Re: (Discrete Math - Induction) 'Formula Differentiation'

On Mar 30, 10:03 pm, "almeidabati...@xxxxxxx"
<almeidabati...@xxxxxxxxx> wrote:
On 31 mar, 02:05, quasi <qu...@xxxxxxxx> wrote:



On Sun, 30 Mar 2008 20:48:11 -0700 (PDT), "almeidabati...@xxxxxxx"

<almeidabati...@xxxxxxxxx> wrote:
On 30 mar, 23:45, quasi <qu...@xxxxxxxx> wrote:
On Sun, 30 Mar 2008 18:30:51 -0700 (PDT), "almeidabati...@xxxxxxx"

<almeidabati...@xxxxxxxxx> wrote:
Hi all! I've got this problem in my set:

'1 + 2q + 3q^2 + ... + nq^(n-1) = [1 - (n+1)q^n + nq^(n-1)]/[(1 -
q)^2], q <> 1.

Estabilish [the formula above] by differentiating the expansion of the
formula for the sum of a geometric progression.'

I've been thinkering with this one all day, no clue on how to start.
ANY hints on how the derivative of the sum of terms of a G.P will get
into this are welcome.

(1) Find an antiderivative of the LHS.

(2) Look at the result -- do you recognize it?

(3) Based on the answer to (2), use a known formula to simplify the
result.

(4) Now differentiate the simplified result,

quasi

Hmmmmmm! This algebric trick would never happen to me wouldn't it be
your answer! Thanks a lot!

Just to be sure, the only 'induction' involved in the solution is
finding the derivative of the summation?

Unless you're required to be ultra-formal, there's no need for
induction. Just show the pattern, making clear that it works, term by
term. If you want, you can also show explicitly what happens to the
k'th term, where k is an arbitrary index variable, left undetermined.

quasi

I guess there's no such need, the doubt just arised on me because this
problem is from the Mathematical Induction chapter of my Discrete
Mathematics book. Thanks for the answer.

I suppose whether or not you need induction depends on what you are
assumed to know already and are allowed to use. For instance, the
derivative of a sum is the sum of the derivatives. One way to show
this is by induction, showing it first for two summands. Also, you
need the formula for the sum of a geometric series. One way to obtain
it is through induction, although there are much better ways of ding
it. You need to know that the derivative of x^n is n*x^(n-1). One say
to obtain this is through induction and the rule about the derivative
of a product of two functions. Are all these things known? Then you
don't need induction.

R.G. Vickson
.

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