My Math Forum Quadratic Taylor series approximation

 Calculus Calculus Math Forum

 April 28th, 2011, 08:26 AM #1 Newbie   Joined: Apr 2011 Posts: 9 Thanks: 0 Quadratic Taylor series approximation I'm not quite sure how to go about a quadratic Taylor series approximation. The question asks: write out the quadratic approximation where x = (L,K)' at the point (1,1)': Q(x) = (L^-0.5 + K^-0.5)^-2 I know the formula is: f(x) = f(a) + f'(a)(x-a) + 0.5f''(a)(x-a)^2 but I just don't know where to start. All I need is a starting block and I'm sure I can work it out from there. Thanks in advance.
 April 28th, 2011, 06:10 PM #2 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Quadratic Taylor series approximation You can use the fact that L=K=1 to set them to a single variable, x. You'll have Q(x)=(2x^(-.5))^(-2). Does this help?
 April 29th, 2011, 08:55 AM #3 Newbie   Joined: Apr 2011 Posts: 9 Thanks: 0 Re: Quadratic Taylor series approximation Well when I use Q(x)=(2x^(-.5))^(-2) and try to compute it comes out looking very complex. But I'm not even sure if I'm doing it right. I was never shown an example of taylor series, all I have is notes on it but I have no idea how to actually work it all out. And as I said, whenever I try to do what I think is supposed to be a taylor series approximation it all comes out looking very complex
 April 29th, 2011, 09:00 AM #4 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Quadratic Taylor series approximation Are you allowed to use a tool to solve this, or must you only solve by hand? Using a Computer Algebra System to do this would make it much easier.
 April 29th, 2011, 09:01 AM #5 Newbie   Joined: Apr 2011 Posts: 9 Thanks: 0 Re: Quadratic Taylor series approximation No, it has got to be by hand. Examination style
 April 29th, 2011, 09:26 AM #6 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Quadratic Taylor series approximation You have Q(x)=(2x^(-.5))^(-2) when x=1, but we'll substitute it in at the end. However, with some simplification that I somehow missed before, $Q(x)=\frac{1}{4}(x^{(-.5\cdot 2)})=\frac{x}{4}$ We have: $Q'(x)=1/4\text{ and} Q''(x)=0 \text{therefore,} Q(x)\approx Q(0)+Q'(0)x+\frac{Q''(0)x^2}{2} Q(x)\approx 0+\frac{x}{4}+0 Q(1)\approx 1/4$ I started typing out a few very long derivatives until I realized how silly I was.
 April 29th, 2011, 09:35 AM #7 Newbie   Joined: Apr 2011 Posts: 9 Thanks: 0 Re: Quadratic Taylor series approximation I've been playing about a bit and also got 1/4 but I obtained it a different way from you though, but I will go back over it. But what if x and a were vectors though? The formula for that is: http://en.wikipedia.org/wiki/Taylor_series#Definition - don't know how to type it with the code :P I'm going to play about a bit more with it now but again it seems very complex
 April 29th, 2011, 09:54 AM #8 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Quadratic Taylor series approximation In that case you'll need to learn multi-variable calculus, and you'd have to use the Taylor series for multiple variables. It gets much harder to use. However, if you have something like $\rightharpoonup_{x}=<1,2=>=$, you can set up Q to be of one variable if you set L=x and K=x+1. At the end you'd set x=1 after you took the quadratic approximation of Q(x).

,

,

,

,

### quadratic approximation formula taylor series

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post RGNIT Calculus 0 March 20th, 2014 11:21 PM g0bearmon Real Analysis 2 May 22nd, 2012 12:10 PM soulnoob Real Analysis 1 May 19th, 2010 05:20 AM cknapp Calculus 2 February 1st, 2008 05:39 AM ^e^ Real Analysis 3 March 8th, 2007 09:05 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top