Taylor's Theorem for finding error for Taylor expansion

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

My answer for (a) and (b) are

(a): $E_2(x) = \sqrt{9} + \frac{1}{2 \sqrt{2}}(x - 9) - \frac{1}{8 \sqrt{9^3}}(x - 9)^2$
(b): $E_2(8) = 2.8287$

However, for (c) does someone please know whether we really need to use Taylors Theorem? For example, why can’t we just do $|\sqrt{8} - 2.8287|$ so error is $-0.00027287525$?

Thanks!

 

[docnet]

The problem statement specifically asks you to use Taylor's theorem, so yes.

 

[Orodruin]

To add to the above, the entire point of that problem is to demonstrate how Taylor's theorem works. Yes, it is a known function and you can compute the error exactly, but that is not the point. It is a good and instructive idea to use Taylor's theorem to estimate the error and then compare to the actual error.

 

[FactChecker]

However, for (c) does someone please know whether we really need to use Taylors Theorem? For example, why can’t we just do $|\sqrt{8} - 2.8287|$ so error is $-0.00027287525$?

This is an exercise to learn how to use Taylor's Theorem in a familiar situation where $\sqrt 8$ is known and easily obtainable with great accuracy. In fact, if you are counting on getting this from your calculator, it uses a numerical method (probably better than a Taylor series approximation). In practice, Taylor's Theorem can be used where the true value is not known. In that case, it is wise to also bound the error using the theorem.

 

[member 731016]

Thank you for your replies @docnet, @Orodruin and @FactChecker !

Ok yeah I prefer to use Taylors theorem since I don't know how to apply it.

But I think it states that a function is of the form:

Function = Taylor polynomial approximation (up to degree n) + Reminder (Lagrange, or other reminder formula)

In limit as n goes to infinity, reminder goes to zero, and taylor polynomial is equal to function.

So for this problem I use the Lagrange form of a remainder.

$R_n(x) = \frac{f^{n + 1}(c)}{(n + 1)!}x^{n + 1}$

$R_2(x) = \frac{f'''(x - 9)^3}{6}$

$| f(x) - T_2 | = | \frac{(x - 9)^3}{24c^{\frac{3}{2}}} |$

However, I don't know how to go from here as it is quite different to what I get without using the Taylor Theorem. Does anybody please know what I should do?

Thanks!

 

[docnet]

The Lagrange form of the remainder is not suitable for this situation because $c$ is not determined.

First try writing out the first 3-4 terms of the remainder $R_2$ and see what you get. With any luck, you could be able to find a formula for the infinite sum and maybe even compute the "exact" value of $R_2$.

 

[member 731016]

The Lagrange form of the remainder is not suitable for this situation because $c$ is not determined.

Thank you for your reply @docnet!

Sorry I don't understand what you mean. What equation for the remainder must we use then?

That is the one I have been taught.

First try writing out the first 3-4 terms of the remainder $R_2$ and see what you get. With any luck, you could be able to find a formula for the infinite sum and maybe even compute the "exact" value of $R_2$.

Sorry which $R_2$?

Thanks!

 

[docnet]

Write out the 'full' Taylor series. The full series with all the terms gives the exact value of $f$. The first 3 terms make up the 2nd degree Taylor polynomial or $E_2$. The remaining terms make up the error term or $R_2$. This is because of the equation
$$f(x)=E_2(x)+R_2(x).$$

If this is unfruitful, you can also use the Lagrange form of error to get a bound on the error because $c$ can be any value between $8$ and $9$.

 

[docnet]

You could also manually add up the first few terms of $R_2$ to get an estimate on the error. The terms contribute smaller and smaller amounts as $n$ gets larger, so the more terms you evaluate, the better.

 

[member 731016]

Thank you for your replies @docnet!

Sorry I'm not familiar with the Lagrange error bound formula. I have only been taught Taylors Theorem ( Taylor’s Remainder Theorem).

I'm slightly confused by what you are saying because I have already wrote up the remainder $R_2(x)$ in post #7, however, as you note my expression is in terms of c which you said is incorrect.

You are saying to get rid of c in my $R_2(x)$ I must put a bound on c to be $8 < c < 9$. However, sorry I am not sure how to use that bound to get rid of c from $R_2(x)$. Do you please know how?

Thanks!

 

[pbuk]

Your first mistake here is not writing down and taking care with your workings: I am sure that you know that the second term of $E_2(x)$ is $f'(a)(x-a) = \frac{1}{2\sqrt a}(x-a)$ and a = 9 so why have you written $\frac{1}{\sqrt 8}(x-9)$?

Next you are getting lost here in the theory (as unfortunately are some of the helpers): the question is not asking for an exact calculation of the remainder using the Lagrange polynomial, it is asking for an estimate of the remainder.

From the expansion of the Taylor series
$$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 \cdots \, $$
it is clear that the first term of $R_2(x)$ is $\frac{f'''(9)}{3!}(9-8)^3$. If you calculate this correctly you will obtain an estimate that is just over 7% less than the exact remainder, and I am fairly sure it is this estimate that the question is looking for.