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Exercise [06.02] b
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variounes Joined: 30 Jun 2008, 22:14 Posts: 25	Exercise [06.02] b Prove by induction $y^{(n)} = \Sigma_{k=0}^{2n} a_k x^{-k}$ Probably same proof as Sameed Zahoor (wouldn't bother reading). Last edited by variounes on 05 Jul 2008, 13:28, edited 2 times in total.
04 Jul 2008, 20:02

variounes Joined: 30 Jun 2008, 22:14 Posts: 25	Re: Excercise [06.02] A quicker proof: $y = e^{-1/x} = \Sigma_{k=0}^\infty \frac{(-1)^k}{k!}x^{-k}$ . It is easy to prove this power series $C^\infty$ -smooth.
05 Jul 2008, 13:14

vasco Supporter Joined: 07 Jun 2008, 08:21 Posts: 182	Re: Exercise [06.02] b Quote: it is easy to prove this power series $C^\infty$ -smooth. OK go on then - I challenge you to do it! $e^{-1/x}$ and all its derivatives are not even defined at the origin and anyway the function in question is as defined in section 6.3 of Penrose's book, and h has no power series centred on the origin since all its derivatives are zero at the origin. Also a necessary but not sufficient condition for a power series to exist is that the function be $C^\infty$ -smooth. You can't use an assumed power series to prove the function $C^\infty$ -smooth.
28 Jul 2009, 16:45

variounes Joined: 30 Jun 2008, 22:14 Posts: 25	Re: Exercise [06.02] b So sad, Vasco. I already gave you two third of the solution. Use of induction to prove the power series C-infinity smooth would be the rest of it. Why'd you bother with maths anyway? And of course I meant h(x) for x > 0.
01 Aug 2009, 20:16

vasco Supporter Joined: 07 Jun 2008, 08:21 Posts: 182	Re: Exercise [06.02] b There is no need for you to feel sad. Just show that h(x) can be differentiated as many times as we like at x=0. There is no power series for this function centred on the origin.
01 Aug 2009, 22:07

variounes Joined: 30 Jun 2008, 22:14 Posts: 25	Re: Exercise [06.02] b h(x) = 0 for x<= 0, it says on the tin, you've got read the label with due care, man. Differentiate 0 as much as you like. Need I say more?
02 Aug 2009, 17:36

vasco Supporter Joined: 07 Jun 2008, 08:21 Posts: 182	Re: Exercise [06.02] b Yes h(x)=0 for x<=0. But that doesn't mean that the derivatives at x=0 are zero. You need to show that the limit as x tends to 0 from left hand side of the origin is equal to the limit as x tends to 0 from right hand side of the origin, for all derivatives. The left hand side is easy because the function is zero for x<= 0 as we already agree on, but you need to show that the limit as x tends to 0 from the right hand side is also zero. That is the definition of the derivative. So that means you need to show that: $\begin{array}{l}\lim~h^{(n)}(x)\\{x\rightarrow0+}\end{array}=\begin{array}{l}\lim~e^{-1/x}\cdot P_n(1/x)\\{x\rightarrow0+}\end{array}=0$ for all $n\ge0$ . where is a polynomial in with $P_{_0}(1/x)$ being defined as equal to for all .
02 Aug 2009, 18:17

variounes Joined: 30 Jun 2008, 22:14 Posts: 25	Re: Exercise [06.02] b Thought you might want to go there with the lims. Was that what your challenge about? Anyway, very good!
03 Aug 2009, 18:13

malinka Joined: 20 Sep 2009, 20:52 Posts: 1	Re: Exercise [06.02] b Hi, If there is a solution of this exercise at x = 0 then sorry for doubling posts. Each of the terms can be written as $x^{-a} e^{-1/x}$ . That is first derivative of $e^{-1/x}$ equals $x^{-2} e^{-1/x}$ and after computing more derivatives we obtain sum of terms that looks like the above one. When x goes to zero then $x^{-a}$ goes to $\inf$ and $e^{-1/x}$ goes to . So we have $\inf * 0$ . However we can notice that $e^{1/x} = 1 + \frac{1}{1! x} + \frac{1}{2! x^2} + ... + \frac{1}{a! x^a} + ...$ . Hence $\frac{1}{x^a e^{1/x}} = \frac{1}{x^a + ... + \frac{1}{a!} + \frac{1}{(a+1)!x} + ... }$ . So if x goes to then $\frac{1}{(a+1)! x}$ goes to $\inf$ and therefore the whole expression goes to .
20 Sep 2009, 21:18

vasco Supporter Joined: 07 Jun 2008, 08:21 Posts: 182	Re: Exercise [06.02] b Yes, there is already a solution, see solution exercise [6.02] c. http://www.roadtoreality.info/viewtopic.php?f=19&t=1401
21 Sep 2009, 15:32

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Exercise [06.02] b

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