The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Courses » ap-calculus

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: geom series/polyn diff
Replies: 0  

Advanced Search

Back to Topic List Back to Topic List  
Doug Kuhlmann

Posts: 3,630
Registered: 12/6/04
geom series/polyn diff
Posted: Feb 17, 1998 10:42 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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)


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?


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
Phillips Academy
180 Main Street
Andover, MA 01810

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.