Derivative of $f(x) = a \sum_{i=1}^x 10^{x-i}$

Question

Consider the following function:

$f(x) = a \sum_{i=1}^x 10^{x-i}= a (10^{x-1} + 10^{x-2} + \cdots + 10^{x-x})$

whose domain are the positive integers greater than 1.

What is its derivative function?

Despite the function being non-continuous I think it should be possible to find its derivative, but I don't know how to do it analytically.

You cannot define the derivative of a function on the integers (unless you are talking $p$-adic numbers, I suppose.) So you'll have to extend this function to all real $x$. — Thomas Andrews, Aug 01 '13 at 15:01
Useful discrete analogues of the derivative are $f(x+1)-f(x)$, or $f(x)-f(x-1)$, or their mean. To explore, google finite differences, or difference equations, — André Nicolas, Aug 01 '13 at 15:13
To put it in different words, I want to find $g'(x)$ where $g(x)$ is a differentiable function such that $g(x) = f(x)$ $\forall x = {x \in \mathbb{N} | x > 1 }$. — Ernest A, Aug 01 '13 at 15:13
@ErnestA, what makes you think that $g(x)$ is unique? (I can think of many such functions.) — George V. Williams, Aug 01 '13 at 15:18
So in Ernest's comment above, $g$ would be related to $f$ in a similar way that the gamma function is related to the factorial, right? — apnorton, Aug 01 '13 at 15:19
@GeorgeV.Williams I don't think $g(x)$ is unique. Does it have to be unique for the problem to have a solution? — Ernest A, Aug 01 '13 at 15:23

George V. Williams · Accepted Answer · 2013-08-01T15:41:10.247

First let me note that your question, as it stands, does not make sense. Assume we are looking for a function $g(x)$ such that:

$g(x)$ is differentiable
$g(x) = f(x)$ for all $x \in \mathbb N$.

This is not unique. Consider the following process to form another function $g^\ast (x)$, take such a $g(x)$ and for all integer $n$, consider the interval $[n + 0.4, n + 0.6]$, and just slightly change your sketch (like add a hump or something). Clearly, this new function $g^\ast(x) $ still satisfies your conditions. We can deform $g(x)$ in different ways and still maintain your properties.

If you are asking for something like $\Gamma(x)$ is to $(x-1)!$, then you still need conditions. There are many such functions which interlope the factorial function, the gamma function just has some useful properties that we like (it is log-convex). The Bohr-Mollerup theorem proves that the gamma function is the only such function satisfying these properties at the integers. That is why we use it.

For other factorial functions, see here.

I'll assume you want to find a closed form for $f(x)$ and then take the derivative. This is far easier and probably more of what you had in mind.

$$ f(x) = a \sum_{i=1}^x 10^{x-i}$$ $$ f(x) = a \sum_{i=1}^x 10^x/10^{i}$$ $$ f(x) = a \cdot 10^x \cdot \sum_{i=1}^x 1/10^{i}$$ $$ f(x) = a \cdot 10^x \cdot \left(\frac19 - \frac{1}{9 \cdot 10^x}\right) $$ $$ f(x) = a \left( \frac{10^x}{9} - \frac19 \right) $$

(I have used the closed form for a finite geometric series in evaluating the sum.)

Finding the derivative is now simple, it is:

$$ f'(x) = \frac{10^x a \log 10}{9} $$

To further stress your first point, one could invoke families of functions such as $g_a:t\mapsto g(t)+a\sin(\pi t)$, which all coincide on the integers. — Did, Aug 01 '13 at 15:33

Derivative of $f(x) = a \sum_{i=1}^x 10^{x-i}$

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