Calc Analysis?

Question

I'm struggling, can you help me please? Thanks a ton.

18. Suppose that f '(n) (a) - (the n-th derivitive of f at a) - and g'(n) (a) exist. Prove Leibniz's formula:
(f * g)'(n) (a) = Sigma, k = 0 to n, of: (n choose k) f '(k) (a) * g'(n-k) (a)

22.a. The number a is called a double root of the polynomial function f if f(x) = (x-a)^2 * g(x) for some polynomial function g. Prove that a is a double root of f if and only if a is a root of both f and f '
b. When does f(x) = ax^2 +bx + c (a not equal to 0) have a double root? What does the condition say geometrically?

27. Suppose f is differentiable at 0, and that f(0) = 0. Prove that f(x) = x*g(x) for some function g which is continuous at 0. Hint: What happens if you try to write g(x) = f(x) / x?

Thanks again!

Answer 1

18. Use induction. If n=0, then it's just (f*g)(a) = f(a)g(a).

Assume that Leibniz' formula holds for some n>=0, and show that it holds for n+1.

(f*g)^(n)(x) = sum k=0 to n (n choose k) f^(k)(x) g^(n-k)(x)

Differentiate this with respect to x:

(f*g)^(n+1)(x) = sum k=0 to n (n choose k) [ f^(k+1)(x) g^(n-k)(x) + f^(k) g^(n-k+1)(x) ]
= sum k=0 to n (n choose k) f^(k)(x) g^(n+1-k)(x) +
sum k=1 to n+1 (n choose k-1) f^(k)(x) g^(n+1-k)(x)
= f^(0)(x) g^(n+1)(x) +
sum k=1 to n [(n choose k)+(n choose k-1)] f^(k)(x) g^(n+1-k)(x)
+ f^(n+1)(x) g^(0)(x)

(n+1 choose 0) = (n+1 choose n+1) = 1, and
(n choose k) + (n choose k-1) = n!/[k!(n-k)!] + n!/[(k-1)!(n-k+1)!]
= [(n-k+1)n! + kn!]/[k!(n+1-k)!] = (n+1)!/[k!(n+1-k)!] =
(n+1 choose k)

putting all the terms together gives the result.

22a.Suppose a is a double root of f. Then f(x)=(x-a)^2*g(x), so a is a root of f. By the product rule, f'(x)=g(x)*2(x-a) + (x-a)^2*g'(x), so f'(a)=0. Suppose f(a)=0 and f'(a)=0. Then f(x) = (x-a)r(x) for some polynomial r(x). f'(x) = r(x) + (x-a)r'(x), and since f'(a)=0, it follows that r(a)=0, so r(x)=(x-a)g(x), meaning f(x)=(x-a)^2*g(x).
b. f'(x) = 2ax + b, so its root is -b/2a. For this to be a root of f(x) as well, must have a(-b/2a)^2 + b(-b/2a) + c = 0, or b^2 - 4ac = 0. Geometrically, the graph is tangent to the x-axis at x=-b/2a.

27. Let L = f'(0), and define g(x) = f(x)/x for nonzero x, and g(0) = L. Then g(0) = L = f'(0) = lim (x->0) f(x)/x = lim (x->0) g(x), so by the definition of continuity, g is continuous at 0.