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If $f_1(x)=\int_0^x f(t)dt$, $f_2(x)=\int_0^x f_1(t)dt$ and $$f_n(x)=\int_0^x f_{n-1}(t)dt$$ for all $n\ge 2$, prove the following:

  1. For $n\in \mathbb N$ $$f_{n+1}(x)=\int \frac{(x-t)^n}{n!}f(t)dt$$
  2. Assume class $C^{n+1}$$$f(x)=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k+\int_0^x f^{n+1}(t)\frac{(x-t)^{n}}{n!}$$

Assume that f is continuous on the interval I containing $0$.

My Work

  1. For this part, I started by using integration by parts on the function $f_{n+1}$, using $v(x)=\frac{(x-t)^n}{n!}$ and $u'(x)=f(t)dt$. I chose these u and v somewhat arbitrarily, but with the idea that I could use the relationship between $f_1$ and $f_2$ to work towards a solution. Unfortunately, while I was able to get some results, the did not seem pertinent to the solution: $$\frac{(x-t)^n}{n!}F(t)|^x_0+\frac1{n(n!)}\int_0^x(x-t)^{n-1}F(t)dt$$ From here, I do not know how to work towards a final solution

  2. To start, I studied how the equation resembled Taylor's Theorem, but honestly, I am pretty intimidated by the problem. I'm not the best at proofs and I really don't know where to even begin to tackle this problem

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  • $\begingroup$ How could I alter it while still keeping all of the pertinent information? The question is simply too long for the title bar $\endgroup$ – Kevin McDonough Dec 7 '15 at 8:11
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1) \begin{align} \int_0^x \frac{(x-t)^n}{n!}f(t)\:dt&=\frac{(x-t)^n}{n!}\left(\int_0^t f(y)\:dy\right)\bigg|_0^x+\int_0^x \frac{(x-t)^{n-1}}{(n-1)!}\left(\int_0^t f(y)\:dy\right)dt \\ &=\int_0^x \frac{(x-t)^{n-1}}{(n-1)!}f_1(t)\:dt \\ &=\cdots \\ &=\int_0^x f_{n}(t)\:dt=f_{n+1}(x) \end{align} 2) Since \begin{align} \int_0^x \frac{(x-t)^n}{n!}f^{(n+1)}(t)\:dt&=\frac{(x-t)^n}{n!}f^{(n)}(t)\bigg|_0^x+\int_0^x \frac{(x-t)^{n-1}}{(n-1)!}f^{(n)}(t)dt \\ &=-\frac{f^{(n)}(0)}{n!}x^n+\int_0^x \frac{(x-t)^{n-1}}{(n-1)!}f^{(n)}(t)\:dt \\ &=\cdots \\ &=-\sum_{k=1}^n\frac{f^{(k)}(0)}{k!}x^k+\int_0^x f^{(1)}(t)\:dt \\ &=-\sum_{k=1}^n\frac{f^{(k)}(0)}{k!}x^k+f(x)-f(0) \end{align} We have $$ f(x)=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k+\int_0^x \frac{(x-t)^n}{n!}f^{(n+1)}(t)\:dt $$

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