Final Exams: Taylor Polynomail Help!!!

**zee** · May 20th 2005, 06:18 AM

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 + ...

Please HELPP!!!!!!!!!!

zee

**paultwang** · May 20th 2005, 07:28 AM

1. Check your typing
2. What is the Taylor approx (the formula)?

**zee** · May 20th 2005, 07:55 AM

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...
thus im in a lost position... please help somehow..
zee

**paultwang** · May 20th 2005, 10:29 AM

Is x a function? If yes, what does x depend on?
Is t a function? If yes, what does t depend on?

**zee** · May 20th 2005, 11:06 PM

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!

**paultwang** · May 21st 2005, 05:58 AM

My guess is that your ODE looks like this:

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

Taylor approximation is:
The nth term of Taylor expansion about t=a is $\frac{1}{(n-1)!}\frac{d^{(n-1)}x}{dt^{(n-1)}}(a) (t-a)^{(n-1)}$

**paultwang** · May 21st 2005, 06:25 AM

If about t=0, then:

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

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

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

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

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

The 7th term is not zero.

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

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

That's all.

In case you are wondering, the exact solution is:

$x(t) = \frac{1}{2} \Gamma(\frac{2}{3}) 3^{(1/6)} AiryBi(-t)+\frac{1}{2} \Gamma(\frac{2}{3}) 3^{(2/3)} AiryAi(-t)$

**zee** · May 26th 2005, 01:08 AM

You were right on the mark!
thanks a bunch!!
zee

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