# Thread: Extended mean value theorem

1. ## Extended mean value theorem

So the first bit of the question gives you the approximation...

$\displaystyle f(a+h)\approx f(a)+f^{(1)}(a)h+\frac{f^{(2)}(a)h^2}{2!}+\frac{f^ {(3)}(a)h^3}{3!}+\frac{f^{(4)}(a)h^4}{4!}$ with $\displaystyle f(x)=e^x$ and $\displaystyle a=0$

...to find the approximate value of $\displaystyle e^1$, which I've done.

The the second part is...

The Extended Mean Value Theorem says that for fixed $\displaystyle f$, $\displaystyle a$ and $\displaystyle h$, there is a $\displaystyle \theta$ in the range $\displaystyle 0\leq\theta\leq1$, such that

$\displaystyle f(a+h)=$$\displaystyle f(a)+f^{(1)}(a)h+\frac{f^{(2)}(a)h^2}{2!}+\frac{f^ {(3)}(a)h^3}{3!}+\frac{f^{(4)}(a)h^4}{4!}+\frac{f^ {(5)}(a+\theta h)h^5}{5!} Use this to show that for some \displaystyle \theta ' with \displaystyle 0\leq\theta '\leq1 we have \displaystyle 4!e^1=65+\frac{3\theta '}{5} ...which I think I've managed okay, but it's this next bit that I'm stuck on... Deduce more generally that \displaystyle \color{red}n!e^1=N_n+\frac{3\theta_n}{(n+1)} ...for some integers \displaystyle \color{red}N_n and numbers \displaystyle \color{red}\theta_n where \displaystyle \color{red}0\leq\theta_n\leq1. Can anyone help? I just can't get my head around it. 2. Originally Posted by chella182 So the first bit of the question gives you the approximation... \displaystyle f(a+h)\approx f(a)+f^{(1)}(a)h+\frac{f^{(2)}(a)h^2}{2!}+\frac{f^ {(3)}(a)h^3}{3!}+\frac{f^{(4)}(a)h^4}{4!} with \displaystyle f(x)=e^x and \displaystyle a=0 ...to find the approximate value of \displaystyle e^1, which I've done. The the second part is... The Extended Mean Value Theorem says that for fixed \displaystyle f, \displaystyle a and \displaystyle h, there is a \displaystyle \theta in the range \displaystyle 0\leq\theta\leq1, such that \displaystyle f(a+h)=$$\displaystyle f(a)+f^{(1)}(a)h+\frac{f^{(2)}(a)h^2}{2!}+\frac{f^ {(3)}(a)h^3}{3!}+\frac{f^{(4)}(a)h^4}{4!}+\frac{f^ {(5)}(a+\theta h)h^5}{5!}$

Use this to show that for some $\displaystyle \theta '$ with $\displaystyle 0\leq\theta '\leq1$ we have

$\displaystyle 4!e^1=65+\frac{3\theta '}{5}$

...which I think I've managed okay, but it's this next bit that I'm stuck on...

Deduce more generally that

$\displaystyle \color{red}n!e^1=N_n+\frac{3\theta_n}{(n+1)}$

...for some integers $\displaystyle \color{red}N_n$ and numbers $\displaystyle \color{red}\theta_n$ where $\displaystyle \color{red}0\leq\theta_n\leq1$.

Can anyone help? I just can't get my head around it.
Take $\displaystyle f(x) := \mathrm{e}^x$ and $\displaystyle h=1$: then, because of $\displaystyle f^{(n)}(x)=\mathrm{e}^x$, we get from the above

$\displaystyle \mathrm{e}^1=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1 }{4!}+\frac{\mathrm{e}^{\theta}}{5!}$

Multiply this equation by $\displaystyle 4!$ and consider that $\displaystyle 0\leq \theta \leq 1$ implies $\displaystyle \theta' := \mathrm{e}^{\theta}$ satisfies $\displaystyle 1\leq \theta'\leq \mathrm{e}<3$.

Similarly for the more general case...

3. Sorry, I don't understand how you get the equation I'm given in the third bit

4. Originally Posted by chella182
Sorry, I don't understand how you get the equation I'm given in the third bit
Ah, I should have known that. Let me try again: so we have, using the same function, the same a=0 and h=1:

$\displaystyle \mathrm{e}^1=\sum_{k=0}^n\frac{1}{k!}+\frac{\mathr m{e}^{\theta}}{(n+1)!}$
Now, you multiply by $\displaystyle n!$ to get

$\displaystyle n!\cdot \mathrm{e}=\sum_{k=0}^n\frac{n!}{k!}+\frac{\mathrm {e}^{\theta}}{n+1}$
Where clearly all the "fractions" $\displaystyle \frac{n!}{k!}$ are natural numbers and, therefore, $\displaystyle N_n := \sum_{k=0}^n\frac{n!}{k!}$ is a natural number. In addition, we have, as before, that $\displaystyle 1\leq \mathrm{e}^{\theta}\leq 3$.

5. Originally Posted by Failure
Ah, I should have known that. Let me try again: so we have, using the same function, the same a=0 and h=1:

$\displaystyle \mathrm{e}^1=\sum_{k=0}^n\frac{1}{k!}+\frac{\mathr m{e}^{\theta}}{(n+1)!}$
Now, you multiply by $\displaystyle n!$ to get

$\displaystyle n!\cdot \mathrm{e}=\sum_{k=0}^n\frac{n!}{k!}+\frac{\mathrm {e}^{\theta}}{n+1}$
Where clearly all the "fractions" $\displaystyle \frac{n!}{k!}$ are natural numbers and, therefore, $\displaystyle N_n := \sum_{k=0}^n\frac{n!}{k!}$ is a natural number. In addition, we have, as before, that $\displaystyle 1\leq \mathrm{e}^{\theta}\leq 3$.
Cheers
Although I'm not sure I like your "I should've known that" comment

6. Originally Posted by chella182
Cheers
Although I'm not sure I like your "I should've known that" comment
I didn't mean to say that I should have known that you would not understand (whatever specific argument). Instead, what I wanted to say (but failed to communicate clearly) is that I should have known from your original question that you were primarily interested in getting some help for the third part (marked red).
Therefore it was not you who failed (to understand) but me who failed to read your question correctly.

7. Haha fair enough