-3
$\begingroup$

If $$f(n)= \sum_{k=1}^n g(k)\space for \space n\in N$$ then can we find $f'(n) , \space i.e. its \space derivative\space wrt. \space 'n'?$

Just give me some reasons & correct me...

  • 1
    You should give more details on your question. I'm not sure what the derivative sign means in this case. – 2017-01-24
  • 0
    You can't differentiate in the usual way with respect to $n$ because $n$ is a discrete variable. – 2017-01-24
  • 1
    A function $f$ needs to be defined in a neighbourhood of $x$ for $f'(x)$ do be defined. A function defined only on $\Bbb N$ (or $\Bbb Z$) as is the case here never satisfies this requirement. – 2017-01-24
  • 0
    Since $n$ can only be an integer, there is no standard measure of infinitismal change here. – 2017-01-24
  • 0
    Look up finite calculus. It covers exactly this and other things. – 2017-01-24

2 Answers 2

1

The derivative only really makes sense when taken with respect to a continuous variable. You normally interpret the upper limit as "$\lfloor n \rfloor$" if $n$ is continuous, so you can see that the derivative would be zero wherever it is defined (because the sum is constant on intervals $[n,n+1)$. Not interesting.

On the other hand, if you can find a closed form of this sum, you might be able to take the derivative of the extended function if it is interestingly differentiable.

For example: suppose $g(n)=n!$ (factorial function). Normally, we only consider discrete values of $n$ ($0,1,2,\ldots$). However, this can be extended to the Gamma function as $n! = \Gamma(n+1)$, which you can differentiate on the positive reals.

0

One sees that, from this, if $f(n)$ can be made a differentiable function $f:\mathbb R\to\mathbb R$, and likewise $g(n)$ is differentiable, then,

$$f(n)=f(n-1)+g(n)$$

Differentiate and you get

$$f'(n)=f'(n-1)+g'(n)$$

And solving this,we get

$$f'(n)=f'(0)+\sum_{k=1}^ng'(k)$$

And without more information, I am unsure if more is concludable.

If $f$ is monotone, for example, you could approximate $f'(0)$ using finite differences.

  • 0
    Some interesting examples of this: http://math.stackexchange.com/questions/1958762/did-i-derive-a-new-form-of-the-gamma-function – 2017-01-24
  • 0
    ??? I can understand downvotes to a degree, but not deletion votes, especially without any explanation. – 2017-01-26