Error bounds on series approximation

In summary: Now, we can use a calculator or estimation techniques to solve for k and find that k \geq 3. Therefore, we can conclude that the minimum value of k required to approximate f(x) within an error of 0.1 on the interval (1
  • #1
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Homework Statement



I want to estimate [itex]f(x)=\ln (\frac{1}{1+x}) [/itex] on the interval [itex](1/10,1)[/itex] with the error on the approximation being no more than [itex]0.1[/itex].

Homework Equations



http://en.wikipedia.org/wiki/Taylor's_theorem#Example

The Attempt at a Solution



Following the example from Wikipedia, I first found that the [itex]n^{\text{th}}[/itex] derivative for [itex]n \geq 1[/itex] of [itex]f(x)[/itex] is:

[tex]f^{(n)}(x)=(-1)^n \frac{(n-1)!}{(1+x)^n}[/tex]

Then the remainder term is:

[tex]R_{k}(x)=\frac{f^{(k+1)}(\xi_{L})}{(k+1)!}(x-a)^{k+1}=(-1)^{k+1}\frac{x^{k+1}}{(1+x)^{k+1} (k+1)}[/tex]

Here's the sketchy part, I think I'm supposed to find an upper bound for [itex]f(x)[/itex] on the interval I want, but since [itex]f(x)[/itex] is a decreasing function, it's maximum value is going to be when [itex]x=1/10[/itex], in which case we have [itex]f(1/10) \approx -0.0953[/itex] so I guess that means I can say [itex]f(x) \leq -0.09[/itex]?

Then:

[tex]|R_{k}(x)| \leq \Big|(-1)^{k+1}\frac{-0.09 x^{k+1}}{(1+x)^{k+1} (k+1)}\Big| = \Big|(-1)^{k+1}\Big| \frac{|-0.09| |x|^{k+1}}{|1+x|^{k+1} |k+1|} = 0.09 \frac{x^{k+1}}{(1+x)^{k+1} (k+1)} \leq \frac{0.09}{(1+x)^{k+1} (k+1)} \leq \frac{0.09}{(k+1)} [/tex]

Finally, impose the condition on the maximum error:

[tex]\frac{0.09}{(k+1)} < 0.1 \implies k>-1/10 \implies k=0[/tex]

That can't be right though. I think it might have something to do with that [itex]\xi_{L}[/itex] I disregarded.
 
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  • #2
Any hints?

Hi there,

Great job on using Taylor's theorem to approximate f(x)! You are correct in thinking that you need to consider the value of \xi_{L} in order to find an upper bound for f(x) on the interval (1/10,1).

To do this, we can use the fact that f(x) is a decreasing function on this interval, meaning that the maximum value of f(x) will occur at the lower bound of the interval, x=1/10. Therefore, we can say that f(x) \leq f(1/10) = -0.0953 on the interval (1/10,1).

Now, let's consider the remainder term R_{k}(x) with this new upper bound for f(x):

|R_{k}(x)| \leq \Big|(-1)^{k+1}\frac{f^{(k+1)}(\xi_{L})}{(k+1)!}(x-a)^{k+1}\Big| = \Big|(-1)^{k+1}\frac{(-0.0953)(x-1/10)^{k+1}}{(k+1)!}\Big|

We can then use the triangle inequality to simplify this expression:

|R_{k}(x)| \leq \Big|(-1)^{k+1}\frac{(-0.0953)(x-1/10)^{k+1}}{(k+1)!}\Big| \leq \frac{0.0953}{(k+1)!}|x-1/10|^{k+1} \leq \frac{0.0953}{(k+1)!}|1-1/10|^{k+1}

Now, we can use the fact that |x-1/10| < 1 on the interval (1/10,1) to simplify this expression even further:

|R_{k}(x)| \leq \frac{0.0953}{(k+1)!}|1-1/10|^{k+1} \leq \frac{0.0953}{(k+1)!}(0.9)^{k+1}

Finally, we can impose the condition on the maximum error:

\frac{0.0953}{(k+1)!}(0.9)^{k+1} < 0
 

Related to Error bounds on series approximation

1. What are error bounds on series approximation?

Error bounds on series approximation refer to the maximum difference between the actual value of a mathematical function and its approximate value obtained through a series approximation. It is a measure of the accuracy of the approximation and helps in understanding the limitations of using series to approximate functions.

2. How are error bounds calculated?

Error bounds are typically calculated using mathematical techniques such as Taylor series expansions and Lagrange remainders. These methods involve using derivatives of the function and evaluating the error at a specific point.

3. Why are error bounds important in scientific research?

Error bounds are important in scientific research because they provide a measure of the accuracy and reliability of the results obtained through series approximation. They also help in identifying the potential sources of error and in determining the level of confidence in the results.

4. Can error bounds be reduced?

Yes, error bounds can be reduced by using more accurate and advanced mathematical techniques for series approximation. Additionally, increasing the number of terms in the series can also improve the accuracy and reduce the error bounds.

5. How are error bounds affected by the choice of series approximation method?

Error bounds can vary depending on the method used for series approximation. Some methods may provide more accurate results with lower error bounds, while others may have higher error bounds. It is important to choose the appropriate method based on the function being approximated and the desired level of accuracy.

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