Question

I think the people who wrote this question made a mistake

Answer 1

In part c, they tell you to show that the error is less than 0.02 on [-0.5, 0.5], but I keep getting an error larger than that... can someone try this problem and see if they get an error less than .02?

Answer 2

P.S. this question is a lot easier if you use this and plug in -3x^2

Answer 3

a.
Refer to the Mathematica attachment.

Answer 4

i get an error less than .02

http://www.wolframalpha.com/input/?i=plot+%28e%5E%28-3x%5E2%29+-+%281-3x%5E2%2B4.5x%5E4-4.5x%5E6%29%2C+.02%29+for+x%3D-.5+to+.5

Answer 5

Yes, as you suggested I would get series for h(x)=e^x and expand the series for h(-3x^2).
If you found an error greater than 0.02, you probably have
1. an incorrect expansion of f(x), or
2. some calculation errors.
You could show your work so we can help you locate the error(s).

To show the maximum error, one of the ways to demonstrate is:
1. f(x) and g(x) are both even, so d(x)=|f(x)-g(x))| is also even. You only need to show the interval \(x\in [0,0.5]\)
2. Within that interval, show that d'(x)>0, which means d(x) increases monotonically.
3. Hence max(d(x))=d(0.5) within that interval.

Answer 6

f(x) = e^(-3x^2)
g(x) = 1 - (6/2!)x^2 + (108/4!)x^4 - (648/6!)x^6
abs( f(0.5) - g(0.5) ) = 0.04482
0.04482 is greater than 0.02

Answer 7

Wait... sorry everyone just realized that the (648/6!) is wrong in g(x)
Thanks for all the helP!

Answer 8

Too bad I only see your post 12 hours after.
Indeed, it is tricky, because if we expanded e^y, we get the first four terms as
\(\large g(x)=1+y+\frac{y^2}{2!}+\frac{y^3}{3!}\)
Substituting y=-3x^2 gives
\(\large 1-3x^2+\frac{(-3x^2)^2}{2!}+\frac{(-3x^2)^3}{3!}\)
\(\large =1-3x^2+\frac{9x^4}{2}-\frac{9x^6}{2}\)
You probably took the n! to match \(x^n\) for the last term.
Glad that the expansion worked out.