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Topic: geom series/polyn diff
Replies: 0

 Doug Kuhlmann Posts: 3,630 Registered: 12/6/04
geom series/polyn diff
Posted: Feb 17, 1998 10:42 AM

While teaching the sum of a (finite) geometric series in precalculus they
other day, I also taught the students how to find the future value of an
account in which P dollars is paid every month paying i interest per month.
It's the same problem and if we just let a=1 (a is the first term) and
P=1, we get these two equivalent formulas:

1 + r + r^2 + ..... + r^(n-1) = (r^n - 1)/(r - 1)

and

1 +(1+i) + (1+i)^2 + ... + (1+i)^n-1) = ((1+i)^n - 1)/i

(The latter formula assumes payments are made at the end of the month.)

In both cases you can look at the expression on the right hand side as an
average rate of change of the function f(x)=x^n either from 1 to r or from
1 to 1+i. Taking limits as r->1 or as i->0 we can use these to find the
derivative of x^n and discover that it is nx^(n-1) without having to use
the binomial thoerem. I'm not saying the binomial theorem is not
important, but it seems that more students have seen the sum of a geometric
series formulas than the binomial theorem. If they have seen neither, the
proof of the gemetric series sum is quicker.

The MVT says that

1 + r + r^2 + ..... + r^(n-1) = nc^(n-1) for some 1<=c<=r or

1 +(1+i) + (1+i)^2 + ... + (1+i)^n-1) = n(1+iav)^(n-1) or some
1<=iav<=i. (I am trying to suggest iav = "average i".

Now here is (are) my question(s): Does anyone have some insight into this?
I.E. can we make this formula seem plausible in the way the the product
rule is made plausible by looking at a rectangle whose sides are expanding
independently of each other? Has anyone derived the formula of x^n this
way in a calculus class? Have you had any success with it?

Doug

P.S. The above limits as r->1 or i->0 actually just shows that if f(x)=x^n
then f'(1) = n. To show the more general do this:

(x^n - a^n)/(x - a) = (a^n)(r^n - 1)/(a(r -1)) = a^(n-1)(r^n - 1)/(r - 1)
where r=x/a.

Doug Kuhlmann
Math Department Chair