# Final Exams: Taylor Polynomail Help!!!

• May 20th 2005, 06:18 AM
zee
Final Exams: Taylor Polynomail Help!!!
Hello!! Boy am I glad that there is such a site!!
I have my final exams approaching in 2 days and im studing
Distance-Ed... so im in dire need of some help!
This is my question!

Determine the first three nonzero terms in the Taylor polynomial
approximations for the given initial value problem:
q) x" + t(x) = 0; x(0)=1, x'(0)=0

I have the final answer, but I want to see the full solution.
The answer is: 1 - (1/6)t^3 + (1/180)t^6 + ...

zee :(
• May 20th 2005, 07:28 AM
paultwang
2. What is the Taylor approx (the formula)?
• May 20th 2005, 07:55 AM
zee
desperate!
Determine the first three nonzero terms in the Taylor polynomial
approximations for the given initial value problem:
x" + t(x) = 0; x(0)=1, x'(0)=0

The question is typed accurately...
I dont have the formula as I have lost my formula sheet...
zee
• May 20th 2005, 10:29 AM
paultwang
Is x a function? If yes, what does x depend on?
Is t a function? If yes, what does t depend on?
• May 20th 2005, 11:06 PM
zee
Taylor Polynomial
well x is definitely a function, but to what?
The question is very unclear, thus im quite confused...
t is not a function.. well atleast it doesnt look like it!
:confused:
• May 21st 2005, 05:58 AM
paultwang
My guess is that your ODE looks like this:

$\displaystyle \frac{d^2x}{dt^2} + t x(t) = 0$
$\displaystyle x(0) = 1$
$\displaystyle \frac{dx}{dt}(0) = 0$

Taylor approximation is:
The nth term of Taylor expansion about t=a is $\displaystyle \frac{1}{(n-1)!}\frac{d^{(n-1)}x}{dt^{(n-1)}}(a) (t-a)^{(n-1)}$
• May 21st 2005, 06:25 AM
paultwang

The first term is x(0), which is 1.

The second term is 0 because x'(0) = 0.

The third term is $\displaystyle \frac{x''(a)}{2!} (t-a)^2$ which is also 0 because $\displaystyle \frac{d^2x}{dt^2} = -t x(t)$ (just plug t=0 in).

The 4th term is $\displaystyle \frac{x'''(a)}{3!} (t-a)^3$ which contains the derivative of the third term:
$\displaystyle \frac{d^3x}{dt^3} = -t x'(t) -x(t)$ (Hint: Use chain rule) We already know x'(0) = 0 and x(0) = 1 so the 4th term is $\displaystyle -\frac{1}{6}t^3$

The 5th and 6th terms are both 0.
$\displaystyle \frac{d^4x}{dt^4} = -2x'(t)-t x''(t)$ (We established that x'(0) = 0, x''(0) = 0.)
$\displaystyle \frac{d^5x}{dt^5} = -3x''(t)-t x'''(t)$ (x''(0) = 0, 0*(stuff) = 0)

The 7th term is not zero.

$\displaystyle \frac{d^6x}{dt^6} = -4x'''(t)-t x''''(t)$
Plug t=0, you get $\displaystyle -4x'''(0)$
Using information from the 4th term:
$\displaystyle \frac{d^3x}{dt^3} = -t x'(t) -x(t)$
$\displaystyle \frac{d^3x}{dt^3}(0) = -x(0) = -1$
$\displaystyle -4x'''(0) = 4$

The 7th term is $\displaystyle \frac{4}{6!}t^6 = \frac{1}{180}t^6$

That's all.

In case you are wondering, the exact solution is:

$\displaystyle x(t) = \frac{1}{2} \Gamma(\frac{2}{3}) 3^{(1/6)} AiryBi(-t)+\frac{1}{2} \Gamma(\frac{2}{3}) 3^{(2/3)} AiryAi(-t)$
• May 26th 2005, 01:08 AM
zee
thankyoU!
You were right on the mark!
thanks a bunch!!
zee