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)
The last one is the 5th derivative.
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.
My real problem lies in doing part c. I guess I'm confused as to the method because my book isn't very clear.
bound is |g^(5)(x)|<=1.03125, and the least upper bound on |g^(5)(x)|
must be greater than or equal to 1-0.03125 = 0.96875, so 1.03125 is
quite a good estimate of an upper bound for |g^(5)(x)|.
I don't get why we're using just 0.03125 sin(x/2), though, instead of the entire g^5(x), though, which is cos(x)-1/32 sin (1/2 x).
Wouldn't we just substitute 4 for x in this equation and set the equation less than or equal to -.68?
Sorry, I'm still confused about the procedure to finding the bound.