# Thread: Problem regarding Taylor's Theorem

1. ## Problem regarding Taylor's Theorem

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.

2. Originally Posted by clockingly
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) = -(1/2) sin(x) sin(x/2) + cos(x) cos(x/2)

.......=3 cos(3x/2)/4 + cos(x/2)/4

RonL

3. Originally Posted by CaptainBlack
g'(x) = -(1/2) sin(x) sin(x/2) + cos(x) cos(x/2)

.......=3 cos(3x/2)/4 + cos(x/2)/4

RonL

Sorry, I forgot to add the plus sign. The expression was actually
g(x) = sin(x) + cos(.5x) on (-4, 4)

4. Originally Posted by clockingly
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.
Since |0.03125 sin(x/2)| <= 0.03125, and |cos(x)|<= 1 on (-4,4) a rough
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)|.

RonL

5. Originally Posted by CaptainBlack
Since |0.03125 sin(x/2)| <= 0.03125, and |cos(x)|<= 1 on (-4,4) a rough
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)|.

RonL

Thanks!

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.