lets-f-continuous-and-diferrenciable-f-x-1-xf-x-proof-that-n-N-0-f-n-x-1-nf-n-1-x-xf-n-x-where-f-n-x-d-n-f-dx-n-

Question Number 4100 by 123456 last updated on 28/Dec/15

lets f continuous and diferrenciable

f (x + 1) = x f (x)

proof that n \in N / {0}

f^{(n)} (x + 1) = {n f}^{(n - 1)} (x) + {x f}^{(n)} (x)

where

f^{(n)} (x) = \frac{d^{n} f}{{d x}^{n}}

Answered by prakash jain last updated on 28/Dec/15

f^{(n)} (x + 1) = {n f}^{(n - 1)} (x) + {x f}^{(n)} (x) \dots (Stmt A)

f (x + 1) = x f (x)

f^{'} (x + 1) = x f^{'} (x) + f (x)

Stmt A is true for n = 1

Let us say A is true for n = k

f^{(k)} (x) = {k f}^{(k - 1)} (x) + {x f}^{(k)} (x)

f^{(k + 1)} (x) = k f^{(k)} (x) + f^{(k)} (x) + {x f}^{(k + 1)} (x)

= (k + 1) f^{(k)} (x) + x f^{(k + 1)} (x)

Hence if A is true for k \Rightarrow A is true for k + 1

Since A is true for k = 1 .

f^{(n)} (x + 1) = {n f}^{(n - 1)} (x) + {x f}^{(n)} (x)

is true for all n \in N, n ⩾ 1