It-is-given-f-x-1-x-n-n-N-Find-f-0-f-0-f-0-2-f-0-3-f-n-0-n-

Question Number 183459 by Matica last updated on 26/Dec/22

I t i s g i v e n f (x) = {(1 + x)}^{n}, n \in N . F i n d

f (0) + f^{^{'}} (0) + \frac{f^{^{″}} (0)}{2!} + \frac{f^{‴} (0)}{3!} + \dots + \frac{f^{(n)} (0)}{n!} .

Answered by mahdipoor last updated on 26/Dec/22

= 1 + n + \frac{n \times (n - 1)}{2} + \frac{n \times (n - 1) \times (n - 2)}{3!} + \dots

\frac{n!}{n!} = (_{0}^{n}) + (_{1}^{n}) + \dots (_{n}^{n}) = 2^{n}

Commented by Matica last updated on 26/Dec/22

p l e a s e m o r e d e t a i l

Commented by mahdipoor last updated on 26/Dec/22

f (0) = {(1 + 0)}^{n} = 1 = (_{0}^{n})

f^{^{'}} (0) = n {(1 + 0)}^{n - 1} = n = (_{1}^{n})

\frac{1}{2!} f^{^{″}} (0) = \frac{n (n - 1)}{2!} {(1 + 0)}^{n - 2} = (_{2}^{n})

\dots .

\frac{1}{n!} f^{n} (0) = \frac{n!}{n!} {(1 + 0)}^{n - n} = (_{n}^{n})

\Rightarrow A = (_{0}^{n}) + (_{1}^{n}) + \dots (_{n}^{n})

B = {(a + b)}^{n} = (_{0}^{n}) a^{n} b^{0} + (_{1}^{n}) a^{n - 1} b^{1} + (_{2}^{n}) a^{n - 2} b^{2} + \dots

+ (_{n}^{n}) a^{n - n} b^{n}

\Rightarrow a = b = 1 \Rightarrow A = B = {(1 + 1)}^{n} = 2^{n}

Commented by Matica last updated on 26/Dec/22

T h a n k y o u!

Answered by mr W last updated on 26/Dec/22

f (x) = {(1 + x)}^{n} = C_{0}^{n} + C_{1}^{n} x + C_{2}^{n} x^{2} + \dots + C_{n}^{n} x^{n} (i)

o n t h e o t h e r s i d e a c c . t a y l o r s e r i e s :

f (x) = f (0) + f^{'} (0) x + \frac{f^{″} (0)}{2!} x^{2} + \frac{f^{‴} (0)}{3!} x^{3} + \dots + \frac{f^{(n)} (0)}{n!} x^{n} + \frac{f^{(n + 1)} (0)}{(n + 1)!} x^{n + 1} + \dots (i i)

c o m p a r e (i) a n d (i i) :

\frac{f^{(k)} (0)}{k!} = C_{k}^{n} f o r 0 ⩽ k ⩽ n a n d

\frac{f^{(k)} (0)}{k!} = 0 f o r k ⩾ n + 1

\Rightarrow f (0) + f^{'} (0) x + \frac{f^{″} (0)}{2!} x^{2} + \frac{f^{‴} (0)}{3!} x^{3} + \dots + \frac{f^{(n)} (0)}{n!} x^{n} = {(1 + x)}^{n}

s e t x = 1,

\Rightarrow f (0) + f^{'} (0) + \frac{f^{″} (0)}{2!} + \frac{f^{‴} (0)}{3!} + \dots + \frac{f^{(n)} (0)}{n!} = {(1 + 1)}^{n} = 2^{n} ✓

Commented by Matica last updated on 26/Dec/22

T h a n k y o u!

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