Register to reply 
Basic taylor expansion... 
Share this thread: 
#1
Oct2803, 08:53 AM

P: 17

Hi, could some one explain how i could use the taylor series to expand out:
f(x)= 1/sqrt(1x^2) Any help would be appreciated, thanks. 


#2
Oct2803, 09:29 AM

Sci Advisor
P: 875

trig substitution, calculate the series, convert back.



#3
Oct2803, 09:32 AM

P: 17

means nothing to me sorry, could u be a bit more specific?
Thanks. 


#4
Oct2803, 10:53 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568

Basic taylor expansion...
Why in the world would one use a trig substitution to find a Taylor's series??
Of course, you could do this "the oldfashioned way": use the definition of Taylor's series. (By the way, you say "Taylor Series" but don't say about what point. I assume you mean the McLaurin series: about x_{0}= 0.) f(x)= 1/(1x^{2}) so f(0)= 1 f'(x)= ((1 x^{2})^{1})' = (1x^{2})^{2}(2x) = (2x)(1x^{2})^{2} so f'(0)= 0 f"(x)= 2(1x^{2})^{2}2(2x)(1x^{2})^{3} so f"(0)= 2, etc. So far we would have f(0)+ f'(0)(x)+ f"(0)/2(x^{2}= 1+ x^{2} but the derivatives quickly become very difficult to find! I would be inclined to remember the formula for sum of a geometric series: If r< 1, then Σr^{n}= 1/(1r). The right hand side looks a lot like 1(1 x^{2}) if we were to take r= x^{2}! Yes, Σ(x^{2})^{n} Σ x^{2n}= 1/(1x^{2}) Since an infinitely differentiable function only has one power series, that IS the Taylor's series. 


#5
Oct2803, 11:01 AM

Sci Advisor
P: 875

A taylor series is a sum of various terms comprised of factors multiplying nth derivitive terms:
f(x)=f(a)+(xa)f'(a)+ [(xa)^2]/2!+[(xa)^3]/31... your function y=f(x)=1/sqrt(1x^2) is well suited to a trigonometric substitution: x=sin(u) (and dx/du=cos(u)) since [sin(u)]^2+[cos(u)]^2=1, y=1/sqrt(1x^2)=sec(u) to get f'(x)=dy/dx for the taylor series, first calculate dy/du then multiply by dx/du dy/du=sec(u)tan(u) multiplying the left by dx/du and the right by cos(u) (remember, dx/du=cos(u) ) yields, f'(x)=dy/dx=tan(u) repeating (differentiating dy/dx=tan(u) and multiplying by dx/du) to get f''(x) yields , f''(x)=sec(u) you will get alternating terms of sec(u) and tan(u) with different multipliers. When you have expressed this as a series, you can solve tan(u) and sec(u) for x, and substitute them back in. OR, skip the whole trig substitution, and keep differentiating 1/sqrt(1x^2) it is more straightforward, but not as satisfying. Njorl 


#6
Oct2803, 12:08 PM

Sci Advisor
P: 875

I believe if yo do it with the trig substitution, you wind up with a series like this:
1/(1x^2)=SUM[ 1/(1x^2) + other nonzero terms] This indicates that the sum of the "other nonzero terms" converges to zero. It is more complicated, but I like the effect. Also, whenever I'm working with 1/(1x^2) I just feel like I'm using the wrong coordinate system. I realize that there is no "wrong" coordinate system, but it feels that way to me. Njorl 


#7
Oct2903, 04:50 AM

P: 17

Thanks guys.
Sorry, I left a few points out... I need to evaluate the taylor series of that equation to second order about x=0. Then I need to compare f(0.1) with my taylor series approximation. Does that make it simpler? Thanks a lot for your help. 


#8
Oct2903, 04:55 AM

P: 17

Halls of Ivy 
I think you missed out the spuare root in my equation when differentiating. Any chance you could show me the working again with the square root in there? ie the differentiation etc. many thanks! 


Register to reply 
Related Discussions  
Taylor expansion for ln(1+z)  Calculus & Beyond Homework  40  
Taylor Expansion  General Math  2  
Taylor expansion...  Advanced Physics Homework  8  
Taylor expansion  Introductory Physics Homework  7  
Taylor expansion  Introductory Physics Homework  6 