2003 3a.) Nice question on use of Rolle's Theorem. No answers provided, this is my candidate. Ignoring the bombastic definition of the theorem, the second part of the question is secondary school math.

Rolle's Theorem. Let a function f R---> R be defined on an interval [a,b]. Then if f(a) = f(b), and f is continuous and differentiable on (a,b), there is a point c in (a,b) such that f'(c) = 0

Suppose that the numbers a(1), a(2).... a(n) satisfy

a(1) + a(2)/2 + a(3)/3 +..... a(n)/n = 0

Prove that there is a a c in (0,1) such that

a(1)c + a(2)c + a(3)c^2 +.... a(n)c^(n-1) = 0

The trick is to see that the first equation can be seen as f(1), where f is the infinite series expansion:

a(1)x + a(2)/2 x^2 + a(3)/3 x^3 +.... a(n)/n x^n

Putting x = 1 into this expansion yields exactly the first equation.

Putting x=0 into this expansion yields 0 = 0.

So we know that f(0) = f(1), and f is an infinite series expansion, hence it is continuous and differentiable as it is the sum of polynomials. So Rolle's Theorem applies, and you can differentiate f with respect to x and notice that all the divisors cancel out nicely, and that f'(c) = 0, where c is in (0,1)

f'(c) = a(1) + a(2)x + a(3) c^2 + .... a(n)c^n = 0

as required.

wah, how to think of this in the pressure of an exam? ok this seems irrelevant now, but it seems to be related to a question in 2002 which I haven't solved yet. But looking at the question it seems like the preliminary of Euler's sine product expansion, which he used to prove that the sum of the inverse squares 1 + 1/2 + 1/4 +1/9 + 1/16 + ... = pi^2/6, which is pretty stunning.

but then again, of course you would think pi is involved, because it's a curve so the area under the curve surely has something to do with a circle. so i will follow this up when i've done that.

## 6/07/2008

Subscribe to:
Post Comments (Atom)

## No comments:

Post a Comment