## Friday, January 16, 2009

### A Simple Lemma Based on Calculus

The lemma in today's proof is used in my proof of Sturm's Problem of the Number of Roots. I will add a link to this proof when it is available.

Lemma 1:

Let F(x) = A*B

Then:

F'(x)/F(x) = A'/A + B'/B

Proof:

(1) Using the Product Rule (see Lemma 4, here):

F'(x) = A'B + B'A

(2) So:

F'(x)/F(x) = (A'B + B'A)/AB = A'B/AB + B'A/AB = A'/A + B'/B

QED

Corollary 1.1:

Let: F(x) = A1*A2*...*An

Then:

F'(x)/F(x) = A'1/A1 + A'2/A2 + ... + A'n/An

Proof:

(1) Let F(x)=AB

Using Lemma 1 above, we know that F'(x)/F(x) = A'/A + B'/B

(2) Assume that this is true up to n-1 so that:

if F(x) = U1*...*Un-1

Then:

F'(x)/F(x) = U'1/U1 + U'2/U2 + ... + U'n-1/Un-1

(3) Let G(x) = Un*(U1*...*Un-1)

(4) Using the Product Rule (see Lemma 4, here):

G'(x) = U'n*(U1*....*Un-1) + Un*(U1*...*Un-1)'

(5) Then G'(x)/G(x) = U'n/Un + (U1*...*Un-1)'/(U1*...*Un-1)

(6) From our assumption in step #2, this gives us:

G'(x)/G(x) = U'n/Un + U'1/U1 + ... + U'n-1/Un-1

(7) By Induction, we are done.

QED

Lemma 2:

Let F(x) = (x - α)a

Then:

F'(x) = a(x - α)a-1

Proof:

(1) Let U = x - α

(2) Using the power rule (see Lemma 2, here):

(d/dx)Ua = aUa-1dU

(3) So, we have:

F'(x) = aUa-1dU = a(x-α)a-1*1 = a(x-α)a-1

QED

Corollary 2.1:

Let F(x) = (x - α)a

Then:

F'(x)/F(x) = a/(x - α)

Proof:

(1) From Lemma 1 above, F'(x) = a(x - α)a-1

(2) Then F'(x)/F(x) = a(x-α)a-1/(x - α)a = a/(x - α)

QED

Corollary 2.2:

Let F(x) = (x - α)a(x - β)b(x - γ)c*...

Then:

F'(x)/F(x) = a/(x - α) + b/(x - β) + c/(x - γ) + ...

Proof:

(1) Let A = (x - α)a, B = (x - β)b, C= (x - γ)c, etc.

(2) So, we have: F(x) = A*B*C*...

(3) Using Corollary 1.1 above, this gives us that:

F'(x)/F(x) = A'/A + B'/B + C'/C + ...

(4) Using Corollary 2.1 and step #1, this gives us:

F'(x)/F(x) = a/(x - α) + b/(x - β) + c/(x - γ) + ...

QED