Taylor Polynomial with Remainder Question

In summary, the minimal degree Taylor polynomial you need to calculate sin(1) to 3 decimal places is 6 decimal places. However, an easier way is to start computing the terms of sin(1) one-by-one, and noting that you have an alternating series. What do you know about the "truncation" (remainder) error in an alternating series?
  • #1
JustinLiang
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0

Homework Statement


What is the minimal degree Taylor polynomial about x=0 that you need to calculate sin(1) to 3 decimal places? 6 decimal places?

Homework Equations


R_nx = f^(n+1)(c)(x-a)^(n+1)/(n+1)(factorial)

The Attempt at a Solution


I have attached my attempt. I am stuck on the last step, how do I solve for n? Did I even do it right up until now?
 

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  • #2
JustinLiang said:

Homework Statement


What is the minimal degree Taylor polynomial about x=0 that you need to calculate sin(1) to 3 decimal places? 6 decimal places?

Homework Equations


R_nx = f^(n+1)(c)(x-a)^(n+1)/(n+1)(factorial)

The Attempt at a Solution


I have attached my attempt. I am stuck on the last step, how do I solve for n? Did I even do it right up until now?

A MUCH easier way is to start computing the terms of sin(1) one-by-one, and noting that you have an alternating series. What do you know about the "truncation" (remainder) error in an alternating series?

RGV
 
  • #3
Ray Vickson said:
A MUCH easier way is to start computing the terms of sin(1) one-by-one, and noting that you have an alternating series. What do you know about the "truncation" (remainder) error in an alternating series?

RGV

But don't you need a calculator for that? You would have to calculate sin1 and compare your approximations to see the difference (remainder).
 
  • #4
JustinLiang said:
But don't you need a calculator for that? You would have to calculate sin1 and compare your approximations to see the difference (remainder).

No, you don't need to know the value of sin(1)---remember, sin(1) is the thing that you are trying to compute!

RGV
 
  • #5
Ray Vickson said:
No, you don't need to know the value of sin(1)---remember, sin(1) is the thing that you are trying to compute!

RGV

Wow, I totally misread the question... Ok so now I have

Pn(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - x^11/11! + ...
Pn(1) = 1 - 1/6 + 1/120 - 1/5040 + 1/362880

How do I know which is 3 decimal and 6 decimal places without a calculator?
 
  • #6
JustinLiang said:
Wow, I totally misread the question... Ok so now I have

Pn(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - x^11/11! + ...
Pn(1) = 1 - 1/6 + 1/120 - 1/5040 + 1/362880

How do I know which is 3 decimal and 6 decimal places without a calculator?

Are you not allowed to use a calculator to do simple addition, subtraction and division? if not, then welcome to the world of manual computation from 50 years ago: this CAN be done by hand, but it is unpleasant.

RGV
 
  • #7
Ray Vickson said:
Are you not allowed to use a calculator to do simple addition, subtraction and division? if not, then welcome to the world of manual computation from 50 years ago: this CAN be done by hand, but it is unpleasant.

RGV

Haha, our prof said we don't need a calculator for his course. But it seems like we do for the assignments.

Back to the question... I am still somewhat clueless. First off, when they said 3 decimals places, would that mean <10^-2? It seems like the 7th derivative at 1/5040 would be a plausible answer but the answer key says 6... What do I do :S
 
  • #8
JustinLiang said:
Haha, our prof said we don't need a calculator for his course. But it seems like we do for the assignments.

Back to the question... I am still somewhat clueless. First off, when they said 3 decimals places, would that mean <10^-2? It seems like the 7th derivative at 1/5040 would be a plausible answer but the answer key says 6... What do I do :S

No. Three-decimal places of accuracy require an |error| < 0.5*10^-4 = 1/2000, so stopping at the term -1/5040 will do (but be sure to INCLUDE that term). Six decimals of accuracy need an |error| < 0.5x10^-7 = 1/20,000,000, so you can figure out where you have to stop the series.

RGV
 
  • #9
JustinLiang said:
Haha, our prof said we don't need a calculator for his course. But it seems like we do for the assignments.

Back to the question... I am still somewhat clueless. First off, when they said 3 decimals places, would that mean <10^-2? It seems like the 7th derivative at 1/5040 would be a plausible answer but the answer key says 6... What do I do :S

No. Three-decimal places of accuracy require an |error| < 0.5*10^-4 = 1/20,000, so stopping at the term -1/362,880 will do (but be sure to INCLUDE that term). Six decimals of accuracy need an |error| < 0.5x10^-7 = 1/20,000,000, so you can figure out where you have to stop the series.

RGV
 

What is a Taylor Polynomial with Remainder?

A Taylor Polynomial with Remainder is a mathematical tool used to approximate a function at a specific point using a polynomial function. It takes into account the derivatives of the function at that point to create a more accurate estimate of the original function.

How is a Taylor Polynomial with Remainder calculated?

A Taylor Polynomial with Remainder is calculated using the Taylor Series, which is an infinite sum of derivatives of the function at a given point. The polynomial is then created by taking a finite number of terms from the Taylor Series and adding them together.

What is the purpose of using a Taylor Polynomial with Remainder?

The purpose of using a Taylor Polynomial with Remainder is to approximate a function at a specific point when the exact value cannot be determined. It provides a close estimate of the function's value and can be used in various mathematical and scientific applications.

How accurate is a Taylor Polynomial with Remainder?

The accuracy of a Taylor Polynomial with Remainder depends on the number of terms used in the polynomial. The more terms that are included, the closer the approximation will be to the actual value of the function. However, there will always be some degree of error, especially when using a finite number of terms.

What is the difference between Taylor Polynomial and Taylor Polynomial with Remainder?

The main difference between Taylor Polynomial and Taylor Polynomial with Remainder is that the latter takes into account the error or remainder term in the approximation. This results in a more accurate representation of the function compared to a regular Taylor Polynomial, which only includes a finite number of terms from the Taylor Series.

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