derivative sum

[kinu]

\(f(x)\) is a function satisfying the following condition....

\(f(x)+f ^{'}(x)+f ^{''}(x)+f ^{'''}(x) ...\) upto n terms =  \(x^n\)

where \(f ^{'}(x)\) = first derivative of \(f(x)\)

\(f ^{''}(x)\) = 2nd derivative of \(f(x)\)  and so on....

Find the value of \(\displaystyle f(x) +\frac{f^{'}(x)}{1!}+\frac{f^{''}(x)}{2!}+\frac{f^{'''}(x)}{3!}+....\) n-terms =

[João P. Ferreira]

Hi

It might be wrong but look at this:

We know by the Taylor serie that:

\(f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\)

if we compute \(f(x)\) at \(x=a+1\) we get

\(f(a+1)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(a+1-a)^n=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(1)^n=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}\)

So we know that

\(f(a+1)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}\)

because you want to compute

\(\sum_{n=0}^{\infty}\frac{f^{(n)}(x)}{n!}\)

you need to find \(f(x+1)\) I suppose...

I don't know if these are the correct steps...

[Rui Carpentier]

Lets assume that \(f\) is a polynomial (I don't know if the solution is unique and how to show it).

For the first equation we get that the degree of \(f\) is \(n\). This mean that \(f^{(n+1)}(x)=0\) and therefore

\(x^n =x^n +f^{(n+1)}(x)=f(x)+f'(x)+\cdots +f^{(n+1)}(x)=f(x)+[f(x)+f'(x)+\cdots +f^{(n)}(x)]'=f(x)+(x^n)'=f(x)+nx^{n-1}\), thus \(f(x)=x^n-nx^{n-1}\).

Now we follow the João's hint. For any polynomial \(f\) of degree \(n\) we have

\(f(t)=f(a)+f'(a)(t-a)+\frac{f''(a)}{2!}(t-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(t-a)^n\)
for any \(a\) and \(t\).

This means, by taking \(t=x+1\) and \(a=x\), that

\(f(x)+f'(x)+\frac{f''(x)}{2!}+\cdots+\frac{f^{(n)}(x)}{n!}=f(x+1)=(x+1)^n-n(x+1)^{n-1}\)

[João P. Ferreira]

Thanks a lot for the support dear Rui

A huge hug...

[kinu]

Thanks Professor and also thanks to Rui Carpenter