1. ## Question about the Taylor series

When approximating trigonometric expressions using Taylor series (like cos 32), should I round up pi to 4 once I acquire R (remainder)? I know you round up sin c and cos c to one, the maximum. My book sometimes change the pi to 4, sometimes not. So I'm kinda confused. Is there an official rule about this?

I hope I was clear.

2. That won't give you an approximation but it can give you an upper bound.

3. What? Aren't they the same?

4. Originally Posted by Kaitosan
What? Aren't they the same?
No, they are not! I suggest you look up the definition of "upper bound". Perhaps "approximation" also.

5. Maybe an example will clarify this up?

"Use a Taylor polynomial to find the S3 approximation of cos 35, and estimate the error in this approximation by using Lagrange's form of the remainder."

After going through the process of setting up the polynomials and numbers, we get.....

cos 35 = sqrt(3)/2 - pi/72 - [sqrt(3)pi^2]/5184 + R

Lagrange's remainder =

R = [sin c (x - pi/6) ]/3!

for some number c between "x" and "a" that maximizes R

Ok, allright?

I actually have two questions. I'd love for someone to clear them up.

1. How is it possible that c=pi/2 if, according to the definition of my book, c is some number between "x" and "a" that maximizes R? By the way, x = pi/6 + pi/36 and a = pi/6.

2. My book sometimes replace the pi sign in the lagrange remainder with 4 as a part of the process of finding the number that is "in error" of cos 35. I find this strange because it'll just enlarge the "potential error."

The thing is, my book does a terrible job explaining approximation topics so I'd really appreciate any help here.

(Don't make me bump this topic lol)

6. Originally Posted by Kaitosan
Maybe an example will clarify this up?

"Use a Taylor polynomial to find the S3 approximation of cos 35, and estimate the error in this approximation by using Lagrange's form of the remainder."

After going through the process of setting up the polynomials and numbers, we get.....

cos 35 = sqrt(3)/2 - pi/72 - [sqrt(3)pi^2]/5184 + R

Lagrange's remainder =

R = [sin c (x - pi/6) ]/3!

for some number c between "x" and "a" that maximizes R

Ok, allright?

I actually have two questions. I'd love for someone to clear them up.

1. How is it possible that c=pi/2 if, according to the definition of my book, c is some number between "x" and "a" that maximizes R? By the way, x = pi/6 + pi/36 and a = pi/6.
Notice that they are not saying this is an approximation. They are looking for an upper bound on the aproximation. They are looking for a number they can be sure is larger than the error. Sin(pi/2)= 1 is the largest possible value for sin(c).

2. My book sometimes replace the pi sign in the lagrange remainder with 4 as a part of the process of finding the number that is "in error" of cos 35. I find this strange because it'll just enlarge the "potential error."
No, it doesn't change the potential error at all. It gives a number that they know is larger than the error so they can say "the potential error is less than this".

The thing is, my book does a terrible job explaining approximation topics so I'd really appreciate any help here.

(Don't make me bump this topic lol)[/QUOTE]

7. Originally Posted by HallsofIvy
Sin(pi/2)= 1 is the largest possible value for sin(c).
That's the problem. It's impossible for c to equal pi/2 because c is supposed to be a number that causes a maximum value, a number that LIES BETWEEN X AND A..... or am I missing something?