MATH1251 Questions HELP (1 Viewer)

1008 · Sep 19, 2016

For this question:

$\\\text{Let }f(x)=\sqrt{1+x}\text{ and suppose that }a_1=3\text{ and }a_{n+1}=f(a_n),\forall n\geq 1\\\text{(a) Prove that }a_n^2-a_n-1>0\forall n\geq 1$

What would be the best approach to do this?

InteGrand · Sep 20, 2016

1008 said:
For this question:

$\\\text{Let }f(x)=\sqrt{1+x}\text{ and suppose that }a_1=3\text{ and }a_{n+1}=f(a_n),\forall n\geq 1\\\text{(a) Prove that }a_n^2-a_n-1>0\forall n\geq 1$

What would be the best approach to do this?

You could use induction. Have you tried this? (The statement given is really just equivalent to saying that the sequence is strictly decreasing.)

leehuan · Sep 20, 2016

$\text{For the series }\sum_{k=2}^{\infty}\frac{(3x-2)^k}{k\ln k}\\ \text{I was able to deduce a radius of convergence of }\frac{1}{3}$

$\text{Hence the interval of convergence is at least }x\in \left(\frac{1}{3},1\right)$

Apparently it is also convergent when x equals to 1/3, or says the answers. But I'm confused as to why this is ok because

$\text{Is this even convergent... }\sum_{k=2}^\infty \frac{(-1)^k}{k \ln k}$

Paradoxica · Sep 20, 2016

leehuan said:
$\text{For the series }\sum_{k=2}^{\infty}\frac{(3x-2)^k}{k\ln k}\\ \text{I was able to deduce a radius of convergence of }\frac{1}{3}$

$\text{Hence the interval of convergence is at least }x\in \left(\frac{1}{3},1\right)$

Apparently it is also convergent when x equals to 1/3, or says the answers. But I'm confused as to why this is ok because

$\text{Is this even convergent... }\sum_{k=2}^\infty \frac{(-1)^k}{k \ln k}$

Conditionally convergent. But yes, convergent.

InteGrand · Sep 20, 2016

leehuan said:
$\text{For the series }\sum_{k=2}^{\infty}\frac{(3x-2)^k}{k\ln k}\\ \text{I was able to deduce a radius of convergence of }\frac{1}{3}$

$\text{Hence the interval of convergence is at least }x\in \left(\frac{1}{3},1\right)$

Apparently it is also convergent when x equals to 1/3, or says the answers. But I'm confused as to why this is ok because

$\text{Is this even convergent... }\sum_{k=2}^\infty \frac{(-1)^k}{k \ln k}$

Yes, convergent by the alternating series test (https://en.wikipedia.org/wiki/Alternating_series_test ).

leehuan · Sep 20, 2016

Oh wait conditional convergence will suffice? Ok thanks, didn't know that
__________________

Best approach to proving this is convergent?

$\sum_{k=1}^\infty \frac{(\ln k)^3}{k^2}$

InteGrand · Sep 20, 2016

leehuan said:
Oh wait conditional convergence will suffice? Ok thanks, didn't know that
__________________

Best approach to proving this is convergent?

$\sum_{k=1}^\infty \frac{(\ln k)^3}{k^2}$

Lemma. For all sufficiently large k, ln k < k^{1/6}. (In fact: for any given positive alpha and beta, (ln k)^{beta} will be less than k^{alpha} for all sufficiently large k. And not just less than, but 'little oh' of as k -> oo, in fact.)

Proof. Exercise. (E.g. L'Hôpital's rule to prove the little oh statement, which implies the "power of ln k < power of k" part.)

Now, we can say that for sufficiently large k, the summand (which is positive) is less than 1/(k^{1.5}) (which converges by p-test), so the given series converges by the comparison test.

leehuan · Sep 20, 2016

InteGrand said:
Lemma. For all sufficiently large k, ln k < sqrt(k).

Proof. Exercise.

Now, we can say that for sufficiently large k, the summand (which is positive) is less than 1/(k^{1.5}) (which converges by p-test), so the given series converges by the comparison test.

Aha guess that'll do. That was basically what I was thinking.
______________________

Also a quickie

$\text{For a sum like this: }\sum_{k=1}^\infty \frac{(-1)^k(x-3)^k}{(k+1)3^k}$

So if I want to use the formula for radius of convergence

$R=\lim_{k\to \infty}\frac{a_k}{a_{k+1}}$

Should I just ignore the (-1)^k or permit R to be negative?

Paradoxica · Sep 20, 2016

leehuan said:
Aha guess that'll do. That was basically what I was thinking.
______________________

Also a quickie

$\text{For a sum like this: }\sum_{k=1}^\infty \frac{(-1)^k(x-3)^k}{(k+1)3^k}$

So if I want to use the formula for radius of convergence

$R=\lim_{k\to \infty}\frac{a_k}{a_{k+1}}$

Should I just ignore the (-1)^k or permit R to be negative?

Negative radius is fine, didn't you deal with polar forms some time ago?

InteGrand · Sep 20, 2016

leehuan said:
Aha guess that'll do. That was basically what I was thinking.
______________________

Also a quickie

$\text{For a sum like this: }\sum_{k=1}^\infty \frac{(-1)^k(x-3)^k}{(k+1)3^k}$

So if I want to use the formula for radius of convergence

$R=\lim_{k\to \infty}\frac{a_k}{a_{k+1}}$

Should I just ignore the (-1)^k or permit R to be negative?

The formula is actually with absolute values around the fraction. So the (-1)^k goes away. The radius of convergence can't be negative, it is by definition either 0, a positive real number, or +oo. (Further reading: https://en.wikipedia.org/wiki/Radius_of_convergence )

Paradoxica · Sep 20, 2016

InteGrand said:
The formula is actually with absolute values around the fraction. So the (-1)^k goes away. The radius of convergence can't be negative, it is by definition either 0, a positive real number, or +∞.

ftfy

leehuan · Sep 29, 2016

InteGrand · Sep 29, 2016

leehuan said:
$\text{Let }f(z)=\int_0^ze^{-\frac{t^2}{2}}dt\text{ and }f(z)=P_n(z)+R_n(z)\text{, where }P_n(z)\text{ is the Maclaurin polynomial of degree }2n+1\text{ for }f$

$\text{Show that }P_n(z)=\sum_{k=0}^n \frac{(-1)^kz^{2k+1}}{k!2^k(2k+1)}\\ \text{and }R_(z)=\frac{(-1)^{n+1}}{(n+1)!2^{n+1}}\int_0^ze^ce^{2n+2}dt$

$\text{where, for each }t,c\text{ lies between }0\text{ and }\frac{-t^2}{2}$

$\noindent The Taylor series for $e^{-\frac{t^2}{2}}$ is $\sum _{k = 0}^{\infty} \frac{\left(-\frac{t^2}{2}\right)^k}{k!}$ (convergent for all $t\in \mathbb{R}$). Just integrate this term by term from $0$ to $z$ ($z\in \mathbb{R}$) to get the desired Taylor series. Truncate to $k=n$ to get the $2n+1$ degree Taylor polynomial.$

leehuan · Sep 29, 2016

InteGrand said:
$\noindent The Taylor series for $e^{-\frac{t^2}{2}}$ is $\sum _{k = 0}^{\infty} \frac{\left(-\frac{t^2}{2}\right)^k}{k!}$ (convergent for all $t\in \mathbb{R}$). Just integrate this term by term from $0$ to $z$ ($z\in \mathbb{R}$) to get the desired Taylor series. Truncate to $k=n$ to get the $2n+1$ degree Taylor polynomial.$

Yeah I figured that'd work for the Taylor polynomial. Sorry, my confusion was mostly in how to get the remainder

1008 · Sep 29, 2016

InteGrand said:
You could use induction. Have you tried this? (The statement given is really just equivalent to saying that the sequence is strictly decreasing.)

Sorry for the late reply, figured that one out, thanks

1008 · Oct 6, 2016

Got more questions...pls help:

$\\\text{(1) Solve }2y'y'' = 1 + (y')^2 \text{ by letting }y' = v\\\text{(2) Use the substitution }v= \frac{\mathrm{d}y }{\mathrm{d} t} \text{ to solve } \frac{\mathrm{d}^2y }{\mathrm{d} t^2}+\omega ^2x=0\text{. Here obtain a differential equation in }v\text{ and }x\\\text{(3) For a twice differentiable function }u,\text{ let }Lu=a \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b \frac{\mathrm{d} u}{\mathrm{d} x} +cu,\text{ where }a,b\text{ and }c\text{ are real numbers.}\\\text{Let }u_1,u_2\text{ be a basis for the solution space of }L(u) = 0.\text{ Find a basis for the solution space of the fourth }\\\text{order equation }L(L(u))=0.\text{ What can you say about the kernels of }L \text{ and }L^2?$

1008 · Oct 6, 2016

These as well, please, if anyone could solve these that would be great. I'll be posting more questions up as exams are roughly in a month...

$\\\text{(1) Two examples involving time explicitly: }\\\text{(a) Seasonal variation: Solve }\\ \frac{\mathrm{d} y}{\mathrm{d} t}=k(1+\text{a} \cos(2\pi t))y.\\\text{(b) It is estimated that the population growth rate of a certain developing country will fall linearly from }\\\text{2\% per year to 1.5\% per year over the next decade. The present population is 10 million. What is the}\\\text{ estimated population in 10 years time? }$

$\\\text{(2) The differential equation } \frac{\mathrm{d} y}{\mathrm{d} x} =\lambda \frac{y}{x}\text{ is employed as a simple model for comparative growth.}\\\text{(a) Solve.}\\\text{(b) For the fiddler crab, the weights of the large claw, y grams, }\\\text{and the body (without claw), x grams, were found to be }$

$\\\text{Show that the above model appears to be reasonable in this example. Determine approximately }\lambda \text{ and the}\\\text{constant of integration.}$

InteGrand · Oct 7, 2016

1008 said:
Got more questions...pls help:

$\\\text{(1) Solve }2y'y'' = 1 + (y')^2 \text{ by letting }y' = v\\\text{(2) Use the substitution }v= \frac{\mathrm{d}y }{\mathrm{d} t} \text{ to solve } \frac{\mathrm{d}^2y }{\mathrm{d} t^2}+\omega ^2x=0\text{. Here obtain a differential equation in }v\text{ and }x\\\text{(3) For a twice differentiable function }u,\text{ let }Lu=a \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b \frac{\mathrm{d} u}{\mathrm{d} x} +cu,\text{ where }a,b\text{ and }c\text{ are real numbers.}\\\text{Let }u_1,u_2\text{ be a basis for the solution space of }L(u) = 0.\text{ Find a basis for the solution space of the fourth }\\\text{order equation }L(L(u))=0.\text{ What can you say about the kernels of }L \text{ and }L^2?$

$\noindent (1) Let $v = y'$. Then the ODE is $2v v' = 1+v^2$. So $v' = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{1+v^2}{2v}$. Thus $\ln \left(1+v^2\right) = t + c$. So $v^2 = Ae^t -1$ for some positive constant $A \equiv e^c$. So $v = y' = \pm \sqrt{A}\sqrt{e^t -B^2}$, where $B= \sqrt{A^{-1}}$. Now try integrating this to get $y(t)$. E.g. let $e^t = B\sec \theta$, so $e^t \, \mathrm{d}t = B \sec \theta \tan \theta \, \mathrm{d}\theta$, and note $ y = \pm \int \frac{\sqrt{A}\sqrt{e^t - B^2}e^t \, \mathrm{d}t}{e^t}$.$

InteGrand · Oct 7, 2016

1008 said:
Got more questions...pls help:

$\\\text{(1) Solve }2y'y'' = 1 + (y')^2 \text{ by letting }y' = v\\\text{(2) Use the substitution }v= \frac{\mathrm{d}y }{\mathrm{d} t} \text{ to solve } \frac{\mathrm{d}^2y }{\mathrm{d} t^2}+\omega ^2x=0\text{. Here obtain a differential equation in }v\text{ and }x\\\text{(3) For a twice differentiable function }u,\text{ let }Lu=a \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b \frac{\mathrm{d} u}{\mathrm{d} x} +cu,\text{ where }a,b\text{ and }c\text{ are real numbers.}\\\text{Let }u_1,u_2\text{ be a basis for the solution space of }L(u) = 0.\text{ Find a basis for the solution space of the fourth }\\\text{order equation }L(L(u))=0.\text{ What can you say about the kernels of }L \text{ and }L^2?$

$\noindent (2) By inspection, the solution is $c_1 \cos \omega t + c_2 \sin \omega t$, if $\omega >0$. (Standard SHM solution.)$

$$\noindent To derive this via the proposed substitution (though it's faster to just use the standard method of solving second-order ODE's with constant coefficients using the characteristic equation etc.; this substitution is basically how a HSC proof of the SHM result would go.), recall that if $v = \dot{y}$, then $\ddot{y} = \frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{1}{2}v^2\right)$ (standard HSC 3U result from motion topic). (Using dots to denote time derivatives.) Then the ODE becomes $\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{1}{2}v^2\right) = -\omega ^2 y$. So $\frac{1}{2} v^2 = -\omega ^2 \cdot \frac{1}{2}y^2 + \frac{1}{2}C$, for some constant $C$. So $v^2 = C -\omega^2 y^2$. Let's just assume $\omega >0$. Note $C$ must be positive in order for the last equation to hold (except in the trivial solution case where $y$ is identically $0$), so we can just write that arbitrary positive constant as $\omega^2 a^2$ for some $a>0$.$

$\noindent Now, $v^2 = C -\omega^2 y$. So $v = \pm \sqrt{a^2 \omega^2 -\omega^2 y^2} = \pm \omega \sqrt{a^2 - y^2}$. Let's look at the positive root (you can show as an exercise that we get the same solution if we also look at the negative root). Writing $v = \frac{\mathrm{d}y}{\mathrm{d}t}$, this DE becomes $\frac{\mathrm{d}y}{\mathrm{d}t} = \omega \sqrt{a^2 -y^2}$. This is now a separable DE, which you should be able to solve.$

InteGrand · Oct 7, 2016

1008 said:
Got more questions...pls help:

$\\\text{(1) Solve }2y'y'' = 1 + (y')^2 \text{ by letting }y' = v\\\text{(2) Use the substitution }v= \frac{\mathrm{d}y }{\mathrm{d} t} \text{ to solve } \frac{\mathrm{d}^2y }{\mathrm{d} t^2}+\omega ^2x=0\text{. Here obtain a differential equation in }v\text{ and }x\\\text{(3) For a twice differentiable function }u,\text{ let }Lu=a \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b \frac{\mathrm{d} u}{\mathrm{d} x} +cu,\text{ where }a,b\text{ and }c\text{ are real numbers.}\\\text{Let }u_1,u_2\text{ be a basis for the solution space of }L(u) = 0.\text{ Find a basis for the solution space of the fourth }\\\text{order equation }L(L(u))=0.\text{ What can you say about the kernels of }L \text{ and }L^2?$

$\noindent (3) Letting $D$ be the differentiation operator, we formally have $L = aD^2 + bD + c$, and $L^2 = \left(aD^2 + bD + c\right)^2$ (remember these operators work just like polynomials; you could expand this out to find $L^2$ in terms of powers of $D$ up to $D^4$). So the characteristic polynomial for the $L^2 u = 0$ equation (call this $p_2 (z)$ say) is the square of that of the $Lu = 0$ equation (say $p_1 (z)$), Hence the characteristic roots of $p_2 (z)$ are just those of $p_1 (z)$ with double the multiplicity. So $u_1$ and $u_2$ are still part of the basis for the solution space of $L^2 u = 0$, coming from the shared roots with $p_1 (z)$. Due to the two additional same roots with increased multiplicity, we also have $x u_1$ and $xu_2$ as part of the basis. So a basis for the kernel of $L^2$ is $u_1, u_2, xu_2, xu_2$. We can say that the kernel of $L$ is a subspace of the kernel of $L^2$.$

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