-3) }=\frac{6(x-3)-3x}{2(x-3)^{\frac{3}{2}}}= \\ =\frac{3x-18}{2(x-3)^{\frac{3}{2}}}=\frac{3(x-6)}{2(x-3)^{\frac{3}{2}}}$$
|
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|
Expressing the area as a function :) Express the area A of an equilateral triangle as a function of the height of the triangle.
Thanks :)
I am not sure where to even start on how to answer this problem.
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$$\text{ Area of an triangle }=\frac{1}{2} \cdot (\text{ base } ) \cdot (\text{ height })$$
The identity of an equilateral triangle is that all sides are equal to $x$.
The height is $AD$. The triangle $ABD$ is a right-angled triangle. Therefore, we can use the Pythagorean Theorem.
$$(AD)^2+(BD)^2=x^2 \Rightarrow (AD)^2+\left ( \frac{x}{2} \right )^2=x^2 \Rightarrow x^2-\frac{x^2}{4}=(AD)^2 \Rightarrow \frac{3}{4}x^2=(AD)^2 \\ \Rightarrow x=\frac{2}{\sqrt{3}}(AD) \Rightarrow x=\frac{2\sqrt{3}}{3}(AD)$$
Since all the sides are equal to $x$, the base $(BC)$ is also equal to $x$.
Therefore, $$\text{ Area of an triangle }=\frac{1}{2} \cdot (\text{ base } ) \cdot (\text{ height })=\frac{1}{2} \cdot x \cdot (AD)=\frac{1}{2} \frac{2\sqrt{3}}{3}(AD) (AD)=\frac{\sqrt{3}}{3}(AD)^2$$
So,$$A(h)=\frac{\sqrt{3}}{3}h^2$$ is the area of an equilateral triangle as a function of the height $h$ of the triangle.
|
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|
Hard Definite integral involving the Zeta function Prove that: $$\displaystyle \int_{0}^{1}\frac{1-x}{1-x^{6}}{\ln^4{x}} \ {dx} = \frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54} $$
I was able to simplify it a bit by substituting ${y = -\ln{x}}$ and some further mathematical manipulation but was not able to get the correct form.
|
The series
$$ 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$
$$ =\frac{24}{6^5}\sum_{k=0}^\infty \left(\frac{1}{(k+1/6)^5}-\frac{1}{(k+2/6)^5}\right)$$
Can be evaluated by the polygamma function $$ =\frac{24}{6^5} \left(\frac{-\psi^4(1/6)}{24} - \frac{-\psi^4(1/3)}{24} \right) $$
$$=\frac{1}{6^5} \left({\psi^4(1/3)} - {\psi^4(1/6)} \right) $$
$$= {\frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54}} $$
|
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|
Secant line slope question I am asked to:
Show the expression for the slope of the secant line through $y=x^2+3x$ at $x=3$ and $x=3+h$ is $msec=9+h$
\begin{align*}
msec &= \dfrac{f(x+h)-f(x)}{h} \\
&= \dfrac{(x+h)^2+3(x+h)-(x^2+3x)}{h}\\
&=\dfrac{(x+h)(x+h)+3x+3h-x^2-3x}{h}\\
&=\dfrac{2xh+h^2+3h}{h}\\
&=\dfrac{h(2x+h+3)}{h}\\
&=2x+h+3
\end{align*}
The first part seems to work:
\begin{align*}
msec&=2(3)+h+3\\
&=9+h
\end{align*}
Second part: \begin{align*}
msec&=2(3+h)+h+3\\
&=6+2h+h+3\\
&=3h+9
\end{align*}
Am I missing something or is this a mistake / flaw in the question?
|
You are missing less that you think. On the line where you conclude with $2x+h+3$, you need to stop, and insert $x=3$ to get $9+h$, and then conclude. There is nothing else left to do. There is no second part.
|
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|
If the equation $2x^2-7x+12=0$ has two roots alpha and beta ,then the value of alpha/beta+beta/alpha is If the equation $2x^2-7x+12 =0$ has two roots $\alpha$ and $\beta$ ,
then the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ is
note $x=\frac{-7+\sqrt{47}}{4},\frac{-7-\sqrt{47}}{4}$
then
$$\frac{\frac{-7+\sqrt{47}}{4}}{\frac{-7-\sqrt{47}}{4}}+\frac{\frac{-7-\sqrt{47}}{4}}{\frac{-7+\sqrt{47}}{4}}$$
so 96/2
|
$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$
since $\alpha+\beta=\frac{7}{2}$ and $\alpha\beta=6$, you can compute the value by substituting.
|
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|
Evaluating $\ln(\cos x))$ using Taylor expansion
Evaluate $\ln(\cos x)$ at $x_0=0$ and with the order of $n=4$.
Noticing that $\ln(\cos x) = \ln(1+ \cos x - 1)$ we can use $\ln(1+x)$ Taylor series.
Now, I've read I should use:
$$\ln(1+x) = x - \frac{x^2}{2} + R_2(x)$$
$$\cos x -1 = -\frac{x^2}{2} + \frac{x^4}{4!} + R_4(x)$$
Questions:
*
*If the demand was $n=4$, why did the author expand $\ln(1+x)$ for only two terms?
*Following the author's way, how to plug $\cos x$ evaluation?
$$\ln(1+\cos x -1) = -\frac{x^2}{2} + \frac{x^4}{4!} + R_4(x) - \frac{\left(-\frac{x^2}{2} + \frac{x^4}{4!} + R_4(x) \right)^2}{2} + R_2(-\frac{x^2}{2} + \frac{x^4}{4!} + R_4(x))$$
Is this what I should do? Because it's getting kinda messy..
|
So far so good---the key observations remaining are:
*
*In the contribution from the $u^2$ term of $\ln (1 + u)$, we can see that the only way to get a term of order $\leq 4$ in $x$ is $-\tfrac{1}{2}\left(-\frac{x^2}{2}\right)^2 = -\frac{x^4}{8}$.
*The term $R_2\left(-\frac{x^2}{2} + \cdots\right)$ is already order $6$ in $x$, and so does not contribute to the Taylor polynomial of order $4$.
Applying these facts to simplify the second and third terms gives that the polynomial you seek is
$\left(-\frac{x^2}{2} + \frac{x^4}{4!}\right) - \left(\frac{x^4}{8}\right)
= -\frac{x^2}{2} - \frac{x^4}{12}$.
|
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|
Is there an injective function such that $f(x^2)-f^2(x)\ge \frac{1}{4}$? The exercise asks me this:
Is there an injective function such that $f(x^2)-f^2(x)\ge \frac{1}{4}$?
ps: $f: \mathbb{R}\to \mathbb{R}$
I really don't know how to start :c, I appreciate hints.
|
There is no such injective function. To solve it we first see that if $x^2 = x$ (which is the case for $x=0$ and $x=1$) then $f(x^2) = f(x)$ and $f(x^2) - f^2(x) \geq \frac{1}{4}$ becomes (for $x=0$ or $x=1$)
$$f(x) - f^2(x) - \frac{1}{4} = -\left(f(x)- \frac{1}{2}\right)^2 \geq 0$$
but this is only possible ($-a^2 \geq 0$ implies $a=0$ since a square cannot be negative) if $f(x) = \frac{1}{2}$. We must therefore have $f(0) = f(1) = \frac{1}{2}$ so $f$ cannot be an injection (since for an injective function $f(x) = f(y)$ implies $x=y$).
|
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|
Prove $\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ if $a^2+b^2+c^2=1$
Ff $a,b,c$ are positive real numbers that $a^2+b^2+c^2=1$ ,Prove: $$\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$$
Additional info: I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done: I tried to change LHS to something more easy to work but I was not successful. For example $$\frac{ab}{1+c^2}=\frac{1}{2}\left(\frac{a^2+b^2+2ab}{1+c^2}-\frac{a^2+b^2}{1+c^2}\right)$$
that was not useful. Any hint for starting step is appreciated.
|
$\displaystyle \frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ is equivalent to
EDIT: Do not pursue this method. It is wrong! Left as warning.
$\displaystyle \frac{a^2+b^2}{1+c^2}+\frac{b^2+c^2}{1+a^2}+\frac{c^2+a^2}{1+b^2}\le\frac{3}{2}$, as $ab\leq\frac{1}{2}(a^2+b^2)$.
This is equivalent to:
$\displaystyle \frac{1-c^2}{1+c^2}+\frac{1-b^2}{1+b^2}+\frac{1-a^2}{1+a^2} \leq \frac{3}{2}$, where $a^2+b^2+c^2=1$.
Let $x=a^2,y=b^2,z=c^2$, for simplicity.
Then the above is equivalent to
$\displaystyle \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z} \leq \frac{9}{4}$, whre $x+y+z=1$.
|
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|
How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4]{xy^3}$$
$$3x+z=x+x+x+z\ge 4\sqrt[4]{x^3z}$$
but this is not true,because not all four equalities can hold at once.
This problem is from china middle school test,so I think have without Lagrange methods,so I think this inequality have AM-GM inequality
|
Hint:
$5y+2 \geq 2\sqrt{10y}$
$2z+5 \geq 2\sqrt{10z}$
$x+3y \geq 2\sqrt{3xy}$
$3x+z \geq 2\sqrt{3xz}$
|
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|
Process to show that $\sqrt 2+\sqrt[3] 3$ is irrational How can I prove that the sum $\sqrt 2+\sqrt[3] 3$
is an irrational number ??
|
If $x=\sqrt{2}+\sqrt[3]3$, then
$$(x-\sqrt{2})^3=x^3-3\sqrt{2}x^2+6x-2\sqrt{2}=3$$
Thus
$$x^3+6x-3=\sqrt{2}(3x^2+2)$$
And
$$\frac{x^3+6x-3}{3x^2+2}=\sqrt{2}$$
But if $x$ is a rational, then so is the left hand side of the above equality. However we know $\sqrt{2}$ is not rational. Contradiction, so $x$ is irrational.
|
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|
Solve the following equation I need help to solve the following equation.
$13x=4x^2$
My attempt:
$13x=4x^2$
$4x^2-13x=0$
$\frac{4x^2-13x}{4}=0$
$x^2-\frac{13x}{4}+0=0$
$x=-\frac{-13}{2} \begin{matrix} + \\ - \end{matrix} \sqrt{(\frac{-13}{2})^2}$
this results to $x1 = 8.125$ $x2 = -4.875$
The correct result is $x1 = 0$ $x2 = 3.25$
Where did I go wrong with this?
Thanks!!
|
$13x=x^2$
If $x\neq 0$ we can divide both sides by $x$ to obtain:
$\frac{13x}{x}=\frac{4x^2}{x}\Rightarrow 13=4x$
Then we divide both sides by $4$ to obtain:
$x = \frac{13}{4}$.
|
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|
Index of Summation Shift? Power Series and Differential Equations I have never had to index shift a summation series before, and it seems relatively straightforward, however, I am looking at an example in my textbook that doesn't make sense. I am wondering if someone might be able to outline the steps that appear to be missing.
Apologies for the formatting issue; perhaps someone can help with that?
Here is how it appears in the textbook:
$(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n$
$= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + \sum_{n=0}^{\infty}[2n(n-1)+6n+2]a_{n}x^n$
So it looks as though the two rightmost terms were combined somehow, but I cannot figure how this took place.
Can someone help with my understanding? I've never done anything like this before, and I'm frustrated I can't understand the rest of the example because of this.
Thanks!
|
Ok, just distribute to see:
\begin{align}
&(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n =\\
&=\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +2x^2\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2}+
\sum_{n=1}^{\infty}6na_{n}xx^{n-1}+\sum_{n=0}^{\infty}2a_{n}x^n \\
&= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty}2n(n-1)a{_{n}}x^2x^{n-2}+
\sum_{n=1}^{\infty}6na_{n}x^{n}+\sum_{n=0}^{\infty}2a_{n}x^n \\
&= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty}2n(n-1)a{_{n}}x^{n}+
\sum_{n=2}^{\infty}6na_{n}x^{n}+\sum_{n=2}^{\infty}2a_{n}x^n +6a_1x+2a_0+2a_1x \\
&= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty} \left( 2n(n-1)+
6n+2 \right)a_nx^n +8a_1x+2a_0\\
\end{align}
Now, notice the rightmost sum begins at $n=0$ in your post, can we see the exceptional terms as the $n=0$ and $n=1$ terms in the text's expression?
Added: I thought it might be useful to show how to group all the terms: continuing,
\begin{align}
&(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n =\\
&= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=0}^{\infty} \left( 2n(n-1)+
6n+2 \right)a_nx^n \\
&= \sum_{j=0}^{\infty}(j+2)(j+1)a{_{j+2}}x^{j} +\sum_{n=0}^{\infty} \left( 2n(n-1)+
6n+2 \right)a_nx^n \\
&= \sum_{j=0}^{\infty}(j+2)(j+1)a{_{j+2}}x^{j} +\sum_{n=0}^{\infty} \left( 2n(n-1)+
6n+2 \right)a_nx^n \\
&= \sum_{n=0}^{\infty} \left( [2n(n-1)+
6n+2]a_n+(n+2)(n+1)a{_{n+2}} \right)x^n
\end{align}
So, then when we have this is zero we find relations between the $n$-th and $n+2$-th coefficients.
|
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|
Simpler closed form for $\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}$ I'm trying to find a closed form of this sum:
$$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}.\tag{1}$$
WolframAlpha gives a large expressions containing multiple generalized hypergeometric functions, that is quite difficult to handle. After some simplification it looks as follows:
$$S=\frac{\pi^{3/2}}{3}-\sqrt{\pi}-\frac{\sqrt{\pi}}{324}\left[9\,_3F_2\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)\\+3\,_4F_3\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)+\,_5F_4\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)\right].\tag{2}$$ I wonder if there is a simpler form. Elementary functions and simpler special funtions (like Bessel, gamma, zeta, polylogarithm, polygamma, error function etc) are okay, but not hypergeometric functions.
Could you help me with it? Thanks!
|
By now, I've found a closed-form by doing some integral evaluation, a lot of hypergeometric, polylogarithm and polygamma manipulation.
$$
S = \sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{1}{192\sqrt3}\psi^{(3)}\left(\tfrac13\right)-\frac{\pi^4}{72\sqrt3}-1\right).
$$
|
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|
Simplify rational expression How do I simplfy this expression?
$$\dfrac{\frac{x}{2}+\frac{y}{3}}{6x+4y}$$
I tried to use the following rule $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$
But I did not get the right result.
Thanks!!
|
$$\begin{align}
\frac{\frac{x}{2} + \frac{y}{3}}{6x+4y} &= \frac{\frac{x}{2} + \frac{y}{3}}{\frac{6x+4y}{1}}\\
&= \frac{\frac{x}{2} + \frac{y}{3}}{\frac{6x+4y}{1}}\cdot\frac{\frac{1}{6x+4y}}{\frac{1}{6x+4y}}\\
&= \frac{\bigg(\frac{x}{2}+\frac{y}{3}\bigg)\cdot\frac{1}{6x+4y}}{\frac{6x+4y}{6x+4y}}\\
&= \frac{\bigg(\frac{x}{2}+\frac{y}{3}\bigg)\cdot\frac{1}{6x+4y}}{1}\\
&= \bigg(\frac{x}{2}+\frac{y}{3}\bigg)\cdot\frac{1}{6x+4y}\\
&= \bigg(\frac{12}{12}\cdot\frac{x}{2}+\frac{12}{12}\cdot\frac{y}{3}\bigg)\cdot\frac{1}{6x+4y}\\
&= \bigg(\frac{1}{12}\cdot\frac{12x}{2}+\frac{1}{12}\cdot\frac{12y}{3}\bigg)\cdot\frac{1}{6x+4y}\\
&= \bigg(\frac{1}{12}\cdot6x+\frac{1}{12}\cdot4x\bigg)\cdot\frac{1}{6x+4y}\\
&= \frac{1}{12}(6x+4y)\cdot\frac{1}{6x+4y}\\
&=\frac{1}{12}\cdot\frac{6x+4y}{6x+4y} \\
&= \frac{1}{12}\cdot 1 \\
&= \frac{1}{12}
\end{align}$$
|
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|
Divisibility rule of 11
Let $M$ be a natural number with $n+1$ digits; represented by $M=a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}$
Show $M$ is divisible by $11$ if and only if
$$(a_{0}+a_{2}+a_{4}+\ldots)-(a_{1}+a_{3}+a_{5}+\ldots)$$ is also a divisible by 11
*
*here is Solution By Mr Américo Tavares
Let
$$M=a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}=a_{n}10^{n}+a_{n-1}10a^{n-1}+\cdots +a_{2}10^{2}+a_{1}10+a_{0}.$$
For $\mod 11$ we have $10^{0}\equiv 1,10^{1}\equiv 10,10^{2}\equiv 1$,
and if $n$ even
$$10^{n}\equiv 1;$$
if $n$ odd $$10^{n}\equiv 10.$$ Thus
$$a\equiv (a_{0}+a_{2}+a_{4}+\cdots )+10(a_{1}+a_{3}+a_{5}+\cdots )\pmod{11}.$$
Since $-10\equiv 1\pmod{11}$, we have $$10\equiv (-(-10))\equiv -1\pmod{11}$$ and
$$a\equiv (a_{0}+a_{2}+a_{4}+\cdots )-(a_{1}+a_{3}+a_{5}+\cdots )\pmod{11}.$$
Could we prove it by the Methods of :
*
*Mr. Mr. JimmyK4542
*Mr Bill Dubuque
from here Particular number is divisible by 11
Or Is There another way to prove it ?
|
$$10=11-1=11k-1 \\100=99+1=9(11)+1=11q+1\\1000=1001-1=11m-1\\10000=9999+1=11n+1\\\text{and so}\\M=a_{n}...a_{3}a_{2}a_{1}a_{0}=a_{0} +10a_{1}+100a_{2}+1000a_{3}+...=\\M=a_{0} +(11-1)a_{1}+(99+1)a_{2}+(1001-1)a_{3}+...=\\M=a_{0} +(-1)a_{1}+(1)a_{2}+(-1)a_{3}+...+11(1a_{1}-9a_{2}+99a_{3}....)=\\M=a_{0} +(-1)a_{1}+(1)a_{2}+(-1)a_{3}+...+11Q\\\text{now divide M by 11}\\M=(a_{0} +(-1)a_{1}+(1)a_{2}+(-1)a_{3}+...)+11Q =M'+11Q\\M/11\rightarrow M'
$$
|
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|
Series representation of function with fractions, logarithms, squares and cosines. I'm looking for a series representation for
$$\dfrac x{x^2+(\log \cos x)^2}$$
Where $x\in(0,\pi/2)$
Note: Both finite and infinite series are accepted.
I have tried taylor series, but it requires the $n$th derivative, which is not trivial. But since it can be used in the taylor series, a formula for $n$th derivative would also be a valid answer.
I know that the $n$th derivative can be found with Faà di Bruno's formula, however I would like to avoid using this formula because of it's complexity. (If you can shorten a Faà di Bruno formula, that would be fine too)
|
For a truncated series, I suppose that we could use $$\cos(x) \simeq 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}$$ then use $$\log(1+y)\simeq y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}$$ Replace $$y=-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}$$ to get $$\log(\cos(x))\simeq-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}$$ $$\Big(\log(\cos(x))\Big)^2\simeq\frac{x^4}{4}+\frac{x^6}{12}+\frac{7 x^8}{240}+\frac{79
x^{10}}{7560}$$ $$x^2+\Big(\log(\cos(x))\Big)^2\simeq x^2+\frac{x^4}{4}+\frac{x^6}{12}+\frac{7 x^8}{240}+\frac{79
x^{10}}{7560}$$ $$x^2+\Big(\log(\cos(x))\Big)^2\simeq x^2\Big(1+\frac{x^2}{4}+\frac{x^4}{12}+\frac{7 x^6}{240}+\frac{79
x^{8}}{7560}\Big)$$ and now perform the long division to arrive to $$\dfrac x{x^2+(\log \cos x)^2} \simeq \frac{1}{x}-\frac{x}{4}-\frac{x^3}{48}-\frac{x^5}{320}-\frac{31
x^7}{48384}$$
May be, a faster solution could use what Greg Martin suggested and write $$\frac {x}{x^2+y^2} = x^{-1} + x^{-3}y^2 + x^{-5}y^4 + \cdots=\frac{1}{x} \Big(1-\frac{y^2}{x^2}+\frac{y^4}{x^4}-\frac{y^6}{x^6}+ \cdots\Big)$$ in which $y$ should be replaced by the development of $\log(\cos(x))$ given above and noticing that $$\frac{y^2}{x^2}=\frac{x^2}{4}+\frac{x^4}{12}+\frac{7 x^6}{240}+\frac{79 x^8}{7560}+O\left(x^9\right)\simeq x^2 \Big(\frac{1}{4}+\frac{x^2}{12}+\frac{7 x^4}{240}+\frac{79 x^6}{7560}\Big)$$
|
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|
Closed form for $1 + 3 + 5 + \cdots +(2n-1)$ What is the closed summation form for $1 + 3 + 5 + \cdots + (2n-1)$ ?
I know that the closed form for $1 + 2 + 3+\cdots + n = n(n+1)/2$ and I tried plugging in $(2n-1)$ for $n$ in that expression, but it didn't produce a correct result:
$(2n-1)((2n-1)+1)/2$
plug in 3
$(2n-1)((2n-1)+1)/2 = 6 != 1 + 2 + 3 = 9$
|
Let $$S_n=1+2+3+...+(2n-1)$$
Clearly
$S_1=1=1^2$
$S_2=1+3=2^2$
$S_3=1+3+5=9=3^2$
Suppose $S_n=n^2,$ then $S_{n+1}=S_n+(2n+1)=(n+1)^2$
By mathematical induction $$S_n=n^2 ,\forall n\in \mathbb{N}$$
|
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|
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
things i have done: first thing to do is to show that $x,y,z$ are non-negative. $$xy+yz+zx=1 \Rightarrow zx=1-yz-xy \Rightarrow zx=1-y(z+x)\Rightarrow zx=1-y(2-y)=(y-1)^2$$
this equality shows that $x$ and $z$ have same sign(both of them are positive or negative).like this we can conclude that $xy=(x-1)^2$ and $yz=(x-1)^2$.So all variables are positive or negative.If all of them are negative then $x+y+z \neq 2$ so all of $x,y,z$ are non negative. I don't know what to do for showing that none of $x,y,z$ will be be bigger than $\frac{4}{3}$.
|
Consider the polynomial:
$$ p(t)=(t-x)(t-y)(t-z) = t^3-2t^2+t+k = t(t-1)^2+k. $$
We know that $p(t)$ has three real roots, hence $-k=xyz$ is bounded between the two values of $t(t-1)^2$ in its stationary points. A stationary point obviously occurs for $t=1$, the other one occurs for $t=1/3$. The three reals roots of $p(t)$ hence belong to the interval $[0,u]$, where $u>1$ is the only real number such that
$$ u(u-1)^2 = \frac{1}{3}\left(\frac{1}{3}-1\right)^2 = \frac{4}{27}. $$
Now it is straightforward to check that $u=\frac{4}{3}$.
For a visual proof:
$\quad$
|
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|
How many integers between $1$ and $10^n$ contain $14$? How can we derive a formula to calculate the number of integers between $1$ and $10^n$ that contain the number $14$ (as a string)?
For example, there are $20$ integers from $1$ to $1000$ that contain at least one $14$:
"14","114","140","141","142","143","144","145","146","147","148","149","214","314","414","514","614","714","814","914"
|
There are $\binom{n-k}{k}$ arrangements of $k$ "$14$"s and $10^{n-2k}$ other digits.
Thus, inclusion-exclusion says there are
$$
\sum_{k=1}^{\lfloor n/2\rfloor}(-1)^{k-1}\binom{n-k}{k}10^{n-2k}\tag{1}
$$
numbers from $1$ to $10^n$ which contain "$14$".
Here are some of the first few values.
$$
\begin{array}{r|l}
n&\text{count}\\
\hline
2&1\\
3&20\\
4&299\\
5&3970\\
6&49401\\
7&590040
\end{array}\tag{2}
$$
Generating Function
Multiply $(1)$ by $x^n$ and sum:
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=1}^\infty(-1)^{k-1}\binom{n-k}{n-2k}10^{n-2k}x^n\\
&=\sum_{n=0}^\infty\sum_{k=1}^\infty(-1)^{n-k-1}\binom{-k-1}{n-2k}10^{n-2k}x^{n-2k}x^{2k}\\
&=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^{2k}}{(1-10x)^{k+1}}\\
&=\frac1{10x-1}\frac{\frac{x^2}{10x-1}}{1-\frac{x^2}{10x-1}}\\
&=\frac{x^2}{1-20x+101x^2-10x^3}\tag{3}
\end{align}
$$
which has the Taylor series $x^2+20x^3+299x^4+3970x^5+49401x^6+590040x^7+\dots$
Closed Form
The denominator in $(3)$ says that $c_n$ satisfies
$$
c_n=20c_{n-1}-101c_{n-2}+10c_{n-3}\tag{4}
$$
The characteristic polynomial for $(4)$ is
$$
x^3-20x^2+101x-10\tag{5}
$$
whose roots are
$$
\left\{10,5+\sqrt{24},5-\sqrt{24}\right\}\tag{6}
$$
We can solve $(4)$ in the standard manner for such recurrences to get
$$
c_n=10^n-\frac{(5+\sqrt{24})^{n+1}}{2\sqrt{24}}+\frac{(5-\sqrt{24})^{n+1}}{2\sqrt{24}}\tag{7}
$$
|
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|
Find the quadratic equation equation of $x_1, x_2$.
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.
By Vieta's theorem:
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have a quadratic equation with the form:
$$-\frac{b}{2}x^2 + bx + b = 0$$
Applying $x_1$ for our equation I get $b=0$. Why?
The final answer is: $x^2-2x-2$.
|
Is $x_1$ and $x_2$ are the solution than you can write
$$(x_1-x)(x_2-x)=0$$
which is
$$x^2-(x_1+x_2)\cdot x+x_1x_2=0$$
Thus $a=1$, $b=-x_1-x_2$ and $c=x_1x_2$
|
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|
Show that the odd prime divisors of $n^2+n+1$ are of form $6k+1$ (exclude 3) Show that the odd prime divisors of $n^2+n+1$ are of form $6k+1$ (exclude 3)
I have started like below:
$n^2+n+1\equiv 0 \pmod {p_i}$
$(n+1)^2\equiv n \pmod {p_i}$
Any hints/help on how to proceed from here ?
|
Let $p$ be an odd prime divisor of $n^2+n+1$
Then $n^2+n+1 \equiv 0 \pmod{p}
\implies n^3-1 \equiv 0 \pmod{p}
\implies n^3 \equiv 1 \pmod{p}$
So $ord_pn=3$ (obvious!) and so $3|\phi(p)=p-1$
Note that $p-1$ is even. So $2|p-1$.
Since $gcd(2, 3)=1$, so $6|p-1\implies p-1=6k, k\in\Bbb{Z}\implies p=6k+1$
|
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|
Inequality $\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$ with weird condition I want to prove the following inequality:
$$\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$$
Where $a,b,c$ are positive reals and with the horrible condition:
$$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2$$
It was a problem of the MEMO 2011. What I tried so far is the following:
Make the substitution $x=\frac{a}{1+a}$, $y=\frac{b}{1+b}$, $z=\frac{c}{1+c}$ so $a=\frac{x}{1-x}$, $b=\frac{y}{1-y}$, $c=\frac{z}{1-z}$. The conditions then change to $x+y+z=2$ and $0<x,y,z<1$ since $\frac{a}{1+a}<1$ and so the other two terms. The original inequality changes to:
$$\frac{\sqrt \frac{x}{1-x}+\sqrt \frac{y}{1-y}+\sqrt \frac{z}{1-z}}{2}\ge\sqrt \frac{1-x}{x}+\sqrt \frac{1-y}{y}+\sqrt \frac{1-z}{z}$$
Now we have $y+z=2-x>1>x$ and similarly $x+y>z$, $z+x>y$. This is equivalent to $x=u+v$, $y=v+w$ and $z=w+u$. So with this change of variables, the condition $0<x,y,z<1$ disappears. The other condition becomes $(u+v)+(v+w)+(w+u)=2 \iff u+v+w=1$. The inequality changes to:
$$\frac{\sqrt \frac{u+v}{1-u-v}+\sqrt \frac{v+w}{1-v-w}+\sqrt \frac{w+u}{1-w-u}}{2}\ge\sqrt \frac{1-u-v}{u+v}+\sqrt \frac{1-v-w}{v+w}+\sqrt \frac{1-w-u}{w+u} \iff \frac{\sqrt \frac{u+v}{w}+\sqrt \frac{v+w}{u}+\sqrt \frac{w+u}{v}}{2}\ge\sqrt \frac{w}{u+v}+\sqrt \frac{u}{v+w}+\sqrt \frac{v}{w+u} \iff \sqrt \frac{u+v}{2w}+\sqrt \frac{v+w}{2u}+\sqrt \frac{w+u}{2v}\ge\sqrt \frac{2w}{u+v}+\sqrt \frac{2u}{v+w}+\sqrt \frac{2v}{w+u}$$
Am I right so far? The last inequality seems kind of "beauty" since each term on the righthand side has its reciprocal on the lefthand side. But I wasn't able to find a solution from there, so any help is highly appreciated! Thanks.
|
Proceeding from your last line: by the power mean inequality (or more simply the AM-HM actually, which is a special case) it is sufficient that $(\sqrt\frac{2w}{u+v}+\sqrt \frac{2u}{v+w}+\sqrt \frac{2v}{w+u})^2 \leq 9$ (left to you! it's not very hard). Motivation: the power mean inequality can give us an inequality between some numbers and their reciprocals, which is just what we were looking for.
Hence we require $\sqrt\frac{2w}{u+v}+\sqrt \frac{2u}{v+w}+\sqrt \frac{2v}{w+u} \leq 3$. By Cauchy-Schwarz, all we then need is
$\frac{2w}{u+v}+\frac{2u}{v+w}+\frac{2v}{w+u} \leq 3$, or
$\frac{2-2(u+v)}{u+v}+\frac{2-2(v+w)}{v+w}+\frac{2-2(w+u)}{w+u} \leq 3$, or
$\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u} \geq \frac{3}{2}$.
$f(x)=\frac{1}{x}$ is convex, so by Jensen's inequality $\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u} \geq 3 \frac{3}{(u+v)+(v+w)+(w+u)}=\frac{3}{2}$, and we're done!
|
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|
Find $ \int \frac{\sin^2x}{1+\sin^2x}\hspace{1mm}\mathrm{d}x$ Find $$\int \dfrac{\sin^2x}{1+\sin^2x}\hspace{1mm}\mathrm{d}x$$
I cannot figure out how start this problem, can anyone explain
|
$$\frac{\sin^2x}{1+\sin^2x}=1-\frac1{1+\sin^2x}=1-\frac{\sec^2x}{2\tan^2x+1}$$
Set $\tan x=u$
Alternatively,
$$\frac{\sin^2x}{1+\sin^2x}=\frac{\tan^2x}{\sec^2x+\tan^2x}$$
$$=\frac{\tan^2x\sec^2x}{(\sec^2x+\tan^2x)\sec^2x}=\frac{\tan^2x\sec^2x}{(\tan^2x+1+\tan^2x)(\tan^2x+1)}$$
Set $\tan x=u$ to find $$\int\frac{\sin^2x}{1+\sin^2x}dx=\int\frac{u^2du}{(2u^2+1)(1+u^2)}$$
$$\frac{u^2}{(2u^2+1)(1+u^2)}=\frac{(2u^2+1)-(1+u^2)}{(2u^2+1)(1+u^2)}$$
|
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|
Find values of $a$ and $b$ that make the function continuous everywhere. I need some help with this question:
Find the values of $a$ and $b$ that make $f$ continuous everywhere.
$$f(x)=\begin{cases}
x^2 − 4/x-2, &\text{if }x < 2\\
ax^2-bx+1, &\text{if } 2 ≤ x ≤ 3\\
4x - a + b, &\text{if } x ≥ 3\end{cases}$$
I started by writing two expressions for $a$ and $b$ based upon the left and right limits of each piece's endpoints. But I ended up with $a=3/2$ and $b=2/3$. Where did I go wrong?
|
If $f$ is continuous then :
$f(2)=\underset{x\rightarrow2}{\lim}f(x)=2^2-\frac{4}2-2=a(2^2)-b(2)+1$
Therefore $4a-2b+1=4-2-2=0$ hence $4a-2b+1=0$.
Same reasoning around $x=3$ :
$a(3^2)-b(3)+1=4(3)-a+b$ (ie) $9a+3b+1=12-a+b$ therefore $10a-4b=11$
You just need to solve $\begin{cases}4a-2b=-1\\10a-4b=11\end{cases}$ to find $a$ and $b$.
|
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|
Comparing the size of square roots How to compare the size of following numbers without using the calculator?
$a=\sqrt{2}+\sqrt{6}+\sqrt{7},$
$b=\sqrt{3}+\sqrt{4}+\sqrt{8},$
$c=\sqrt{5}+\sqrt{5}+\sqrt{5}$
|
Looking at the three expressions, one easily notices that the sum of the radicants is $15$ in all three cases. This together with the concavity of the square root ($\sqrt{\lambda a+(1-\lambda)b}\ge\lambda\sqrt{a}+(1-\lambda)\sqrt{b}$ for all $0\le\lambda\le1$) helps to compare the numbers. Note that the concavity of the square root is strict, so in the above equation, if $a\neq b$ and $\lambda\notin\{0,1\}$, the left hand side is strictly greater.
Clearly $c$ is the largest of these numbers, because here all three radicants are equal. So what remains is to compare $a$ with $b$.
It is also obvious that for a and b, additionally the product of the first two terms is equal. Therefore it is useful to square the numbers:
$$\begin{aligned}
a^2 &= 2 + 6 + 7 + 2 \sqrt{12} + 2 \sqrt{14} + 2 \sqrt{42}\\
b^2 &= 3 + 4 + 8 + 2 \sqrt{12} + 2 \sqrt{24} + 2 \sqrt{32}
\end{aligned}$$
Obviously the leading terms are equal, so ultimately we only have to compare $a' := 2 \sqrt{14} + 2 \sqrt{42} = \sqrt{8}(\sqrt{7} + \sqrt{21})$ with $b' := 2\sqrt{24}+2\sqrt{32} = \sqrt{8}(\sqrt{12}+\sqrt{16})$. Here we see that again, the sum of the radicants is the same, but for $a'$ their difference is larger, and therefore due to concavity $a'<b'$.
So putting everything together, we get
$$a < b < c.$$
|
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|
Geometry question with three triangles Any help on this question would be great, I'm not sure how to solve it.
Thanks in advance!
Given $\triangle ABC$, let $A'$ be the point $\frac{1}{3}$ of the way from $B$ to $C$, as shown. Similarly, $B'$ is the point $\frac{1}{3}$ of the way from $C$ to $A$, and $C'$ is the point $\frac{1}{3}$ of the way from $A$ to $B$. In this way, we have constructed a new triangle starting with an arbitrary triangle. Now apply the same procedure to $\triangle A'B'C'$, thereby creating $\triangle A''B''C'' $. Show that the sides of $\triangle A''B''C'' $ are parallel to the appropriate sides of $\triangle ABC$. What fraction of the area of $\triangle ABC$ is the area of $\triangle A''B''C'' $?
|
This is nicely done in barycentric coordinates. We have:
$$ A' = \frac{2B+C}{3},\quad B'=\frac{2C+A}{3},\quad C'=\frac{2A+B}{3} $$
hence:
$$ A''=\frac{2B'+C'}{3} = \frac{4A+B+4C}{9},\quad B''=\frac{4A+4B+C}{9},\quad C''=\frac{A+4B+4C}{9} $$
The last relation implies:
$$ A''-B'' = \frac{1}{3}(C-B),\quad B''-C''=\frac{1}{3}(A-C),\quad C''-A''=\frac{1}{3}(B-A) $$
so $B''A''\parallel BC,C''B''\parallel CA,A''C''\parallel AB$ and the ratio between the area of $ABC$ and the area of $A''B''C''$ is $9$.
|
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|
Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ and $x < - \sqrt{6}$
But the solution says that $x > -\sqrt{6}$, what am I not getting?
|
In this case, we have $ |x^2 - 4| < 2 $, which leads to $ (x^2 - 4)^2 < 4 $, because $ |a| = \sqrt{a^2} $ where the root is the positive result.
Continuing, $ (x^2 - 4)^2 - 4 < 0$ and $ x^4 - 8x^2 + 12 < 0 $ or $ (x^2 - 6)(x^2 - 2) < 0 $.
This is only satisfied when $ x^2 > 2 $ and $ x^2 < 6 $. This is possible when $ \sqrt{2} < x < \sqrt{6} $ or $ -\sqrt{2} > x > -\sqrt{6} $.
To figure this out, break it into cases. When is $x^2 > 2$? One case is when $ x > \sqrt{2} $. Then you compare this with the two possibilities of the other inequality, namely $x < \sqrt{6}$ and $x > -\sqrt{6}$. You will notice that one result will be eliminated as impossible. Then do the two cases for when $ x < -\sqrt{2} $. Same thing will happen.
|
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|
Combinations of $6$-digit natural numbers In each of the following 6-digit natural numbers: $333333,225522,118818,707099$,
every digit in the number appears at least twice. Find the number of such 6-digit natural numbers.
This is how I'm intending to do.
1) Find the total number of 6-digit combinations.
2) Subtract the number of times there's 0 and 1 repetition of digits so that I can get at least 2 repetitions.
Total Combinations = $9 \times 10^5$
Repeat 0 Times = $9 \times 9 \times 8 \times 7 \times 6 \times 5$
Repeat 1 Time = Not sure how to calculate
The answer is 11754 but I'm struggling to get it!
|
First let us also accept numbers starting with a $0$.
Let $d$ denote the number of distinct digits in the number. For $d>3$ there are $0$ possibilities.
For $d=1$ there are $10$ choices of the digit and every choice leads to $1$ possibility.
For $d=3$ there are $\binom{10}{3}=120$
choices for the digits and each choice leads to $\frac{6!}{2!2!2!}=90$
possibilities.
For $d=2$ we have two split-ups.
One of them is $6=3+3$
with $\binom{10}{2}=45$ choices for the digits and each choice leads
to $\binom{6}{3}=20$ possibilities.
The other is $6=4+2$. Here the
chosen digits are distinghuisable. One of them is used $4$ times
and the other $2$ times. So we have $10\times9=90$ choices and each
choice leads to $\binom{6}{2}=\binom{6}{4}=15$ possibilities.
Adding up we find $10\times 1+120\times90+45\times20+90\times15=13060$ possibilities.
Subtracting the numbers that start with a $0$ we find $\frac{9}{10}\times13060=11754$
possibilities.
|
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|
Square in Interval of Primes Denote by $a_n$ the sum of the first $n$ primes. Prove that there is a perfect square between $a_n$ and $a_{n+1}$, inclusive, for all $n$.
The first few sums of primes are $2$, $5$, $10$, $17$, $28$, $41$, $58$, $75$. It seems there is a perfect square between each pair of successive sums. In addition, we can put a bound on $a_n$, namely $a_n \le 2+3+5+7+9+11+...+(2n+1)=n^2+2n+5$.
|
(waving my hands frantically to keep me aloft:)
$a_n
\approx \sum_{k=1}^n k\ln k
\approx \int_{1}^n k \ln k\ dk
\approx n^2\ln n
$
and
$a_{n+1}-a_n
\approx n\ln n
$
so if
$b_n
= \lfloor \sqrt{a_n}\rfloor
$,
$b_n
\approx n\sqrt{\ln n}
$.
The next square
after $b_n^2$
is
$(b_n+1)^2
=b_n^2+2b_n+1
\approx b_n^2+2n\sqrt{\ln n}
$
and since
$a_{n+1}-a_n
\approx n\ln n
$,
$(b_n+1)^2
< a_{n+1}
$
for large anough $n$.
Note that
$(b_n+k)^2
=b_n^2+2kb_n+k^2
\approx b_n^2+2kn\sqrt{\ln n}
$,
so,
by choosing
$k \approx \frac12 \sqrt{\ln n}$,
it looks like
there are about
$\frac12 \sqrt{\ln n}$
squares between
$a_n$ and
$a_{n+1}$.
|
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|
What is the limit of this specific function? Please evaluate the following limit for me:
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$
I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the denominator so it always stay in the indeterminate form.
|
Problem:
$$\lim_{x\to-1}\frac{\sqrt{x^2+8}-3}{x+1}$$
Rationalize the numerator by multiplying by its conjugate
$$=\lim_{x\to-1}\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
Multiply out the numerator
$$=\lim_{x\to-1}\frac{\sqrt{x^2+8}^2+\boxed{3\sqrt{x^2+8}-3\sqrt{x^2+8}}+-3*3}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
Simplify
$$=\lim_{x\to-1}\frac{x^2+8-9}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
$$=\lim_{x\to-1}\frac{x^2-1}{\left(x+1\right)\left(\sqrt{x^2+8}+3\right)}$$
Factor the numerator
$$=\lim_{x\to-1}\frac{\left(x-1\right)\boxed{\left(x+1\right)}}{\boxed{\left(x+1\right)}\left(\sqrt{x^2+8}+3\right)}$$
Cancel
$$=\lim_{x\to-1}\frac{x-1}{\sqrt{x^2+8}+3}$$
Plug in $-1$
$$=\frac{\left(-1\right)-1}{\sqrt{\left(-1\right)^2+8}+3}$$
Simplify
$$=-\frac{1}{3}$$
|
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|
Evaluate the integral $\int_0^{1/4}\frac{x-1}{\sqrt{x}-1}\mathrm dx$ so I have this Integral I have to solve without a calculator.
$$\int_0^{1/4}\dfrac{x-1}{\sqrt{x}-1}\mathrm dx.$$
How would I go about finding the antiderivative of that fraction?
|
Problem:
$\int_0^\frac{1}{4}\frac{x-1}{\sqrt{x}-1}dx$
For the integrand $\frac{x-1}{\sqrt{x}-1}$, substitute $u=\sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}dx$. This gives a new lower bound $u=\sqrt{\frac{1}{4}}=\frac{1}{2}$:
$=2\int_0^\frac{1}{2}\frac{u\left(u^2-1\right)}{u-1}du$
For the integrand $\frac{u\left(u^2-1\right)}{u-1}$, cancel common terms in the numerator and denominator:
$=2\int_0^\frac{1}{2}u\left(u+1\right)du$
For the integrand $u\left(u+1\right)$, substitute $s=u+1$ and $ds=du$. This gives a new lower bound $s=1+0=1$ and upper bound $s=1+\frac{1}{2}=\frac{3}{2}$:
$=2\int_1^\frac{3}{2}s\left(s-1\right)ds$
Expanding the integrand $s\left(s-1\right)$ gives $s^2-s$:
$=2\int_1^\frac{3}{2}\left(s^2-s\right)ds$
Integrate the sum term by term and factor out constants:
$=2\int_1^\frac{3}{2}s^2ds-2\int_1^\frac{3}{2}sds$
Apply the fundamental theorem of calculus. The antiderivative of $s^2$ is $\frac{s^3}{3}$:
$=\frac{2s^3}{3}\Big|_1^\frac{3}{2}-2\int_1^\frac{3}{2}sds$
Evaluate the antiderivative at the limits and subtract. $\frac{2s^3}{3}\Big|_1^\frac{3}{2}=\frac{2}{3}\left(\frac{3}{2}\right)^3-\frac{2*1^3}{3}=\frac{19}{12}$:
$=\frac{19}{12}+\left(-s^2\right)\Big|_1^\frac{3}{2}$
Evaluate the antiderivative at the limits and subtract. $\left(-s^2\right)\Big|_1^\frac{3}{2}=\left(-\left(\frac{3}{2}\right)^2\right)-\left(-1^2\right)=-\frac{5}{4}$:
$=\frac{1}{3}$
|
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|
Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
|
Hint:
\begin{align}
(a + b + c + \ldots )^2 & = (a+b+c+\ldots) \times (a+b+c+\ldots) \\
& = a^2 + ba + ca + \ldots + a b + b^2 + cb + \ldots ac + bc +c ^2 + \ldots
\end{align}
Can you take it from here?
|
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|
Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following
$u = \sqrt{x^2+4 }$ , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$ , $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck here
$\int \dfrac{1}{4 x^4} \dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck here please help
|
Use substitution method
First, let $u = x^2$
Then you will have $du = 2x dx$
$$ \int\,x^3\sqrt{x^2+4}\,dx = \dfrac{1}{2}\int\,u\sqrt{u+4}\,du $$
Then let $s = u+4$, which implies $ds=du$
You should be able to get the answer
If you are right, you should get the following:$\frac{ 1}{5} (x^2+4)^\frac {5}{2} -\frac{ 4}{3} (x^2+4)^\frac {3}{2} +C $, where $C$ is a constant.
|
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|
Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$
I rewrote the function to the form
$$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
and figured that the answer would be $0$, but apparently this is wrong.
The correct answer is $\displaystyle{{1 \over 2}\left(\,a - b\,\right)}$.
|
You could also use the basic Taylor Series expansion $\left( 1 + \frac{1}{t}\right)^{\lambda} = 1+\frac{\lambda}{t} + O(1/t^2)$ as follows:
$$x^2 \Big( \underbrace{ \sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} }_{1+\frac{a}{2x^2} + O(\frac{1}{x^3})} - \underbrace{ \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}} }_{1+\frac{b}{2x^2}+O(\frac{1}{x^3})} \Big) = x^2 \Big( \frac{a-b}{2x^2} + O\big(\frac{1}{x^3}\big) \Big) = \frac{a-b}{2} + O\big( \frac{1}{x} \big)$$
since the extra term $\frac{1}{x^4}$ in each square root is negligble compared to $\frac{a}{x^2}$ and $\frac{b}{x^2}$ for large $x$.
|
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|
Sum of squares of two integers divisible by five Supposing $x,y$ are natural numbers,
what is the probability that the sum of their squares are divisible by 5?
I am getting $1/3$ as squares can only end with $0,1,4,5,6,9$. So $36$ pairs are possible. And $12$ pairs { $(0,0),(1,4),(4,1),(4,6),...$} have sums which end in $...0$ or $...5$.
But the answer is given as $\frac{9}{25}$. Can anyone posit a solution?
|
Given $x, y$, only the last digits are relevant in determining if $x^2 + y^2$ is divisible by $5$. The last digit can be $0, 1, \ldots, 9$ and all of them are equally likely. Taking the last two digits of $x, y$, we get a sample space of size $10 \times 10 = 100$, where each of them are equally likely.
Now, we count the number of last digit combinations that lead to $x^2 + y^2$ being divisible by 5. The unordered list of pairs is: $$\{ \{0, 0\}, \{0, 5\}, \{1, 2\}, \{1, 3\}, \{1, 7\}, \{1, 8\}, \{2, 4\}, \{2, 6\}, \{2, 9\}, \{3, 4\}, \{3, 6\}, \{3, 9\}, \{4, 7\}, \{4, 8\}, \{5,5 \}, \{6, 7\}, \{6,8\}, \{7, 9\}, \{8, 9\} \},$$ which is $19$ in all. This leads to $17*2 + 2 = 36$ ordered pairs.
Therefore, the probability is $\frac{36}{100} = \frac{9}{25}$.
|
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|
Integrating $\int \frac {dx}{\sqrt{4x^{2}+1}}$ $\int \dfrac {dx}{\sqrt{4x^{2}+1}}$
I've been up to this one for quite a while already, and have tried several ways to integrate it, using substituion, with trigonometric as well as hyperbolic functions. I know(I think) I'm supposed to obtain:
$\dfrac {1}{2}\ln \left| 2\sqrt {x^2 +\dfrac {1}{4}}+2x\right|+c$
However, whatever idea I come up with to try to solve it, it seems I always obtain:
$\dfrac {1}{2}\ln \left| x + \sqrt {x^2 + 1} \right| + c$
Thanks for reading, and hopefully answering too.
Cheers
Yann
|
Let $x = \frac{1}{2} \tan\theta$ so that $dx=\frac{1}{2}\sec^2\theta d\theta$. Then
$\begin{align*}
\int\frac{dx}{\sqrt{4x^2+1}} &= \frac{1}{2} \int \frac{\sec^2 \theta}{\sec\theta} d\theta \\
&= \frac{1}{2} \int\sec\theta d\theta \\
&= \frac{1}{2} \log|\sec\theta + \tan\theta| + C
\end{align*}$
for some real constant $C$. Note that we assumed $\tan\theta = 2x$, so $\sec\theta = \sqrt{4x^2 - 1}$. Hence,
$$\int\frac{dx}{\sqrt{4x^2+1}}= \frac{1}{2}\log|\sqrt{4x^2-1} + 2x| + C.$$
|
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|
Is my outcome wrong? (Evaluating a logarithm) $$log\sqrt [ 4 ]{ x^2+y^2 } $$
$$log\sqrt { x+y } $$
$$logx^{ 1/2 }+log^{ 1/2 }$$
$$\frac { 1 }{ 2 }log (x+y)$$
The answer key saids:
$$\frac { 1 }{ 4 } log(x^2+y^2)$$
|
I'm afraid your answer is incorrect.
You should get this:
\begin{align*}
&\log \sqrt[4]{x^2+y^2}\\
&=\log(x^2+y^2)^{\frac{1}{4}}\\
&=\dfrac{1}{4}\log(x^2+y^2)
\end{align*}
You should note that $(a+b)^n\neq a^n+b^n$ in general, so $(a^2+b^2)^{\frac{1}{4}}\neq a^{\frac{1}{2}}+b^{\frac{1}{2}}$
|
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|
Least value of an Expression? Find the least value of $\dfrac {3a}{b+c} + \dfrac{4b}{a+c}+ \dfrac{5c}{a+b}$ for positive $a, b, c$.
I tried using the Cauchy-Schwarz inequality, but could not proceed after a bad equation which couldn't be further solved .
|
You can minimize it by using Cauchy-Schwarz, but yes, it gets pretty bad. Firstly you can modify the expression as follows:
$$\frac {3a}{b+c} + \frac{4b}{c+a}+ \frac{5c}{a+b}=\frac {3(a+b+c)}{b+c} + \frac{4(a+b+c)}{c+a}+ \frac{5(a+b+c)}{a+b}-12=(a+b+c)\cdot \left(\frac {3}{b+c} + \frac{4}{c+a}+ \frac{5}{a+b}\right)-12=\frac{1}{2}\cdot ((b+c)+(c+a)+(a+b))\cdot \left(\frac {3}{b+c} + \frac{4}{c+a}+ \frac{5}{a+b}\right)-12$$
Now, by Cauchy-Schwarz we have:
$$((b+c)+(c+a)+(a+b))\cdot \left(\frac {3}{b+c} + \frac{4}{c+a}+ \frac{5}{a+b}\right)\ge(\sqrt3+\sqrt4+\sqrt5)^2$$
With equality if:
$$\frac{b+c}{\sqrt3}=\frac{c+a}{\sqrt4}=\frac{a+b}{\sqrt5}$$
So, if we use the modified expression from above, we get:
$$\frac {3a}{b+c} + \frac{4b}{c+a}+ \frac{5c}{a+b}\ge\frac{1}{2}\cdot(\sqrt3+\sqrt4+\sqrt5)^2-12=\sqrt{12}+\sqrt{15}+\sqrt{20}-6\approx 5.809$$
To find a triplet $(a,b,c)$ for wich equality holds, the system of equations from above can be, after big effort, simplified to:
$$a=\frac{\sqrt5+\sqrt4-\sqrt3}{\sqrt5-\sqrt4+\sqrt3}\cdot b=\frac{\sqrt4+\sqrt5-\sqrt3}{\sqrt4-\sqrt5+\sqrt3}\cdot c$$
And indeed, if we set $b=\frac{\sqrt5-\sqrt4+\sqrt3}{\sqrt5+\sqrt4-\sqrt3}\cdot a$ and $c=\frac{\sqrt4-\sqrt5+\sqrt3}{\sqrt4+\sqrt5-\sqrt3}\cdot a$ in the original expression, we get:
$$\frac {3\cdot a}{\frac{\sqrt5-\sqrt4+\sqrt3}{\sqrt5+\sqrt4-\sqrt3}\cdot a+\frac{\sqrt4-\sqrt5+\sqrt3}{\sqrt4+\sqrt5-\sqrt3}\cdot a} + \frac{4\cdot\frac{\sqrt5-\sqrt4+\sqrt3}{\sqrt5+\sqrt4-\sqrt3}\cdot a}{\frac{\sqrt4-\sqrt5+\sqrt3}{\sqrt4+\sqrt5-\sqrt3}\cdot a+a}+ \frac{5\cdot\frac{\sqrt4-\sqrt5+\sqrt3}{\sqrt4+\sqrt5-\sqrt3}\cdot a}{a+\frac{\sqrt5-\sqrt4+\sqrt3}{\sqrt5+\sqrt4-\sqrt3}\cdot a}=\frac{1}{2}\cdot (\sqrt3 \cdot(\sqrt5+\sqrt4-\sqrt3)+\sqrt4 \cdot(\sqrt5-\sqrt4+\sqrt3)+\sqrt5 \cdot(-\sqrt5+\sqrt4+\sqrt3))=\sqrt{12}+\sqrt{15}+\sqrt{20}-6$$
So we're done.
I'm sorry for all the mistakes in my English, I'm not a native speaker.
|
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|
Why is $1 \times 3 \times 5 \times \cdots \times (2k-3) = \frac{(2k-2)!}{2^{(k-1)}(k-1)!}$ In order to find out the Catalan numbers from their generating function you have to evaluate the product above.
Here is what I thought:
\begin{align*}
1 \times 3 \times 5 \times...\times (2k-3) &= \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{2 \times 4 \times \cdots \times (2k - 2)} \\
&= \frac{(2k - 3)!}{2^{k - 1}(k - 1)!} \\
\end{align*}
But a Mathematica session quickly proved me wrong, instead the result in the title is true. Why is that true, what mistake did I make?
|
Note that all even numbers in the numerator cancel out with the denominator. But in your calculation, $2k-2$ from denominator doesn't cancel out.
The correct way to arrive at the answer is:
$$\begin{align}1 \times 3 \times 5 \cdots \times (2k-3) &= 1 \times \dfrac{2}{2} \times 3 \times \dfrac{4}{4} \times \cdots \times (2k-3) \times \dfrac{2k-2}{2k-2} \\&= \dfrac{1\times 2\times \cdots \times (2k-3) \times (2k-2)}{2\times 4 \times \cdots \times (2k-2)} \\&= \dfrac{(2k-2)!}{2^{k-1}(k-1)!}\end{align}$$
|
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|
Missing root of equation $\tan(2x)=2\sin(x)$ I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong?
$$
\tan(2x)=2\sin x
$$
$$
\frac{(\sin 2x)}{(\cos 2x)}=2\sin x
$$
$$
\frac{(2\sin x \cos x)}{((\cos x)^2-(\sin x)^2)}=2\sin x
$$
$$
2\sin x\cos x=2\sin x\cos^2 x-2\sin^3 x
$$
$$
\sin x\cos x=\sin x(\cos^2 x-\sin^2x)
$$
$$
\cos x=\cos^2 x-\sin^2 x
$$
$$
\cos x=\cos^2 x+\cos^2x-1
$$
$$
2\cos^2x-\cos x-1=0
$$
Solving this quadratic equation gives $\cos x=1$ or $\cos x=-\frac 12$
$$
\cos x=1=\cos \theta
$$
$$
x=n2\pi
$$
$$
\cos x=-\frac 12=\cos(\frac{2\pi}3)
$$
$$
x=\pm\frac{2\pi}3+n2\pi
$$
|
You can't just cancel out $\sin(x)$ in your steps because $\sin(x)$ can also be zero so you have to factor out even $\sin(x)$. $\sin(x)=0$ implies $x=n \pi $ is also the solution.
|
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|
Evaluate $\int\frac{x^3}{\sqrt{81x^2-16}}dx$ using Trigonometric Substitution $$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$
I started off doing $$u =9x$$ to get $$\int\frac{\frac{u}{9}^3}{\sqrt{u^2-16}}dx$$
which allows for the trig substitution of $$x = a\sec\theta$$
making the denomonator $$\sqrt{16\sec^2\theta-16}$$
$$\tan^2\theta + 1 = \sec^2\theta$$
$$\Rightarrow4\tan\theta$$
to give $$\int\frac{\frac{4\sec}{9}^3}{4\tan\theta}dx$$
Am I doing this correctly?
|
Set $9x=4\sec\theta\implies81x^2-16=(4\tan\theta)^2$
$$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$
$$=\int\frac{4^3\sec^3\theta}{9^3\cdot4\tan\theta\cdot(\text{sign of}\tan\theta)}\frac49\sec\theta\tan\theta\ d\theta$$
$$=\frac{4^3}{9^4\cdot(\text{sign of}\tan\theta)}\int(1+\tan^2\theta)\sec^2\theta\ d\theta$$
Hope you can take it home from here
|
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|
Integrate by partial fraction decomposition $$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx$$
Here's what I have so far...
$$\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)} = \frac{\mathrm A}{x+1}+\frac{\mathrm Bx+\mathrm C}{x^2+2x+5}\\$$
$$5x^2 + 9x + 16 = \mathrm A(x^2+2x+5) + (\mathrm Bx+\mathrm C)(x+1)=\\$$
$$\mathrm A(x^2+2x+5) + \mathrm B(x^2+x)+\mathrm C(x+1)=\\$$
$$(\mathrm A+\mathrm B)x^2 + (2\mathrm A + \mathrm B + \mathrm C)x + (5\mathrm A+\mathrm C)\\$$
$$\mathrm A=-3,\;\mathrm B=8,\;\mathrm C = 31$$
$$$$
$$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx = \int\bigg(-\frac{3}{x+1}+\frac{8x+31}{x^2+2x+5}\bigg)dx\Rightarrow$$
$$\int-\frac{3}{x+1}dx +\int\frac{8}{x^2+2x+5}dx+\int\frac{31}{x^2+2x+5}dx $$
Hopefully I've got it correct until this point (if not, someone point it out please!). I can do the first integration by moving the -3 out and using $u=x+1$ to get
$$-3 \ln(x+1)$$
but I'm stuck on the next two.
|
It should be $A=3,B=2,C=1$ and you get:
$$\int \frac{3}{x+1} dx + \int \frac{2x+2}{x^2+2x+5} dx -\int \frac{1}{x^2+2x+5}dx$$
The second is easy by using $u=x^2+2x+5$ and other answerers have covered the last.
The key is writing $\frac{2x+1}{x^2+2x+5} = \frac{2x+2}{x^2+2x+5} - \frac{1}{x^2+2x+5}$.
It's probably easier to just set $u=x+1$ at the start. Then you get:
$$\int\frac{5(u-1)^2+9(u-1)+16}{u(u^2+4)}\,du=\int\frac{5u^2-u+12}{u(u^2+4)}dx$$
Then write:
$$\frac{5u^2-u+12}{u(u^2+4)} = \frac{A}{u} + \frac{Bu+C}{u^2+4}$$
Giving $$A=3,B=2,C=-1$$, and the integrals:
$$\int\left(\frac{3}{u} + \frac{2u}{u^2+4} - \frac{1}{u^2+4}\right)\,du$$
|
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|
If $ab+bc+ca=1$, then $\frac{((a+b)^2+1)}{(c^2+2)}+\frac{((b+c)^2+1)}{(a^2+2)}+\frac{((c+a)^2+1)}{(b^2+2)} \geq 3$ Let $\displaystyle a, b, c> 0, ab+bc+ca=1$. Prove that the following inequality holds:
$$\frac{((a+b)^2+1)}{(c^2+2)}+\frac{((b+c)^2+1)}{(a^2+2)}+\frac{((c+a)^2+1)}{(b^2+2)} \geq 3.$$
I tried using the Cauchy-Schwarz inequality, but can't work it out.
|
Hint: By CS inequality,$$\sum_{cyc} \frac{(a+b)^2}{(c^2+2)}+\sum_{cyc} \frac1{(c^2+2)}\ge \frac{4(a+b+c)^2+9}{(a^2+b^2+c^2)+6}$$
and then its enough to show
$$4(a+b+c)^2+9 \ge 3(a^2+b^2+c^2)+18 $$
which should be easy using the constraint given.
|
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|
How would you solve $\int \frac{x}{x^2 - 4x + 5} dx$ What is the tip for integrating that integral? I completed the square on the bottom to make it $$\frac{x}{(x-2)^2 + 1}$$
but it doesn't seem helpful. Any tips?
Thanks.
|
You can continue by letting $u=x-2$, $x=u+2$, $dx=du$ to get
$\displaystyle\int\frac{u+2}{u^2+1} du=\frac{1}{2}\int\frac{2u}{u^2+1} du+2\int\frac{1}{u^2+1} du=\frac{1}{2}\ln(u^2+1)+2\arctan u + C$
$\;\;=\frac{1}{2}\ln(x^2-4x+5)+2\arctan(x-2)+C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/952908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Number of solutions of this trigonometric equation.
Q. Find the number of solutions of the equation $\sin(x) + 2\sin(2x) - \sin(3x) = 3$, in the interval $x\in (0,\pi)$.
I tried clubbing the $\sin(x)$ and $\sin(3x)$ terms together but got nothing. I also tried the $\sin(x)$ with $\sin(2x)$ and $\sin(2x)$ with $\sin(3x)$. How do i do it?
|
We have
\begin{align}
2\sin 2x+\sin x-\sin 3x &=4\sin x\cos x-2\sin x \cos 2x=\\
&=\sin x(4\cos x -4\cos^2 x+2)=\\
&=\sin x \left[3-(2\cos x-1)^2\right].
\end{align}
Since $\sin x\ge 0$ in the interval $[0,\pi]$ and the second factor is bounded above by $3$, we must have simultaneously $\sin x=1$, $2\cos x=1$, which is not possible. Therefore the equation has no solutions in $[0,\pi]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/955509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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|
Expressing a summation using matrix algebra Consider the $r \times n$ matrix
$$\begin{pmatrix}
X_{11} & X_{12} & \cdots & X_{1n} \\
X_{21} & X_{22} & \cdots & X_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
X_{r1} & X_{r2} & \cdots & X_{rn}
\end{pmatrix}\text{.}$$
Define
$$\begin{align*}
&\bar{X} = \dfrac{\sum\limits_{i=1}^{r}\sum\limits_{j=1}^{n}X_{ij}}{nr} \\
&\bar{X}_{i} = \dfrac{\sum\limits_{j=1}^{n}X_{ij}}{n}\text{.}
\end{align*}$$
I am interested in knowing if there is a possible way to write the summations
$$\begin{align*}
\hat{v}^{S} &= \sum\limits_{i=1}^{r}\sum\limits_{j=1}^{n}\left(X_{ij}-\bar{X}_{i}\right)^{2} \\
\hat{a}^{S} &= \sum\limits_{i=1}^{r}\left(\bar{X}_i - \bar{X}\right)^{2}
\end{align*}$$
in terms of matrix operations (anything one would learn in a first course in linear algebra, such as multiplication of matrices, inverses of matrices, determinants, eigenvalues, etc.).
The reason why is because I have to memorize these formulas for the actuarial exam I will be taking soon, and I am not interested in memorizing summations if there is a way to express them in matrix form.
There may not be an answer to what I seek, and I might just have to memorize these summations as is, but I thought I would ask in case there is.
ETA: I did pass this exam (at least 93% scored) but am still interested in knowing if there is a solution to this problem.
|
Here are some expressions, although not necessarily easier to memorize!
Let $\mathbf{1}_{n}$ denote the column vector of all ones and length $n$,
and $\mathbf{I}$ the $n\times n$ identity matrix. Then,
$$
\hat{v}^{S} = \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right),
$$
where $\text{Tr}(\mathbf{M})$ denotes the trace of $\mathbf{M}$, and
\begin{align*}
\hat{a}^{S}
&=
\frac{1}{n^{2}}
\left(
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \mathbf{X}\mathbf{1}_{n} - \frac{1}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
\right).
\end{align*}
Moving things around, you can also alternatively write:
\begin{align*}
\hat{a}^{S}
&=
\frac{1}{n^{2}}
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \left( \mathbf{I} - \frac{1}{r}
\mathbf{1}_{r}\mathbf{1}_{r}^{T}
\right)
\mathbf{X}\mathbf{1}_{n}.
\end{align*}
It is useful to note that $\mathbf{1}_{n}\mathbf{1}_{n}^{T}$ is the $n \times n$ all-ones matrix, and $\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n}$ is the sum of all entries in $\mathbf{X}$.
Why are the above true?
Note that ${\overline{\mathbf{X}}}_{i}$ is the sum of all entries of the $i$th row of $\mathbf{X}$ normalized by $n$, which can be written as $\frac{1}{n}\mathbf{X}_{i,:}\mathbf{1}_{n}$, where $\mathbf{X}_{i,:}$ is the $i$th row of $\mathbf{X}$.
Let $\mathbf{R}$ denote the column vector of length $r$ obtained by vertically stacking the ${\overline{\mathbf{X}}}_{i}$'s. Then,
$$
\mathbf{R} = \frac{1}{n}\mathbf{X}\mathbf{1}_{n}.
$$
Also, note that $\mathbf{R}\mathbf{1}_{n}^{T}$ is an $r \times n$ matrix whose entire $i$th row is equal to $R_{i} = {\overline{\mathbf{X}}}_{i}$.
Then, $\hat{v}^{S}$ is the sum of the squared entries of the matrix
$$
\mathbf{X} - \mathbf{R}\mathbf{1}_{n}^{T},
$$
which is also called the squared Frobenius norm of the matrix, denoted by $\|\|_{F}^{2}$,
that is,
\begin{align*}
\hat{v}^{S}
&= \| \mathbf{X} - \mathbf{R}\mathbf{1}_{n}^{T} \|_{F}^{2} \\
%= \sum\limits_{i=1}^{r}\sum\limits_{j=1}^{n}\left(X_{ij}-\bar{X}_{i}\right)^{2}
&= \| \mathbf{X} - \frac{1}{n}\mathbf{X}\mathbf{1}_{n}\mathbf{1}_{n}^{T} \|_{F}^{2} \\
&= \| \mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right) \|_{F}^{2}\\
&= \text{Tr}\left( \mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)^{T}\mathbf{X}^{T}\right)\\
&= \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{2}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T} +\frac{1}{n^{2}}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right)\\
&= \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{2}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T} +\frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right)\\
&= \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right).
\end{align*}
For the second part, $\overline{\mathbf{X}}$ is a scalar, equal to the sum of all entries in $\mathbf{X}$ normalized by $rn$, i.e.,
$$
\overline{\mathbf{X}}
=
\frac{1}{rn}
\mathbf{1}_{r}^{T}
\mathbf{X}
\mathbf{1}_{n}.
$$
Then,
\begin{align*}
\hat{a}^{S} &= \sum\limits_{i=1}^{r}\left(\bar{X}_i - \bar{X}\right)^{2}\\
&=
\|\mathbf{R} - \overline{\mathbf{X}} \cdot \mathbf{1}_{r} \|_{2}^{2}\\
&=
\|\frac{1}{n}\mathbf{X}\mathbf{1}_{n} - \frac{1}{rn}
\mathbf{1}_{r}^{T}
\mathbf{X}
\mathbf{1}_{n} \cdot \mathbf{1}_{r} \|_{2}^{2} \\
&=
\frac{1}{n^{2}}
\|\mathbf{X}\mathbf{1}_{n} - \frac{1}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})
\cdot \mathbf{1}_{r} \|_{2}^{2} \\
&=
\frac{1}{n^{2}}
\left(
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \mathbf{X}\mathbf{1}_{n} - \frac{2}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})
\cdot \mathbf{1}_{n}^{T}\mathbf{X}^{T}\mathbf{1}_{r}
+
\frac{1}{r^{2}}(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
\mathbf{1}_{r}^{T}\mathbf{1}_{r}
\right)\\
&=
\frac{1}{n^{2}}
\left(
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \mathbf{X}\mathbf{1}_{n} - \frac{2}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
+
\frac{1}{r}(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
\right)\\
&=
\frac{1}{n^{2}}
\left(
\mathbf{1}_{n}^{T}\mathbf{X}^{T} \mathbf{X}\mathbf{1}_{n} - \frac{1}{r}
(\mathbf{1}_{r}^{T}\mathbf{X}\mathbf{1}_{n})^{2}
\right),
\end{align*}
which completes the proof.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/956254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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|
Prove that $\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$ What ways would you propose for getting the inequality below?
$$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}> \frac{3\sqrt{3}}{2}(\zeta(2s)-1),\space s>1$$
The left side may be written as
$$\sum_{k=2}^{\infty} \frac{k^s}{k^{2s}-1}>\zeta(s)-1$$
but how do we prove then that
$$\zeta(s)-1> \frac{3\sqrt{3}}{2}(\zeta(2s)-1)$$
?
Can we do it without using integrals?
|
From $n^s\ge3^s>2^s+1$ when $n\ge3$, we have
\begin{align}
&(\zeta(s)-1)-(2^s+1)(\zeta(2s)-1)\\&=\sum_{n=2}^\infty\frac1{n^s}\left(1-\frac{2^s+1}{n^s}\right)\\&=\sum_{k=1}^\infty\frac{1}{2^{ks}}\left(1-\frac{2^s+1}{2^{ks}}\right)+\sum_{\substack{n\ge3\\n\ne2^k}}\frac{1}{n^{s}}\left(1-\frac{2^s+1}{n^s}\right)\\&>\sum_{k=1}^\infty\frac{1}{2^{ks}}\left(1-\frac{2^s+1}{2^{ks}}\right)+\sum_{\substack{n\ge3\\n\ne2^k}}\frac{1}{n^{s}}\left(1-\frac{n^s}{n^s}\right)\\&=\sum_{k=1}^\infty\frac{1}{2^{ks}}\left(1-\frac{2^s+1}{2^{ks}}\right)\\&=\sum_{k=1}^\infty\left(\frac{1}{2^{ks}}-\frac{1}{2^{(2k-1)s}}-\frac{1}{2^{2ks}}\right)=0
\end{align}
So$$\frac{\zeta(s)-1}{\zeta(2s)-1}>2^s+1>3>\frac{3\sqrt{3}}{2}$$and we are done.
We can also prove $\zeta(s)>\zeta(2s)^3$ using $\displaystyle\zeta(s)=\prod_p\left(1-\frac1{p^s}\right)^{-1}$ and finish by $\displaystyle\frac{\zeta(s)-1}{\zeta(2s)-1}>\zeta(2s)^2+\zeta(2s)+1>3$ but I think the above way is more cleaner. To add some more info on this direction, Mathematica plot suggests $\zeta(s)>\zeta(2s)^5$ hold.
Proving the first proposed inequality is much more easier... We just need to prove $$\frac{x}{x^2-1}>\frac{3\sqrt{3}}{2}\frac{1}{x^2}$$ for $x>2$. $\frac{x^3}{x^2-1}$ has minimum at $x=\sqrt{3}$ so $\frac{x^3}{x^2-1}>\frac{3\sqrt{3}}{2}$. Perhaps that's where the constant $\frac{3\sqrt{3}}{2}$ came from?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/958142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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|
Integrating $\cos^3(x)$ My attempt at integrating $\cos^3(x)$:
$$\begin{align}\;\int \cos^3x\mathrm{d}x &= \int \cos^2x \cos x \mathrm{d}x
\\&= \int(1 - \sin^2 x) \cos x \mathrm{d}x
\\&= \int \cos x dx - \int \sin^2x \cos x \mathrm{d}x
\\&= \sin x - \frac {1}{3}\sin^3x + C\end{align}$$
My question is how does integrating $\sin^2(x) \cos x $ become $\frac {1}{3}\sin^3(x) + C$? What mathematical process is being done? Why does the $\cos x$ disappear?
|
You can use this bit of trickery.
$\cos 3x = 4\cos^3 x - 3\cos x\\
\cos^3x = \frac 14 \cos 3x + \frac 34\cos x$
And that is easy to integrate.
But, to your actual question. $\int \sin^2x\cos x \ dx$
let $u = \sin x, du = \cos x$
$\int u^2 \ du$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/958586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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|
Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$ How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is the Euler–Mascheroni constant.
But I could not manipulate this series.
|
Notice that $$S=\sum^{\infty}_{k=1}\Big(\frac{1}{2k}-\frac{1}{2}\log(1+\frac{1}{k})+\frac{1}{2}\log(1+\frac{1}{k})-\frac{2}{2}\log(1+\frac{1}{2k})\Big)$$
Therefore
$$S=\frac{1}{2}\gamma+\frac{1}{2}\sum^{\infty}_{k=1}\log\Big(\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)$$
In other words
\begin{align}
S=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\prod^{\infty}_{k=1}\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)&=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\prod^{N}_{k=1}\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)\\&=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\prod^{N}_{k=1}\frac{k+1}{k}\frac{(2k)^2}{(2k+1)^2}\Big)\\
&=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\frac{\pi\Gamma(N+1)\Gamma(N+2)}{4\Gamma^2(N+3/2)}\Big)\\
&=\frac{1}{2}\gamma+\frac{1}{2}\log(\frac{\pi}{4})
\end{align}
We have used the limiting value $$\lim_{N\to\infty}\frac{\Gamma(N+1)\Gamma(N+2)}{\Gamma^2(N+3/2)}=1$$
|
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"url": "https://math.stackexchange.com/questions/959413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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|
A closed form for $\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx$? I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx = \frac{\pi^2}{12}$$ using power series expansion.
Thank you.
|
First note that
\begin{align}
\int \frac{\ln^{2}(1+x)}{x} \, dx = - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x)
\end{align}
for which
\begin{align}
I_{2} = \int_{0}^{1} \frac{\ln^{2}(1+x)}{x} \, dx = \frac{\zeta(3)}{4}.
\end{align}
Now,
\begin{align}
\int \frac{\ln^{3}(1+x)}{x} \, dx = \int \ln(1+x) \, \frac{\ln^{2}(1+x)}{x} \, dx
\end{align}
can be integrated by parts. This leads to
\begin{align}
\int \frac{\ln^{3}(1+x)}{x} \, dx &= \ln(1+x) \left( - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \right) \\
& - \int \frac{- 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x)}{1+x} \, dx \\
&= \ln(1+x) \left( - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \right) \\
& \hspace{10mm} + 6 Li_{4}(1+x) + Li_{2}(1+x) \, \ln^{2}(1+x) - 4 Li_{3}(1+x) \, \ln(1+x)
\end{align}
This leads to
\begin{align}
I_{3} &= \int_{0}^{1} \frac{\ln^{3}(1+x)}{x} \, dx \\
&= 6 Li_{4}(2) + \ln^{2}(2) \, Li_{2}(2) - 4 \ln(2) \, Li_{3}(2) - 6 \zeta(4) - \frac{7}{4} \, \ln(2) \, \zeta(3) \\
&= \frac{\pi^{4}}{15} + \frac{\pi^{2}}{4} \, \ln^{2}(2) - \frac{21}{4} \, \zeta(3) \, \ln(2) - \frac{1}{4} \, \ln^{4}(2) - 6 Li_{4}\left(\frac{1}{2} \right)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/965504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 1
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|
Evaluating $ \sum\frac{1}{1+n^2+n^4} $ How to evaluate following expression?
$$ \sum_{n=1}^{\infty}\frac{1}{1+n^2+n^4}$$
I doubt it is a telescopic Sum.
|
First, we have
$$
\begin{align}
\frac1{z^4+z^2+1}
&=\frac1{12}\left(
\frac{-3-i\sqrt3}{z-e^{\pi i/3}}
+\frac{3+i\sqrt3}{z-e^{4\pi i/3}}
+\frac{3-i\sqrt3}{z-e^{2\pi i/3}}
+\frac{-3+i\sqrt3}{z-e^{5\pi i/3}}
\right)\tag{1}
\end{align}
$$
Let $\gamma$ be the rectangle
$$
[-1-i,1-i]\cup[1-i,1+i]\cup[1+i,-1+i]\cup[-1+i,-1-i]\tag{2}
$$
then the integral
$$
\frac1{2\pi i}\int_{(n+\frac12)\gamma}\frac{\pi\cot(\pi z)}{z^4+z^2+1}\,\mathrm{d}z\tag{3}
$$
tends to $0$ since along the horizontal paths, $|\pi\cot(\pi z)|\to\pi$ and along the vertical paths, $|\pi\cot(\pi z)|\lt\pi$.
Since $\pi\cot(\pi z)$ has residue $1$ at each integer, we get that $(3)$ is the sum of the residues of $(1)$ times $\pi\cot(\pi z)$ at the singularities of $(1)$ plus
$$
1+2\sum_{n=1}^\infty\frac1{n^4+n^2+1}\tag{4}
$$
The sum of the residues of $(1)$ times $\pi\cot(\pi z)$ at the singularities of $(1)$ is
$$
-4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\cot\left(\pi\tfrac{1+i\sqrt3}2\right)\right)\tag{5}
$$
Since the sum of $(4)$ and $(5)$ is $0$, we have
$$
\begin{align}
1+2\sum_{n=1}^\infty\frac1{n^4+n^2+1}
&=4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\cot\left(\pi\tfrac{1+i\sqrt3}2\right)\right)\\
&=-4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\tan\left(\pi\tfrac{i\sqrt3}2\right)\right)\\
&=\tfrac{\pi\sqrt3}3\tanh\left(\pi\tfrac{\sqrt3}2\right)\tag{6}
\end{align}
$$
Therefore,
$$
\sum_{n=1}^\infty\frac1{n^4+n^2+1}=\frac{\pi\sqrt3}6\tanh\left(\pi\frac{\sqrt3}2\right)-\frac12\tag{7}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/966942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
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|
How to calculate the value of the special integral I get
$${\left. \frac{\partial ^2}{\partial n^2} \left( \frac{\partial ^2}{\partial m^2} B(m,n) \right) \right|_{m = \frac{1}{2},n = 0}} = \int_0^1 \frac{\ln^2 x \ln^2 (1 - x)}{\sqrt x (1 - x)} \, dx =\text{ ?}$$
but how to calculate the value of beta function derivative.
where $B(m,n)$ is beta function.
|
For the integral:
$${\left. \frac{\partial ^2}{\partial n^2} \left( \frac{\partial ^2}{\partial m^2} B(m,n) \right) \right|_{m = \frac{1}{2},n = 0}} = \int_0^1 \frac{\ln^2 x \ln^2 (1 - x)}{\sqrt x (1 - x)} \, dx $$
Since
\begin{align}
\partial_{m} B(m,n) = B(m,n) \left[ \psi(m) - \psi(m+n) \right]
\end{align}
then
\begin{align}
\partial_{m}^{2} B(m,n) = B(m,n) \left[ \psi'(m) - \psi'(m+n) + \left( \psi(m) - \psi(m+n) \right)^{2} \right]
\end{align}
and
\begin{align}
\partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} = B(1/2,n) \left[ \psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} \right].
\end{align}
Now
\begin{align}
\partial_{n}\partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} &= B(1/2,n) \left[ - \psi''(n+1/2) -2 \psi'(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\
& \left. + (\psi(n) - \psi(n+1/2))(\psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} )\right].
\end{align}
Also
\begin{align}
\partial_{n}^{2} \partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} &= B(1/2,n) \left[ - \psi'''(n+1/2) -2 \psi''(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\
& \left. + 2 \left( \psi'(n+1/2)\right)^{2}
+ (\psi(n) - \psi(n+1/2))( - \psi''(n+1/2) -2 \psi'(n+1/2) \cdot \right. \\
& \left. \left( \psi(1/2) - \psi(n+1/2) \right) )\right] \\
& + B(1/2,n) (\psi(n) - \psi(n+1/2)) \left[ - \psi''(n+1/2) -2 \psi'(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\
& \left. + (\psi(n) - \psi(n+1/2))(\psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} )\right]
\end{align}
When $n=0$ this becomes
\begin{align}
\partial_{n}^{2} \partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}}^{n=0} &= B(1/2,0) \left[ - \psi'''(1/2) + 2 \left( \psi'(1/2)\right)^{2}
+ 2(\psi(0) - \psi(1/2))( - \psi''(1/2)) \right].
\end{align}
This is value trying to be sought. Notice that $\Gamma(0) = \infty$ and thus the result is invalid.
Correct Process
Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{\ln^2 x \ln^2 (1 - x)}{(1-x)\sqrt{x}} \, dx
\end{align}
for which integration by parts, in the form,
\begin{align}
I &= \int_{0}^{1} \frac{\ln^2 x}{\sqrt{x}} \cdot \frac{\ln^2 (1 - x)}{(1-x)} \, dx \\
&= \left[ - \frac{1}{3} \ln^{3}(1-x) \cdot \frac{\ln^2 x}{\sqrt{x}} \right]_{0}^{1} + \frac{1}{3} \int_{0}^{1} \ln^{3}(-x) \, D\left( x^{-1/2} \, \ln^{2}(x) \right) \, dx \\
&= \frac{1}{3} \int_{0}^{1} \ln^{3}(-x) \, D\left( x^{-1/2} \, \ln^{2}(x) \right) \, dx \\
&= \frac{2}{3} \int_{0}^{1} x^{-3/2} \ln(x) \, \ln^{3}(1-x) \, dx - \frac{1}{6} \int_{0}^{1} x^{-3/2} \ln^{2}(x) \, \ln^{3}(1-x) \, dx \\
&= \frac{2}{3} \partial_{x} \partial_{y}^{3} \left. B(x,y) \right|_{x=-1/2}^{y=1} - \frac{1}{6} \partial_{x}^{2} \partial_{y}^{3} \left. B(x,y) \right|_{x=-1/2}^{y=1}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $ $$ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $$
If I pick $ x = 0$ I get:
$$ \lim_{(x,y) \to (0,0)} \frac{0}{2y^2} = 0$$
So if the limit exists it must be $0$
Now for ${(x,y) \to (0,0)}$ I have $xy \to 0$
So I can use the Taylor series of $sin(t) = t + o(t)$ where $t \to 0$
$$0 \leq \frac{\sin^2(xy)}{3x^2+2y^2} = \frac{x^2y^2 + 2o(x^2y^2) + o (x^2y^2)}{3x^2 + 2y^2} =$$
$$\frac{x^2y^2 + o (x^2y^2)}{3x^2 + 2y^2} = $$
Now I can use the polar coordinates:
$$\frac{\rho^4 \cos^2(\theta)\sin^2(\theta) + o (\rho^4 \cos^2(\theta)\sin^2(\theta))}{3\rho^2 \cos^2(\theta) + 2\rho^2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 \cos^2(\theta) + 2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 - 3 \sin^2(\theta) + 2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 - \sin^2(\theta)}\leq$$
$\frac{\rho^2}{2} \to 0$ for $\rho \to 0$
I would like to know if it is solved in the right way
|
Or $\sin^2(xy)\leq(xy)^2$ and $3x^2\geq 2x^2$ thus $\frac{\sin^2(xy)}{3x^2+2y^2}\leq \frac {(xy)^2}{2(x^2+y^2)}=xy\cdot \frac {xy}{2(x^2+y^2)}\leq xy\cdot \frac {1}{4}\to 0$ because $\frac {xy}{x^2+y^2}\leq \frac {1}{2}\Leftrightarrow 2xy\leq x^2+y^2\Leftrightarrow 0\leq (x-y)^2$.
|
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|
$n^{3} + 2n$ is divisible by $3$. Is this induction proof correct? Question: Prove by means of the principle of induction that for every $n ∈ N$ the
number $n^{3} + 2n$ is divisible by $3$.
Proof
Denote "$n^{3} + 2n$ is divisible by 3" by $P(n)$. Check $P(n)$ for an arbitrary $n$, for instance $n=1$. $1^{3}+2*1=3*1$ and thus divisible by three, therefore $P(1)$ holds.
Induction step: Assume $P(n)$ is is true, let $n ∈ N$. Then $(n+1)^{3}+2(n+1) =(n+1)(n^{2}+2n+1)+2(n+1)=(n+1)(n^{2}+2n+3)=n^{3}+2n^{2}+3n+n^{2}+2n+3=n^{3}+2n+3(n^{2}+n+1).$
We assumed $P(n)$ holds, thus the $n^{3}+2n$ part of the induction holds. We can obviously see that $3(n^{2}+n+1)$ is divisible by 3 and that concludes the proof.
|
I suggest a slight rewording.
Proof
Denote the statement
$n^{3} + 2n$ is divisible by 3
by $P(n)$. We check $P(n)$ for an $n=1$: $1^{3}+2*1=3*1$. Thus $P(1)$ holds.
Induction step: Assume $P(n)$ is is true, let $n ∈ N$. Then $(n+1)^{3}+2(n+1) =(n+1)(n^{2}+2n+1)+2(n+1)=(n+1)(n^{2}+2n+3)=n^{3}+2n^{2}+3n+n^{2}+2n+3=n^{3}+2n+3(n^{2}+n+1).$
We assumed $P(n)$ holds, thus $n^{3}+2n$ is divisible by $3$. We can obviously see that $3(n^{2}+n+1)$ is divisible by 3 and that concludes the proof.
|
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|
No. of integral solutions of $x_1+x_2+x_3+x_4=20.$ I've to solve a no. of questions of this type but don't get how to do it:
Determine the no. of integral solutions of $x_1+x_2+x_3+x_4=20.$ given the constraint that
$$1\leq x_1\leq 6,0\leq x_2\leq7,4\leq x_3\leq 8 ,2\leq x_4\leq 6.$$
I made the following expansions :
$(x+x^2+x^3+x^4+x^5+x^6)(1+x+x^2+\cdots +x^7)(x^4+\cdots +x^8)(x^2+\cdots+ x^6)$
now we'll find the coeffiecient of $x^{20}$ in the above expression .
but I can't understand how...
|
One method that leads to a general solution of these types of problems is to sum each geometric series and then deal with the resulting product. In this case, your generating function would be
$$\frac{x(1-x^6)}{1-x}\cdot
\frac{(1-x^8)}{1-x}\cdot
\frac{x^4(1-x^5)}{1-x}\cdot
\frac{x^2(1-x^5)}{1-x}.$$
Then, the coefficient of $x^{20}$ of this would coincide with the coefficient of $x^{13}$ in $\frac{(1-x^6)(1-x^8)(1-x^5)^2}{(1-x)^4}$. I think you're stuck with expanding the numerator, but the denominator is just $\frac1{(1-x)^4}=\sum_n\binom{n+3}3x^n$. This leads to $$(1-2x^5-x^6-x^8+x^{10}+2x^{11}+2x^{13}+\cdots)\cdot\sum_n\binom{n+3}3x^n.$$ So, the number of solutions to your original equation would be
$$\binom{16}3-2\binom{11}3-\binom{10}3-\binom83+\binom63+2\binom53+2\binom33=96.$$
|
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|
Computing the limit of an expression I have the following question. Determine the limit of the following expression.
$\lim_{x \rightarrow 0} \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4}$.
My attempt to this question is the following. Let the function $f(x) = \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4} = \frac{\cos(x \sqrt{2})}{x^4} - \frac{1}{(1+x^2) x^4}.$ Then $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\cos(x \sqrt{2})}{x^4} - \lim_{x \rightarrow 0} \frac{1}{(1+x^2) x^4} = 0 - Undefined$. So the limit does not exists.
But my professor's answer is $-\frac{5}{6}$. He used Taylor expansion. What am I doing wrong??
|
We have :
$$\mathop {\lim }\limits_{x \to {x_0}} \left( {f(x) + g(x)} \right) = \mathop {\lim }\limits_{x \to {x_0}} f(x) + \mathop {\lim }\limits_{x \to {x_0}} g(x) ~~~~ \left(*\right) $$
ONLY IF:
$$\mathop {\lim }\limits_{x \to {x_0}} f(x) ~ and
\mathop {\lim }\limits_{x \to {x_0}} g(x) ~ exist $$
Back to this problem, $\mathop {\lim }\limits_{x \to 0} {{\cos \left( {x\sqrt 2 } \right)} \over {{x^4}}} \to \infty , \mathop {\lim }\limits_{x \to 0} {1 \over {(1 + {x^2}){x^4}}}$
So $\left(*\right)$ is invalid here.
Using the Taylor Expansion:
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} {{\cos \left( {x\sqrt 2 } \right) - {1 \over {1 + {x^2}}}} \over {{x^4}}} \cr
& = \mathop {\lim }\limits_{x \to 0} {{(1 + {x^2})\cos \left( {x\sqrt 2 } \right) - 1} \over {{x^4}(1 + {x^2})}} \cr
& = \mathop {\lim }\limits_{x \to 0} {{(1 + {x^2})\left( {1 - {1 \over 2}{{\left( {x\sqrt 2 } \right)}^2} + {1 \over {24}}{{\left( {x\sqrt 2 } \right)}^4} + o\left( {{x^4}} \right)} \right) - 1} \over {{x^4}}} \cr
& = \mathop {\lim }\limits_{x \to 0} {{ - {5 \over 6}{x^4} + o\left( {{x^4}} \right)} \over {{x^4}}} \cr
& = - {5 \over 6} \cr} $$
|
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|
When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
|
This is correct:
$$
\sqrt{a^2+4}=a\sqrt{1+\frac{4}{a^2}}=a+a\left(\sqrt{1+\frac{4}{a^2}}-1\right)=a+a\frac{1+\frac{4}{a^2}-1}{\sqrt{1+\frac{4}{a^2}}+1}\\=a+\frac{4}{a}\cdot\frac{1}{\sqrt{1+\frac{4}{a^2}}+1}\approx a+\frac{2}{a}
$$
|
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|
Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\sqrt{6}-5)$$
Could you show me the steps to get the final result?
Thank you.
|
I think your question is about $\sqrt{x}(\sqrt{3}+\sqrt{2})=17$. Then
$$
\begin{array}{rcl}
\sqrt{x}(\sqrt{3}+\sqrt{2}) & = & 17 \\
(\sqrt{x}(\sqrt{3}+\sqrt{2}))^2 & = & 17^2 \\
x(\sqrt{3}^2+2\sqrt{3}\sqrt{2}+\sqrt{2}^2) & = & 289 \\
x(3+2\sqrt{3}\sqrt{2}+2) & = & 289 \\
x(5+2\sqrt{3\times2}) & = & 289\\
x(5+2\sqrt{6}) & = & 289 \\
x & = & \frac{289}{5+2\sqrt{6}} \\
x & = & \frac{289}{5+2\sqrt{6}} \times \frac{5-2\sqrt{6}}{5-2\sqrt{6}} \\
x & = & \frac{289 \times (5-2\sqrt{6})}{5^2+4\times 6} \\
x & = & \frac{289 \times (5-2\sqrt{6})}{5^2-4\times 6} \\
x & = & \frac{289 \times (5-2\sqrt{6})}{25 - 24} \\
x & = & 289 \times (5-2\sqrt{6})
\end{array}
$$
|
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|
Double Integration Inequality I've been trying to work out the following. Could anyone please show me how to do this?
Let $D$ be the domain bounded by $y=x^2+1$ and $y=2$. Prove the inequality
$$\frac{4}{3}\le \iint_D(x^2+y^2)\,\mathrm{d}A\le\frac{20}{3}.$$
Thank you.
|
First notice that in $D$ we have $1 \le x^2 + y^2 \le 5$. This implies $$ |D| \le \iint_Dx^2 + y^2 \le 5|D|$$
Now we only need to find $|D|$ ( = area of $D$), which is of course equal to $$4 - \int_{-1}^1(x^2 + 1) = 4 - 2\int_0^1(x^2 + 1) = 4 - \frac{2}{3} - 2 = \frac{4}{3}$$
|
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Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.
"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.
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Observe that
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2} \iff (\frac{a}{b+c}+1)+(\frac{b}{c+a}+1)+(\frac{c}{a+b}+1)= (a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})\ge \frac{9}{2}.$
By Titu's lemma
$\frac{1^2}{b+c}+\frac{1^2}{c+a}+\frac{1^2}{a+b}\ge \frac{(1+1+1)^2}{(a+b)+(b+c)+(c+a)} = \frac{9}{2(a+b+c)}$.
So
$(a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})\ge (a+b+c)\frac{9}{2(a+b+c)} = \frac{9}{2}.$
|
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|
Evaluating a series of hypergeometric functions I would like to prove (or disprove) the following statement:
$$
\sum_{n=0}^\infty \left[\frac{{}_2{\rm F}_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};1\right)}{n!}\right] = \frac{\pi}{2} \left[ \sum_{a=0}^\infty \frac{1}{4^a (a!)^2} + \frac{1}{2} \sum_{b=0}^\infty \frac{\left(\frac{1}{2}\right)^{2b}}{\Gamma\left(b+\frac{3}{2}\right)^2}\right]
$$
also can be seen as
$$
\sum_{n=0}^\infty \left[\frac{{}_2{\rm F}_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};1\right)}{n!}\right] = \frac{\pi}{2} \bigg[ I_0(1) + L_0(1) \bigg]
$$
Where $I_0(1)$ is the modified bessel function of the first kind and $L_0(1)$ is the modified struve function.
Major brownie points on the line here!
|
Semiclassical has told me to post an answer to my own question using the technique i have explained to him, now for me to go any further i should state that the original function i am dealing with is:
$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x))
$
therefore using the power series of a hyper geometric function i can say:
$$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x)) = \sum_{a=0}^{\infty} \frac{(\frac{1}{2})_a (\frac{1-n}{2})_a \cos^{2a}(x)}{(\frac{3}{2})_a a!}
$$
Having $(f(x))_a be the rising factorial therefore with the definition of a rising factorial as using gamma functions i can state the following aswell:
$$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x)) = \sum_{a=0}^{\infty} \frac{\Gamma(\frac{1}{2}+a) \Gamma(\frac{1-n}{2}+a) \Gamma(\frac{3}{2}) \cos^{2a}(x)}{\Gamma(\frac{3}{2}+a)\Gamma(\frac{1}{2}) \Gamma(\frac{1-n}{2}) a!}
$$
using the following identities i can simplify the function more
$\Gamma(\frac{1}{2}) = \sqrt{\pi}$
$\Gamma(\frac{3}{2}) = \sqrt{\pi}/2$
$ \Gamma(\frac{1}{2} + n) = \frac{(2n)! \sqrt{\pi}}{4^{n} n!}$
$$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x)) = \frac{1}{2}\sum_{a=0}^{\infty}\frac{\pi(2a)!4^{a+1}(a+1)! \Gamma(\frac{1-n}{2}+a)}{ \pi 4^{a}(a!)^2 (2a+2)! \Gamma(\frac{1-n}{2})}
$$
$$
= \frac{1}{2}\sum_{a=0}^{\infty}\frac{4 \cos^{2a}(x) (a+1) \Gamma(\frac{1-n}{2}+a)}{ a!(2a+2)(2a+1) \Gamma(\frac{1-n}{2})}
$$
$$
= \frac{4}{2}\sum_{a=0}^{\infty}\frac{\cos^{2a}(x) \Gamma(\frac{1-n}{2}+a)}{ 2 a!(2a+1) \Gamma(\frac{1-n}{2})}
$$
$$
= \sum_{a=0}^{\infty}\frac{\cos^{2a}(x)\Gamma(\frac{1-n}{2}+a)}{ a!(2a+1) \Gamma(\frac{1-n}{2})}
$$
Which i know looks very ugly, but when i graphed this on Mathematica on the interval $ [0,2\pi]$ it was the same result as the function given by Semiclassical (in my specific scenario). therefore you can say that they are equal i suppose?
|
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|
One Step in an Integration I was doing the integration
$$
\int\frac{1}{(u^2+a^2)^2}du
$$
and I had a look at the lecturer's steps where I got stuck in the following step:
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
I guess it is integrating this by parts, but I could't see the trick. Could somebody help me with this please?
I had some work after I asked this question here. Here is my approach.
First note
$$
d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{2a^2u}{(u^2+a^2)^2}du,
$$
then we have
$$
\int\frac{1}{(u^2+a^2)^2}du=\int\left(-\frac{1}{2a^2u}\right)\left(-\frac{2a^2u}{(u^2+a^2)^2}\right)du=\frac{1}{2a^2}\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right).
$$
Then integrate by parts, we have
$$
\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{a^2}{u(u^2+a^2)}-\int\frac{a^2}{u^2(u^2+a^2)}du.
$$
But notice
$$
\frac{a^2}{u^2(u^2+a^2)}=\frac{1}{u^2}-\frac{1}{u^2+a^2},
$$
then we get
$$
\int\frac{a^2}{u^2(u^2+a^2)}du=-\frac{1}{u}-\int\frac{1}{u^2+a^2}du,
$$
then
$$
\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{a^2}{u(u^2+a^2)}+\frac{1}{u}+\int\frac{1}{u^2+a^2}du.
$$
But
$$
\frac{1}{u}-\frac{a^2}{u(u^2+a^2)}=\frac{u}{u^2+a^2},
$$
then
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
Is this approach okay? I am positive with this but not sure it is comprehensive.
|
The trick is using the substitution $u = a\tan x $, Then we have
$$ \int \frac{du}{(u^2 +a^2)^2 } = \int \frac{du}{(a^2 \tan^2 x + a^2)^2}= \int \frac{du}{(a \sec x)^2}$$
and $ du = a \sec^2 x dx $
|
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}
|
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result when I test (using www.WolframAlpha.com) for various values of $a$ where $0<a<1$.
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$
D_2
\, =\,
\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
How could I go about proving:-
(1) $D_1$ = $D_2$,
(SOLVED, I think, by my two answers below, but using WolframAlpha to obtain integral solutions)
$$$$
(2) $D_1$ = $R$ or $D_2$ = $R$.
(MOVED to a separate question:
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$).
UPDATE 1
You can see how WolframAlpha
produces these results by inputting the following input texts:-
For Eqtn 1 with a=0.1 input: integrate (3*0.1(sinx)^2)/((1-0.1*cosx)^4) from x=0 to 2*pi
For Eqtn 2 with a=0.1 input: integrate (cosx)/((1-0.1*cosx)^3) from x=0 to 2*pi
for Result with a=0.1 input: evaluate 3 0.1 pi/(1-0.1^2)^(5/2)
UPDATE 2
WolframAlpha also computes expressions for the indefinite integrals as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
$$$$
$$$$
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression are equivalent to each other. This could be useful. For example we can define a difference function $f_3 = f_1-f_2$ whose indefinite integral $I_3 = I_1-I_2$ will exclude the common awkward third term.
Let us assume that $f_3$ is continuously integrable over the range $0,2\pi$ (we cannot be sure by inspection alone, but it can be shown, see my answer below). Then, if $D_1=D_2$ over the range $0,2\pi$ then $D_1-D_2=0$ and so $D_3$ (=$\int_0^{2\pi}f_3\,d\theta$) should have value zero. This is expanded on in my answer below.
|
Hint:
Observe that,
$$\frac{\partial}{\partial a}\left[\frac{1}{\left(1-a\cos{\theta}\right)^2}\right]=\frac{2\cos{\theta}}{\left(1-a\cos{\theta}\right)^3}.$$
Thus, we can simplify the integral we have to compute via the technique of differentiating under the integral sign:
$$\begin{align}
I(a)
&=\int_{0}^{2\pi}\frac{\cos{\theta}}{\left(1-a\cos{\theta}\right)^3}\,\mathrm{d}\theta\\
&=\frac12\int_{0}^{2\pi}\frac{\partial}{\partial a}\left[\frac{1}{\left(1-a\cos{\theta}\right)^2}\right]\,\mathrm{d}\theta\\
&=\frac12\frac{\partial}{\partial a}\int_{0}^{2\pi}\frac{\mathrm{d}\theta}{\left(1-a\cos{\theta}\right)^2}.\\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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|
Condition for trigonometric inequality I want to prove the following statement:
Suppose $\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2\leq 1$ holds for all $\theta_1,\theta_2\in[-\pi,\pi]$, then we should have $\lambda^2(a^2+b^2)\leq \frac{1}{2}$.
This problem appears when I want to find the stability condition for a numerical scheme. I tried to use Lagrange multiplier, but it turns out to be very complicated. I have also tried to find some specific $\theta_1,\theta_2$, so that the first inequality can imply the second, but I failed to do so.
|
You want some bound for the expression $\lambda^2(a^2 + b^2)$.
We have
$$
\frac{1}{4}(\cos(\theta_1)+\cos(\theta_2))^2+\lambda^2(a\sin(\theta_1)+b\sin(\theta_2))^2\leq 1
$$
and with two other angles
$$
\frac{1}{4}(\cos(\theta_3)+\cos(\theta_4))^2+\lambda^2(a\sin(\theta_3)+b\sin(\theta_4))^2\leq 1
$$
Let us choose the angles such that $\sin(\theta_1)\sin(\theta_2) + \sin(\theta_3)\sin(\theta_4) = 0$. This is general, since it is necessary to remove terms $a \, b$.
Then we can add the two inequalities to give
$$
\frac{1}{4}((\cos(\theta_1)+\cos(\theta_2))^2 + (\cos(\theta_3)+\cos(\theta_4))^2) +\lambda^2(a^2(\sin^2(\theta_1) + \sin^2(\theta_3))+b^2 (\sin^2(\theta_2) + \sin^2(\theta_4)))\leq 2
$$
Let us now also impose $\sin^2(\theta_1) + \sin^2(\theta_3) - \sin^2(\theta_2) - \sin^2(\theta_4) = 0$. This is is general, since it is necessary to obtain the term $(a^2 +b^2)$.
Then we have
$$
\lambda^2(a^2 + b^2) < \frac{2 - \frac{1}{4}((\cos(\theta_1)+\cos(\theta_2))^2 + (\cos(\theta_3)+\cos(\theta_4))^2)}{\sin^2(\theta_1) + \sin^2(\theta_3) }
$$
So the tightest bound will be obtained by minimizing the RHS subject to the two conditions $\sin^2(\theta_1) + \sin^2(\theta_3) - \sin^2(\theta_2) - \sin^2(\theta_4) = 0$ and $\sin(\theta_1)\sin(\theta_2) + \sin(\theta_3)\sin(\theta_4) = 0$.
The two conditions can be combined, replacing $\sin^2(\theta_1)$:
$\left[\sin^2(\theta_2) - \sin^2(\theta_3) \right]\left[ \sin^2(\theta_4) +\sin^2(\theta_2) \right]= 0$
This can only be observed for a) $\theta_2 = \theta_3$ or b) $\theta_2 = \pi \pm \theta_3$ or c) $\theta_2 = n \pi $ and $\theta_4 = m \pi$ $(n,m \in \cal Z)$. Let's look at these cases.
Case a) gives, for the second condition, $\sin(\theta_1) + \sin(\theta_4) = 0$, which is $\theta_1 = - \theta_4$, or $\theta_1 = \pi + \theta_4$. Inserting into the desired bound gives
$$
\lambda^2(a^2 + b^2) < \frac{2 - \frac{1}{2}(\cos(\theta_1)+\cos(\theta_2))^2 }{\sin^2(\theta_1) + \sin^2(\theta_2) }
$$
The smallest value that the RHS can take is 1.
Case b) gives, for the second condition at $\theta_2 = \pi + \theta_3$ , $\sin(\theta_1) - \sin(\theta_4) = 0$, which is $\theta_1 = \theta_4$, or $\theta_1 = \pi - \theta_4$. Inserting into the desired bound gives
$$
\lambda^2(a^2 + b^2) < \frac{2 - \frac{1}{2}(\cos^2(\theta_1)+\cos^2(\theta_2)) }{\sin^2(\theta_1) + \sin^2(\theta_2) }
$$
The smallest value that the RHS can take is 1.
Case c) gives, for the first condition, $\sin^2(\theta_1) + \sin^2(\theta_3) = 0$, which is $\theta_1 = n \pi $ and $\theta_3 = m \pi$ $(n,m \in \cal Z)$. Inserting into the desired bound gives a diverging RHS.
So in total, the tightest bound one can obtain for $\lambda^2(a^2+b^2)$ is
$$
\lambda^2(a^2+b^2)\leq 1
$$
So by adding specific inequalities, a tighter bound could not be found.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
How to show that the integral of bivariate normal density function is 1? How to show the following?
$$\large \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{x^2+y^2-2 \rho x y}{2(1-\rho^2)}} dx\ dy=1$$
|
Complete the square of the exponent $x^2+y^2-2\rho xy = \left(x-\rho y\right)^2+y^2(1-\rho^2)=:u^2+y^2(1-\rho^2)$
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{x^2+y^2-2 \rho x y}{2(1-\rho^2)}} dx dy=\\
=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{u^2}{2(1-\rho^2)}} du \; e^{-\frac{y^2(1-\rho^2)}{2(1-\rho^2)}} dy = \\
=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} e^{\frac{-y^2}{2}} dy=1$$
PS. Using the well known formula that $$\int_{-\infty}^{\infty}e^{-\frac{u^2}{a}} du=\sqrt{a\pi}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How can I find the complex numbers satisfying this condition? For a given complex number $a$ with $|a|\ge1,$ I want to find the all complex numbers on the unit circle such that $$\dfrac{z}{(a-z\bar a)^2}\in\mathbb{R}$$ and satisfying the condition $$\dfrac{z}{(a-z\bar a)^2}\le-\dfrac{1}{4}.$$
I express $z$ as $x+iy,$ but resulting expression is not easy to solve.
How can we solve these kind of problems?
|
Write $a = re^{i\varphi}$ with $r > 0$ and $\varphi \in \mathbb{R}$. An easy computation shows that
$$f(z) = \frac{z}{(a-\overline{a}\cdot z)^2} = \frac{1}{r^2}\frac{e^{-2i\varphi}z}{(1-e^{-2i\varphi}z)^2} = \frac{1}{r^2} K(e^{-2i\varphi}z),$$
where
$$K(w) = \frac{w}{(1-w)^2}$$
is the Koebe function. It is relatively well-known (and if you don't know it, it's a good exercise to prove it) that the Koebe function is a Schlicht function minimising the disk around $0$ contained in its image, and that
$$K\colon \mathbb{D} \to \mathbb{C} \setminus \left(-\infty, -\tfrac{1}{4}\right]$$
is biholomorphic.
So all points on the unit circle except for $1$, where $K$ has a pole, are mapped to $\left(-\infty,-\frac{1}{4}\right]$ by $K$, and if $r = \lvert a\rvert = 1$, then the same is true of $f$ [up to rotation, the pole is $e^{2i\varphi}$]. If $r > 1$, we want to find the points on the unit circle that $K$ maps to $\left(-\infty,-\frac{r^2}{4}\right]$ (and then rotate by $e^{2i\varphi}$). So let's invert $K$ (suppose $w\neq 0$):
\begin{align}
&& w &= \frac{z}{(1-z)^2}\\
&\iff& w(z^2-2z+1) &= z\\
&\iff& z^2 - 2\left(1 + \tfrac{1}{2w}\right)z + 1 &= 0\\
&\iff& \left(z - \left(1 + \tfrac{1}{2w}\right)\right)^2 &= \frac{4w+1}{4w^2}\\
&\iff& z &= \frac{2w+1\pm \sqrt{4w+1}}{2w}.
\end{align}
Note that the possible values of $z$ are reciprocals of each other, so if one of them lies on the unit circle, both do.
Now insert $w = -\frac{t}{4}$ for $t \geqslant r^2$ and obtain
$$K^{-1}\left(-\tfrac{t}{4}\right) = \left\{ \frac{1-t/2 \pm \sqrt{1-t}}{-t/2}\right\} = \left\{ 1-\tfrac{2}{t} \pm \frac{2i}{t}\sqrt{t-1}\right\}.$$
The two values describe two arcs on the unit circle, one in the upper, and one in the lower half-plane, approaching $1$ for $t\to\infty$, and $-1$ for $t\to 1$.
Inserting specifically $t = r^2$ yields the end points of the arc,
$$1 - \frac{2}{r^2} \pm \frac{2i}{r^2}\sqrt{r^2-1}.$$
If we want to find the angle, which seems most convenient, we obtain
$$\alpha_{r^2} = \operatorname{arccot} \frac{r^2-2}{\sqrt{r^2-1}},$$
and
$$K^{-1}\left(\left(-\infty, -\tfrac{r^2}{4}\right]\right) = \left\{ z \in \partial \mathbb{D} : 0 < \lvert \arg z\rvert \leqslant \operatorname{arccot} \frac{r^2-2}{\sqrt{r^2-1}}\right\}.$$
Hence
$$f^{-1}\left(\left(-\infty,-\tfrac{1}{4}\right]\right) = \left\{ z \in \partial \mathbb{D} : 0 < \lvert \arg z - 2\arg a\rvert \leqslant \operatorname{arccot} \frac{\lvert a\rvert^2-2}{\sqrt{\lvert a\rvert^2-1}}\right\}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Does $\sum_{n=0}^\infty (-1)^n (e-(1+\frac{1}{n})^n)$ converge absolutely, conditionally, or diverge? I've been given the hint to use the binomial theorem and show that $e-\left(1+\frac{1}{n}\right)^n > \frac{1}{2n}$ for $n \geq 2$. So I've written
\begin{align*}
e-\left(1+\frac{1}{n}\right)^n = e - \sum_{k=0}^n {n \choose k} \frac{1}{n^k}
\end{align*}
But I'm not sure where to go from here to assess convergence. What are the next steps?
|
Converge absolutely? WRONG. Because as you said, $e-\left(1+\frac{1}{n}\right)^n > \frac{1}{2n}$.
Let us prove this inequality.
\begin{align*}e-\left(1+\frac{1}{n}\right)^n>{}&\left(1+\frac{1}{2n}\right)^{2n}-\left(1+\frac{1}{n}\right)^n \\
={}&\left(1+\frac{1}{n}+\frac{1}{4n^2}\right)^{n}-\left(1+\frac{1}{n}\right)^n\\
={}&\left(\left(1+\frac{1}{n}+\frac{1}{4n^2}\right) -\left(1+\frac{1}{n}\right)\right)\sum_{i=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{4n^2}\right)^i\left(1+\frac{1}{n}\right)^{n-1-i}\\
>{}&\left(\left(1+\frac{1}{n}+\frac{1}{4n^2}\right) -\left(1+\frac{1}{n}\right)\right)\left(1+\frac{1}{n}\right)^{n-1}n\\
={}&\frac{1}{4n}\left(1+\frac{1}{n}\right)^{n-1}\\
>{}& \frac{1}{2n}
\end{align*}
The last inequality holds for $n$ large enough.
Converge conditionally? YES. Since $e-\left(1+\frac{1}{n}\right)^n$ decreases monotonically and goes to $0$. (Leibniz criterion)
|
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|
Find $a^2 + b^2+c^2$ Given
$a^2+2b = 7$
$b^2+4c = -7$
$c^2+6a = -14$
Find $a^2 + b^2 + c^2$
The answer was an Integer
I tried to solve it by making $a$ the subject of the equation and substituting in others but equations became too complex(got $a^8$!) and difficult to solve.
|
HINT:
Completing the square,
$$(a+3)^2+(b+1)^2+(c+2)^2=7-7-14+9+1+4=0$$
Now if $a,b,c$ are real, what can we derive from here?
|
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|
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.
Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.
|
Approach 1: Write the desired inequality as
$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9\tag{i}
$$
and use the AM-GM inequality $x+y+z\geq 3(xyz)^{1/3}$ to each sum in the parentheses above.
Edit: just a few more approaches (among potentially many others). Hope you'll find this useful.
Approach 2:
Multiply out the LHS of (i) as
$$
1+1+1+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)
$$
and use $x+y\geq 2\sqrt{xy}$ 3 times.
Approach 3: Use Cauchy-Schwarz like Macavity suggested below:
$$
\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right)\left(\sqrt{\frac{1}{a}}^2+\sqrt{\frac{1}{b}}^2+\sqrt{\frac{1}{c}}^2\right)\geq(1+1+1)^2=9
$$
Approach 4: Use Jensen's inequality: the function $f(x)=\frac{1}{x}$ is convex for $x>0$ so:
$$
\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{3}f(a)+\frac{1}{3}f(b)+\frac{1}{3}f(c)\\
\geq f\left(\frac{1}{3}(a+b+c)\right)=\frac{3}{a+b+c}
$$
from which you can rearrange to obtain your claim.
Approach 5: Please also look up the Chebyshev's sum inequality from which your inequality is an immediate consequence.
|
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|
Can we express the following ordinary generating function? I wish to express the following power series
$$ \sum_{k \ge 0} \binom{n-k}{m} x^k$$
where $n,m$ are positive integer such that $0< m \le n$
|
$$
\begin{align}
\sum_{k=0}^\infty\binom{n-k}{m}x^k
&=\sum_{k=0}^n\binom{n-k}{m}x^k+\sum_{k=n+1}^\infty\binom{n-k}{m}x^k\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+\sum_{k=1}^\infty\binom{-k}{m}x^{k+n}\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+(-1)^m\sum_{k=1}^\infty\binom{m-1+k}{m}x^{k+n}\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+(-1)^m\sum_{k=1}^\infty\binom{m-1+k}{k-1}x^{k+n}\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+(-1)^mx^{n+1}\sum_{k=1}^\infty(-1)^{k-1}\binom{-m-1}{k-1}x^{k-1}\\
&=\underbrace{\sum_{k=0}^n\binom{n-k}{m}x^k}_{\text{polynomial}}-\frac{x^{n+1}}{(x-1)^{m+1}}
\end{align}
$$
If $n\lt m$, the polynomial part is $0$.
If $n\ge m$, the polynomial part has degree $n-m$:
$$
\begin{align}
\sum_{k=0}^n\binom{n-k}{m}x^k
&=\sum_{k=0}^{n-m}\binom{n-k}{m}x^k\\
&=\sum_{k=m}^n\binom{k}{m}x^{n-k}
\end{align}
$$
|
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|
How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$? How to prove this inequality ? $$\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$$
for $a,b,c>0 $ and $a+b+c=\frac 1a+\frac 1b+\frac 1c$.
I do not know where to start. I need some idea and advice on this problem.Thanks
|
Let $\sum\limits_{cyc}\frac{1}{2+a}>1$ and $a=ka'$ such that $k>0$ and
$$\frac{1}{2+a'}+\frac{1}{2+b}+\frac{1}{2+c}=1.$$
Hence,
$$\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}>1=\frac{1}{2+a'}+\frac{1}{2+b}+\frac{1}{2+c}$$ or
$$\frac{1}{2+ka'}>\frac{1}{2+a'},$$
which gives $k<1$.
In another hand,
$$a'+b+c-\frac{1}{a'}-\frac{1}{b}-\frac{1}{c}=\frac{a}{k}+b+c-\frac{1}{\frac{a}{k}}-\frac{1}{b}-\frac{1}{c}>a+b+c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=0,$$
which is contradiction because we'll prove now that
$$a'+b+c-\frac{1}{a'}-\frac{1}{b}-\frac{1}{c}\leq0.$$
Indeed, let $a'=\frac{2x}{y+z}$ and $b=\frac{2y}{x+z}$, where $x$, $y$ and $z$ are positives.
Hence, the condition $\frac{1}{2+a'}+\frac{1}{2+b}+\frac{1}{2+c}=1$ gives $c=\frac{2z}{x+y}$ and we need to prove that
$$\sum_{cyc}\frac{y+z}{2x}\geq\sum_{cyc}\frac{2x}{y+z}$$ or
$$\sum_{cyc}x\left(\frac{1}{y}+\frac{1}{z}\right)\geq\sum_{cyc}\frac{4x}{y+z},$$
which is C-S:
$$\frac{1}{y}+\frac{1}{z}\geq\frac{(1+1)^2}{y+z}=\frac{4}{y+z}.$$
Done!
|
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|
Help with epsilon delta proof $x^3$ is near $27$ when $x$ is near $3$ but $x$ is not equal to $3$.
So I have
$$0<|x-3|<\delta \implies 3-\delta<x<3+\delta$$
$$|x^3-27|<\epsilon \implies|(x-3)(x^2+3x+3^2)|<\epsilon$$
$$=(x^2+3x+9)|(x-3)|<\epsilon \implies28|(x-3)|<\epsilon$$
$$=|(x-3)|<\epsilon/28$$
How do I prove $x^2+3x+9$ is less than $28$?
My attempt
$|x-3|<\delta$
so $$|x^2+3x+9|=|x^2+3x-18+27|\leq|x^2+3x-18|+27 \\ =|(x-3)(x+6)|+27
=|(x-3)(x-3+9)|+27$$
I'm not really sure where to go from here.
|
You can continue by saying that $|(x-3)(x-3+9)|+27\le|x-3|(|x-3|+9)+27<\delta(\delta+9)+27$,
so now you just have to make sure that $\delta(\delta+9)\le1$ by, say, choosing $\delta$ so that it satisfies $\delta\le\frac{1}{18}$
since then $\delta^2\le\frac{1}{2}$ and $9\delta\le\frac{1}{2}$
(along with any other conditions required to ensure that $|x^3-27|<\epsilon$).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integration of $ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $ How can I integrate this by changing variable?
$$ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $$
Thanks.
|
Let $a$ be such that $(2a)^2+a^2=1$. Let $\theta$ be such that $\cos(\theta)=a$ and $\sin(\theta)=2a$. Then you have
$$
\begin{align}
\int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx
&=\int \frac{2a\sin x + a\cos x}{a\sin x + 2a\cos x} dx\\
&=\int \frac{\sin(\theta)\sin x + \cos(\theta)\cos x}{\cos(\theta)\sin x + \sin(\theta)\cos x} dx\\
&=\int \frac{\cos(x-\theta)}{\sin(x+\theta)} dx\\
&=\int \frac{\cos(u-2\theta)}{\sin(u)} du\\
&=\int \frac{\cos(u)\cos(2\theta)+\sin(u)\sin(2\theta)}{\sin(u)} du\\
&=\int \left(\cot(u)\cos(2\theta)+\sin(2\theta)\right) du\\
\end{align}
$$
And it should be easy from here.
|
{
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"url": "https://math.stackexchange.com/questions/1004318",
"timestamp": "2023-03-29T00:00:00",
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|
The limit: $\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)}$ $$\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)}$$
My steps
$\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)} = \lim_{x\rightarrow0}\frac{1-\cos2x}{2x}\cdot2x\cdot\frac{3x}{\sin3x}\cdot\frac{1}{3x}\cdot\frac{1}{\ln(1+\sin4x)}\cdot\frac{4x\sin4x}{4x\sin4x}\\= \lim_{x\rightarrow0}\frac{2x}{3x\cdot4x}\cdot\frac{1-\cos2x}{2x}\cdot\frac{3x}{\sin3x}\cdot\frac{4x}{\sin4x}\cdot\frac{\sin4x}{\ln(1+\sin4x)} = \lim_{x\rightarrow0}\frac{2}{12x}$
I'm so close to the right answer $\frac{1}{6}$.
What did I do wrong?
Edit 1: So apparently, $\lim_{x\rightarrow0}\frac{1-\cos2x}{2x} = 0 \not=1$ and so the limit would be $\lim_{x\rightarrow0}\frac{2x}{3x\cdot4x}\cdot\frac{1-\cos2x}{2x}\cdot\frac{3x}{\sin3x}\cdot\frac{4x}{\sin4x}\cdot\frac{\sin4x}{\ln(1+\sin4x)} = 0\cdot\frac{2}{12\cdot0}$
And so now the limit is $[\frac{0}{0}]$. I'm not sure how to fix this.
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We have $1-\cos(2x)\sim \frac{(2x)^2}{2}=2x^2$ as $x\to 0$; $\sin(3x)\sim 3x$, $\ln(1+\sin(4x))\sim \sin(4x)\sim 4x$ as $x\to 0$. Thus
$$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {2x} \right)}}{{\sin \left( {3x} \right).\ln \left( {1 + \sin \left( {4x} \right)} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{x^2}}}{{3x.4x}} = \frac{1}{6}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1005961",
"timestamp": "2023-03-29T00:00:00",
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|
Solution for equation with inclusion-exclusion principle. Using the principle of inclusions and exclusions count how many solutions to the equation
$$ x + y + z = 12 $$
$$ 1 \le x \le 5$$ $$ -2 \le y \le 4$$ $$ 0 \le z \le 5 $$
$$ x,y,z \in \mathbb{Z}$$
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Let $a = x - 1, b = y + 2, c = z$. Then our equation becomes $a+b+c=13 \ (\star)$ with constraints: $$\begin{align*}
0&\le a \le 4 \\
0&\le b \le 6 \\
0&\le c \le 5
\end{align*}$$
Let $U$ be the set of all nonnegative integer solutions of $(\star)$. This is $|U| = \binom{15}{2}.$
Let the following sets be subsets of $U$.
$$\begin{align*}
A &= \{s \in U : a > 4\} \\
B &= \{s \in U : b > 6\} \\
C &= \{s \in U : c > 5\}
\end{align*}$$
In other words, $A$ is the set of all solutions of $(\star)$ where $a > 4$, etc.
Since we wish to count the number of solutions of $(\star)$ and satisfy our constraints, then we want to find the size of $A^c \cap B^c \cap C^c.$
By the Inclusion-Exclusion principle, this is
$$|A^c \cap B^c \cap C^c| = |U| - |A| - |B| - |C| + |A\cap B| + |B \cap C| + |A\cap C| -|A\cap B\cap C|. $$
We can count each one of the above sets. For example, since $A$ is the set of all solutions of $(\star)$ where $a>4$, we're guaranteed $a \ge 5$, so we just have to look at the number of nonnegative integer solutions of $a+b+c=8$, i.e. $|A| = \binom{10}{2}$.
Similarly $|B|$ is the number of solutions of $a+b+c=6$, i.e. $|B| = \binom{8}{2}$. And $|C|$ is the number of solutions of $a+b+c=7$, i.e. $|C| = \binom{9}{2}$.
Also $|A\cap B|$ is just the number of solutions where $a\ge 5$, and $b\ge 7$, so we just look at the number of solutions of $a+b+c=1$, i.e. $|A \cap B| = \binom{3}{2}$. Through similar reasoning, we can find the others. We arrive at
$$|A^c \cap B^c \cap C^c| = \binom{15}{2} - \binom{10}{2} - \binom{8}{2} - \binom{9}{2} + \binom{3}{2} + \binom{2}{2} + \binom{4}{2} - 0,$$
which turns out to be a measly $6$. So small it makes us mad!
$$\begin{align*}
3+4+5=12 \\
4+3+5=12 \\
4+4+4=12 \\
5+2+5=12 \\
5+3+4=12 \\
5+4+3=12 \\
\end{align*}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1006780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
What's wrong with the calculation with polar coordinates here? Suppose $x=r\cos t$ and $y=r\sin t$. I did the following calculation:
$$
\begin{align}
&x^4+y^4=(r\cos t)^4+(r\sin t)^4\\
=&r^4(\sin^4t+\cos^4t)\\
=&r^4[(\sin^2t+\cos^2t)^2-2\sin^2t\cos^2t]\\
=&r^4(1-2\sin^2t\cos^2t)=r^4-2r^4\sin^2t\cos^2t
\end{align}
$$
On the other hand
$$
\begin{align}
x^4+y^4&=(x^2+y^2)^2-2x^2y^2\\
&=r^4-2r^2\sin^2t\cos^2t
\end{align}
$$
What is wrong with the calculation?
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Notice that $2x^2y^2 = 2(r\sin t)^2(r\cos t)^2 = 2r^4\sin^2t\cos^2t$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1007903",
"timestamp": "2023-03-29T00:00:00",
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|
Find $a$, $b$ and $c$ in $\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3$
Find the values of the positive constants $a$, $b$ and $c$ given that when $x$ is sufficiently small for terms in $x^4$, and higher powers of $x$, to be neglected then:
$$
\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3 \space\space\text{(assume $|bx| < 2$)}
$$
Expanding $e^{ax}$ first, we get:
$$
1+ax+\frac{a^2x^2}{2}+\frac{a^3x^3}{6}
$$
and expanding $(2+bx)^{-1}$ we get:
$$
\frac{1}{2}-\frac{bx}{4}+\frac{b^2x^2}{8}-\frac{b^3x^3}{16}
$$
I tried dividing the expressions and comparing the coefficients, but didn't really get anywhere. Is there an easier way to do this?
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Hint: Don't divide, Multiply!
$$\frac{e^{ax}}{2+bx}=e^{ax}(2+bx)^{-1}=\left(1+ax+\frac{a^2x^2}{2}+\frac{a^3x^3}{6}+\cdots\right)\left(\frac{1}{2}-\frac{bx}{4}+\frac{b^2x^2}{8}-\frac{b^3x^3}{16}\cdots\right)$$
Can you take it from here?
|
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"url": "https://math.stackexchange.com/questions/1013253",
"timestamp": "2023-03-29T00:00:00",
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|
Related Rates - Growth of the side of a triangle recently I saw a question on a book but I personally don't agree with the answer. I would appreciate if you could help me achieve the final result. It follows:
Two sides of a triangle measure $12m$ and $15m$.
The angle between the two sides grow at the rate of $2^o/min$. How fast is the third side growing when the angles between the sides that have constant size is $60^o$ ?
Thank you,
Best Regards,
Bruno
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Let $x=x(t)$ be the third side, and $\theta$ the angle between the first two sides. Using the Law of Cosines,
when $\theta = 60^\circ$
$$x^2 = 15^2+12^2- 2 \cdot 15 \cdot 12 \cdot \cos(60^\circ)$$
$$x = 3 \cdot \sqrt{21}$$
For the Related Rate,
$$[x^2]' = 0-2 \cdot 15 \cdot 12 [ \cos \theta]'$$
$$2x \cdot x' = 2 \cdot 15 \cdot 12 \cdot \sin(\theta) \cdot [\theta]'$$
Since $[\theta]' = 2 \cdot \pi /180$
$$2 (3 \cdot \sqrt{21}) \cdot x' = 2 \cdot 15 \cdot 12 \cdot (\sqrt{3}/2) \cdot (2 \cdot \pi /180) $$
$\therefore x' = \pi \cdot \frac{\sqrt{7}}{21}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1014051",
"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.