Extended mean value theorem

**chella182** · August 5th 2009, 07:30 AM

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

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

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

The the second part is...

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

$f(a+h)=$ $f(a)+f^{(1)}(a)h+\frac{f^{(2)}(a)h^2}{2!}+\frac{f^ {(3)}(a)h^3}{3!}+\frac{f^{(4)}(a)h^4}{4!}+\frac{f^ {(5)}(a+\theta h)h^5}{5!}$

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

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

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

Deduce more generally that

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

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

Can anyone help? I just can't get my head around it.

**Failure** · August 5th 2009, 09:29 AM

Originally Posted by chella182

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

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

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

The the second part is...

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

$f(a+h)=$ $f(a)+f^{(1)}(a)h+\frac{f^{(2)}(a)h^2}{2!}+\frac{f^ {(3)}(a)h^3}{3!}+\frac{f^{(4)}(a)h^4}{4!}+\frac{f^ {(5)}(a+\theta h)h^5}{5!}$

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

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

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

Deduce more generally that

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

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

Can anyone help? I just can't get my head around it.

Take $f(x) := \mathrm{e}^x$ and $h=1$ : then, because of $f^{(n)}(x)=\mathrm{e}^x$ , we get from the above

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

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

Similarly for the more general case...

**chella182** · August 5th 2009, 09:36 AM

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

**Failure** · August 5th 2009, 09:47 AM

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:

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

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

**chella182** · August 5th 2009, 09:53 AM

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:

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

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

Cheers

Although I'm not sure I like your "I should've known that" comment

**Failure** · August 5th 2009, 10:01 AM

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.

**chella182** · August 5th 2009, 11:30 AM

Haha fair enough

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