Undergrad Regarding Error Bound of Taylor Series

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SUMMARY

The discussion centers on determining the upper bound for the error term in a Taylor series, specifically when given a sigma summation form of an unknown function. Participants clarify that to find the Lagrange error term, one must evaluate the (n+1)th derivative of the function and identify its maximum value over the interval [a, x]. An example using the Maclaurin series for sin(x) illustrates that without knowledge of the original function, it is impossible to accurately compute the error bound. The consensus is that the (n+1)th term cannot be assumed to be the largest error term in all cases.

PREREQUISITES
  • Understanding of Taylor series and their general expression
  • Knowledge of Lagrange error term in Taylor series
  • Familiarity with Maclaurin series, particularly for functions like sin(x) and cos(x)
  • Ability to compute derivatives of functions
NEXT STEPS
  • Study the properties of Taylor series and their convergence
  • Learn how to derive and apply the Lagrange error formula
  • Explore examples of Maclaurin series for various functions
  • Investigate the behavior of derivatives over specific intervals
USEFUL FOR

Mathematicians, students studying calculus, and anyone interested in understanding the error bounds of Taylor series expansions.

Aldnoahz
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Hi all, I am very confused about how one can find the upper bound for a Taylor series.. I know its general expression, which always tells me to find the (n+1)th derivative of a certain function and use the equation f(n+1)(c) (x-a)n+1/(n+1)! for c belongs to [a,x]

However, there are numerous cases in which I am only given a sigma summation form of an unknown function. If I want to find the nth degree error term (assume nth degree is the nth term in the summation), do I simply take the (n+1)th term? I personally do not agree with this since the n+1 th term is not necessarily the biggest. In normal situations I always re-evaluate the n+1 th derivative and find its max value, but with unknown functions I am not able to do that.

Any explanation on how to find the error bound? Do I need to find the n+1 th derivative every time?
 
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Aldnoahz said:
Hi all, I am very confused about how one can find the upper bound for a Taylor series.. I know its general expression, which always tells me to find the (n+1)th derivative of a certain function and use the equation f(n+1)(c) (x-a)n+1/(n+1)! for c belongs to [a,x]

However, there are numerous cases in which I am only given a sigma summation form of an unknown function.
Can you give one example of the kind you're talking about? It seems to me that you might be talking about a problem where you are given a series, and are asked to find an upper bound for the remainder if you use the first n terms in the series.
Aldnoahz said:
If I want to find the nth degree error term (assume nth degree is the nth term in the summation), do I simply take the (n+1)th term? I personally do not agree with this since the n+1 th term is not necessarily the biggest. In normal situations I always re-evaluate the n+1 th derivative and find its max value, but with unknown functions I am not able to do that.

Any explanation on how to find the error bound? Do I need to find the n+1 th derivative every time?
Yes, if you are working with a Taylor series, but all you need is the largest value of this derivative on the interval [a, x], as you said. An example would be very helpful for me to understand what you're asking.
 
Mark44 said:
Can you give one example of the kind you're talking about? It seems to me that you might be talking about a problem where you are given a series, and are asked to find an upper bound for the remainder if you use the first n terms in the series.
Yes, if you are working with a Taylor series, but all you need is the largest value of this derivative on the interval [a, x], as you said. An example would be very helpful for me to understand what you're asking.

A simple example would be maclaurin expansion of f(x)=sin(x) given the interval [0, 0.5]. Assume I don't know that this is the expansion for sin(x), all I know is the series. Now I am asked to write down the expansion to degree 3, and find the approximate value of this function at f(0.5). Looking for the approximate value part is obvious to me, but when it asks me to find the upper bound to its error, I can't just take the next term since in the given interval the 4th derivative of sin(x) will not reach 1. So in this case I need to find the 4th derivative of the original function. But as I assumed, if I do not know the original function but only know its maclaurin form, how can I find the lagrange error?
 
Aldnoahz said:
A simple example would be maclaurin expansion of f(x)=sin(x) given the interval [0, 0.5]. Assume I don't know that this is the expansion for sin(x), all I know is the series.
Which is an alternating series, right? There's a theorem that you can use to find the remainder when you use the first n terms of an alternating series.
Aldnoahz said:
Now I am asked to write down the expansion to degree 3, and find the approximate value of this function at f(0.5). Looking for the approximate value part is obvious to me, but when it asks me to find the upper bound to its error, I can't just take the next term since in the given interval the 4th derivative of sin(x) will not reach 1. So in this case I need to find the 4th derivative of the original function. But as I assumed, if I do not know the original function but only know its maclaurin form, how can I find the lagrange error?
 
Mark44 said:
Which is an alternating series, right? There's a theorem that you can use to find the remainder when you use the first n terms of an alternating series.

I don't think it's an alternating series.. I know what you are talking about though. I think you are talking about truncation error where I can just take the next term as the error bound for an alternating series. But I am talking about Lagrange error of nth degree expansion.. I don't know if this is clear.
 
Aldnoahz said:
I don't think it's an alternating series.. I know what you are talking about though. I think you are talking about truncation error where I can just take the next term as the error bound for an alternating series. But I am talking about Lagrange error of nth degree expansion.. I don't know if this is clear.
The Maclaurin series for both sin(x) (=## x - \frac{x^3}{3!} + \frac{x^5}{5!} -+ \dots##) and cos(x) (=## 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -+ \dots##)are alternating series.

Regarding your question, if you're expected to find the Lagrange error term, you would need to know the function whose derivatives appear in the Taylor expansion. Without the underlying function, I don't see how you would be able to get ##f^{(n)}(c)##.
 
Mark44 said:
The Maclaurin series for both sin(x) (=## x - \frac
{x^3}{3!} + \frac{x^5}{5!} -+ \dots##) and cos(x) (=## 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -+ \dots##)are alternating series.

Regarding your question, if you're expected to find the Lagrange error term, you would need to know the function whose derivatives appear in the Taylor expansion. Without the underlying function, I don't see how you would be able to get ##f^{(n)}(c)##.

So Lagrange error always require me to find the derivative. I can't just take the next term (if the maclaurin or taylor expansion is not alternating) as my answer?
 
Aldnoahz said:
So Lagrange error always require me to find the derivative. I can't just take the next term (if the maclaurin or taylor expansion is not alternating) as my answer?
Right on both counts.
 
Ok! Thanks a lot!
 

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