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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)(xa) + 0.5f''(a)(xa)^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, We have: 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 multivariable 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 , 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). 

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