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I am reading through Finite Calculus: A Tutorial for Solving Nasty Sums by David Gleich and on page 9 he computes the general derivative of an exponent:

$$ \triangle(c^x) = c^{x+1} - c^x = (c-1)c^x $$

This part is clear to me. However then the text goes on to say

Because $c$ is a constant in this expression, we can then immediately compute the anti-derivative as well

$$ \sum (c^x)\delta x = \frac{c^x}{c-1} + C $$

How is this immediate computation done? Are the rules of infinite calculus being applied?

Friedrich 'Fred' Clausen
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The idea is that the sum and the difference cancel each other, hence

$$c^x=\sum\triangle(c^x) \delta x= \sum (c-1)c^x\delta x=(c-1)\sum c^x\delta x,$$

which results in

$$\sum (c^x)\delta x = \frac{c^x}{c-1} + C.$$

We need to add the constant because functions differing in a constant have the same difference.

Patricio
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  • Ah I see, the $(c-1)$ can be "taken outside" the sum by way of the distributive law and then we are rearranging to reveal the anti-derivative of $\sum(c^x)\delta x$ – Friedrich 'Fred' Clausen May 03 '21 at 10:46