Taylor's Theorem: g(x) = sin(x) + cos(.5x) on (-4, 4)

[mooshupork34]

g(x) = sin(x) + cos(.5x) on (-4, 4)

Okay, so for the above expression, I have to:
a) Calculate an expression for the 5th derivative g^(5)(x).
b) Find a bound for the absolute value of g^(5)(x) on the interval indicated.
c) Use Taylor's Theorem to find a cap for the error; that is, a cap for the difference between the function and the Maclaurin polynomial of degree four on the interval.

I did part a and this is what I got:

g'(x) = cos(x)-.5sin(.5x)
g''(x) = -.25cos(.5x) - sin(x)
g'''(x) = .125sin(.5x) - cos(x)
g^(4)(x) = .0625cos(.5x) + sin(x)
g^(5)(x) = cos(x) - .03125sin(.5x)

I attempted part b and basically what I did was substitute 4 for x in the expression for the 5th derivative of g(x). From this I got approximately .68, which I rounde up to .7. Then for a final answer, I got:

the absolute value of cos(x) - .03125sin(.5x) <= .7

I'm not sure if the above is right, but that's what I got.

I think I know how to do part c, but in order to do it, I need to know how to do part b. If anyone could help, I would REALLY appreciate it!

 

[Count Iblis]

You can easily see that your upper bound is not correct. Just substitute x = 0 where g^(5) is 1. If you make x a litlle negative, cos almost stays the same, while the contribution form the sin term becomes positive...

To find the maximum you can add up the sin and the cos and express it in the form A cos(3/2 x + phi). If you don't know how to do it, don't look it up, but derive it from first principles. The maximum will be in the interval (-4,4), so the maximum value is A, which is sqrt(1 + 2^(-10)) = about 1 + 2^(-11).

Divide this by factorial 5 to obtain the error bound.