Remainder Theorem and Error Help Why are these 2 examples different?

[Cloudless]

So we had two examples in class, but I don't understand why they're different. And the professor is away today, which means I won't see him until the entire weekend has passed (a nightmare for students like me who obsess over a problem).

1. For which x is the approximation sin(x) ≈ x - (x^3)/6 correct within 1/100?


The Taylor series for sin(x):

x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...

The answer was, For every x not equal to 0, this is an alternating series. Thus the error in sin(x) ≈ x - (x^3)/3! is no larger than |(x^5)/5!|. So |(x^5)/5!| < 1/100. 2. For which x is the approximation sin(x) = x - x^3/6 correct within 1/1000?

My professor took the 3 order polynomial for the Taylor Series of sin(x) and used the Remainder Theorem:
R₃(x) = (f⁴c)(x-0)⁴/4!

f(x) = sin(x)
f¹(x) = cos(x)
f²(x) = -sin(x)
f³(x) = -cos(x)
f⁴(x) = sin(x)

So |sin(c)| ≤ 1
|R₃(x)| ≤ |x|⁴/4!

Answer: |x|⁴/4! ≤ 1/1000I don't understand why they used Remainder Theorem for the second one. Yes, I know the 1/100 and 1/1000 are different But why is the approach different?

 

[Mark44]

So we had two examples in class, but I don't understand why they're different. And the professor is away today, which means I won't see him until the entire weekend has passed (a nightmare for students like me who obsess over a problem).

1. For which x is the approximation sin(x) ≈ x - (x^3)/6 correct within 1/100?


The Taylor series for sin(x):

x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...

The answer was, For every x not equal to 0, this is an alternating series. Thus the error in sin(x) ≈ x - (x^3)/3! is no larger than |(x^5)/5!|. So |(x^5)/5!| < 1/100.


2. For which x is the approximation sin(x) = x - x^3/6 correct within 1/1000?

My professor took the 3 order polynomial for the Taylor Series of sin(x) and used the Remainder Theorem:
R₃(x) = (f⁴c)(x-0)⁴/4!

f(x) = sin(x)
f¹(x) = cos(x)
f²(x) = -sin(x)
f³(x) = -cos(x)
f⁴(x) = sin(x)

So |sin(c)| ≤ 1
|R₃(x)| ≤ |x|⁴/4!

Answer: |x|⁴/4! ≤ 1/1000


I don't understand why they used Remainder Theorem for the second one. Yes, I know the 1/100 and 1/1000 are different But why is the approach different?

In the first, your professor is using a result that applies only to alternating series, in which the error is less than the absolute value of the first unused term. In the second, he/she is using a more general result that applies to Taylor series, in which the error is estimated by a formula involving the next term in the series.

 

[Cloudless]

In the first, your professor is using a result that applies only to alternating series, in which the error is less than the absolute value of the first unused term. In the second, he/she is using a more general result that applies to Taylor series, in which the error is estimated by a formula involving the next term in the series.

So if the second question was instead the same as the first ( For which x is the approximation sin(x) = x - (x^3)/6 correct within 1/100? ), I should be able to find an answer identical to the first answer?

Steps: Same procedure for the 2nd example above. Instead, |R₃(x)| ≤ 1/100, so

|x|^4/4! ≤ 1/100.

But according to the answer in the first example, it should be |x|^5/5! ≤ 1/100, I thought.

 

[Mark44]

If I'm understanding your question correctly, then no, the two techniques would not give the same answer for an error less than 1/100.

In both techniques we are approximating sin(x) by x - x³/6
The alternating series remainder technique calculates the error from the first unused term, which is fifth degree.
The Taylor remainder theorem calculates the error from the next term, which is fourth degree. For values of x that are reasonably close to zero, x⁵ < x⁴.