1. ## Taylors Theorem

I need to calculate the upper bound for the error (and the exact error) in the following approximation using Taylor Polynomials:

$e\approx 1+1+\frac{1^2}{2!}+\frac{1^3}{3!}+\frac{1^4}{4!}+\ frac{1^5}{5!}=P_5(1)$

Where $P_5(x)$ is the 5th degree Taylor Polynomial for $e^x$

Taylors Theorem states that the error is $R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-c)^{n+1}$ where $c

So $R_5(x)=\frac{e^z}{6!}x^6$ where $0

This is where I'm lost, because $e^1=e$ is the max value on this interval. So is the upper bound for the error just $R_5(1)=\frac{e}{6!}$

???

It just seems strange because e is the number I'm approximating. And how do I calculate exact error?

2. At this point, you might have to use the well-known inequality $2.

As for exact error, I don't think that it is easy to compute. It isn't a good use of your time.

I need to calculate the upper bound for the error (and the exact error) in the following approximation using Taylor Polynomials:

$e\approx 1+1+\frac{1^2}{2!}+\frac{1^3}{3!}+\frac{1^4}{4!}+\ frac{1^5}{5!}=P_5(1)$

Where $P_5(x)$ is the 5th degree Taylor Polynomial for $e^x$

Taylors Theorem states that the error is $R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-c)^{n+1}$ where $c

So $R_5(x)=\frac{e^z}{6!}x^6$ where $0

This is where I'm lost, because $e^1=e$ is the max value on this interval. So is the upper bound for the error just $R_5(1)=\frac{e}{6!}$

???

It just seems strange because e is the number I'm approximating. And how do I calculate exact error?
What is wrong with leaving it like that? Or if you don't like that, use what you have so far to get a numeric bound on e which you feed back into what you have to get a bound that does not involve e explicitly.

$e\le P_5(1)+\dfrac{e}{6!}$

so

$e\le \dfrac{P_5(1)}{1-\frac{1}{6!}}$

So:

$e = P_5(1)+R_5$

with:

$|R_5|\le \dfrac{\left(\frac{P_5(1)}{1-\frac{1}{6!}}\right)}{6!}$

CB

4. Originally Posted by CaptainBlack
What is wrong with leaving it like that? Or if you don't like that, use what you have so far to get a numeric bound on e which you feed back into what you have to get a bound that does not involve e explicitly.

$e\le P_5(1)+\dfrac{e}{6!}$

so

$e\le \dfrac{P_5(1)}{1-\frac{1}{6!}}$

So:

$e = P_5(1)+R_5$

with:

$|R_5|\le \dfrac{\left(\frac{P_5(1)}{1-\frac{1}{6!}}\right)}{6!}$

CB
The problem was worded like this: Use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error.

I like the way you set the bounds for the error. That seems smart than just leaving the error with the "e" in it, because if we're calculating the error of an approximation for "e." Technically, we aren't supposed to know the value...

But your method helps with that dilemma, but I still don't see how to find the "exact value" of the error.

5. But your method helps with that dilemma, but I still don't see how to find the "exact value" of the error.
Correct me if I wrong...

You can't find the exact value of error, because then e will be rational.

The problem was worded like this: Use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error.

I like the way you set the bounds for the error. That seems smart than just leaving the error with the "e" in it, because if we're calculating the error of an approximation for "e." Technically, we aren't supposed to know the value...

But your method helps with that dilemma, but I still don't see how to find the "exact value" of the error.
The exact value of the error is:

$e-P_5(1)$

(you are allowed to use a calculator to get the approximate value for this to 14 decimal places)

Code:
(%i1) p5(x):=1+x+x^2/2!+x^3/3!+x^4/4!+x^5/5!  /*define the Taylor polynomial*/;
(%o1) p5(x):=1+x+x^2/2!+x^3/3!+x^4/4!+x^5/5!

(%i2) p5(1)  /*evaluate the approximation to e: p5(1)*/;
(%o2) 163/60

(%i3) (p5(1)/(1-1/6!))/6!  /*compute the error bound*/;
(%o3) 163/43140

(%i4) float(%), numer  /*convert the error bound to a float*/;
(%o4) 0.0037783959202596

(%i6) %e-p5(1)  /*compute the actual error*/;
(%o6) %e-163/60

(%i7) float(%), numer  /*convert actual error to float approximation*/;
(%o7) 0.0016151617923783
CB

7. Originally Posted by CaptainBlack
The exact value of the error is:

$e-P_5(1)$

(you are allowed to use a calculator for this)

CB
Oh yeah, duh!

Taylor's Theorem provides the bounds, not the error. Thanks!