My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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.
Paul4763 is offline  
 
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?
The_Fool is offline  
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
Paul4763 is offline  
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.
The_Fool is offline  
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
Paul4763 is offline  
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.
The_Fool is offline  
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
Paul4763 is offline  
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 , 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).
The_Fool is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
approximation, quadratic, series, taylor



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Linear approximation of Non linear system by Taylor series RGNIT Calculus 0 March 20th, 2014 11:21 PM
In need of help disk, series test, taylor, and power series g0bearmon Real Analysis 2 May 22nd, 2012 12:10 PM
taylor approximation soulnoob Real Analysis 1 May 19th, 2010 05:20 AM
Taylor Approximation cknapp Calculus 2 February 1st, 2008 05:39 AM
Taylor error approximation for continuous nth derivative ^e^ Real Analysis 3 March 8th, 2007 09:05 PM





Copyright © 2019 My Math Forum. All rights reserved.