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Is there any way I can change a summation, say from $k=1$ to $n$ of the derivative of order $k$ of a function into closed form, or some form that would be more manageable? Ex. From $k=1$ to $3$ of $d(2x^2)/dx$ would be $4x + 4 + 0 = 4x + 4$ But how can I simplify this if I am running from $1$ to $n$ of an unspecified function $f(x)$? What about the same but with integrals?

user3810965
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  • Without knowing anything about $f$, not really. In certain special cases (trig functions, exponentials, polynomials, ...) there are things that can be done. – Ian May 13 '15 at 14:29

2 Answers2

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It seems that you want $D(f, j, k) =\sum_{i=j}^k f^{(i)}(x) $.

There might be some analogy with the partial sums of a geometric series.

What occurs to me is the Taylor (or MacLaurin) series $f(x+h) =\sum_{n=0}^{\infty} \dfrac{h^n f^{(n)}(x)}{n!} $.

There's a way of getting rid of the $n!$ I'm trying to remember.

Multiply by $e^{-h}$, integrate from $0$ to $\infty$, and use $\int_0^{\infty} x^n e^{-x}dx = n! $.

We get

$\begin{array}\\ \int_0^{\infty} f(x+h)e^{-h}dh &=\int_0^{\infty} e^{-h}dh \sum_{n=0}^{\infty} \dfrac{h^n f^{(n)}(x)}{n!}\\ &=\sum_{n=0}^{\infty}f^{(n)}(x)\int_0^{\infty} e^{-h}dh \dfrac{h^n }{n!}\\ &=\sum_{n=0}^{\infty}f^{(n)}(x)\\ \end{array} $

This is all I can think of at the moment.

marty cohen
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Let $$g(x)=\sum_{k=1}^n f^{(k)}(x).$$

Then $$g'(x)-g(x)=f^{(n+1)}(x)-f'(x).$$

You can integrate this differential equation with

$$\left(g'(x)-g(x)\right)e^{-x}=\left(g(x)e^{-x}\right)'=\left(f^{(n+1)}(x)-f'(x)\right)e^{-x},$$ then $$g(x)=e^x\int\left(f^{(n+1)}(x)-f'(x)\right)e^{-x}\,dx.$$

Maybe not simpler, but the sum over $k$ has disappeared.

  • I understand how to integrate the differential equation, but how you got to that equation from the original summation I don't understand. If someone could explain that step in particular that removes the summation I would be most grateful. – user3810965 May 15 '15 at 16:53
  • Compute $g'-g$ and see the cancellation. (Take $n=4$ for instance.) –  May 15 '15 at 16:55
  • Do you want $\sum_{k=1}^n f^{(n)}(x)$ or $\sum_{k=1}^n f^{(k)}(x)$? – marty cohen May 19 '15 at 23:43
  • @martycohen: fixed. –  May 20 '15 at 06:16