Differntiation...and a bit of series

Calculus Level 3

If f ( x ) = ( 1 − x ) n f(x)=(1-x)^n , then find the value of the expression below:

f ( 0 ) + f ′ ( 0 ) + f ′ ′ ( 0 ) 2 ! + f ′ ′ ′ ( 0 ) 3 ! + ⋯ + f ( n ) ( 0 ) n ! \large f(0)+f'(0)+\frac {f''(0)}{2!} + \frac {f'''(0)}{3!} + \cdots + \frac {f^{(n)}(0)}{n!}

Notation: f ( k ) f^{(k)} denotes the k k th derivative of f ( x ) f(x) .

0 None of the others 2 n − 1 2^{n-1} 2 n 2^n

Aakhyat Singh by Aakhyat Singh , 20, India

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2 solutions

g ( x ) = f ( x ) + f ′ ( x ) + f ′ ′ ( x ) 2 ! + f ′ ′ ′ ( x ) 3 ! + ⋯ + f ( n ) ( x ) n ! = ( 1 − x ) n 0 ! + − n ( 1 − x ) n − 1 1 ! + n ( n − 1 ) ( 1 − x ) n − 2 2 ! + − n ( n − 1 ) ( n − 2 ) ( 1 − x ) n − 3 3 ! + ⋯ + ( − 1 ) n n ( n − 1 ) ⋯ 3 ⋅ 2 ⋅ 1 ( 1 − x ) 0 n ! = ∑ k = 0 n ( − 1 ) k n ! k ! ( n − k ) ! ( 1 − x ) n − k = ∑ k = 0 n ( − 1 ) n ( n k ) ( 1 − x ) n − k = ∑ k = 0 n ( − 1 ) n ( n k ) ( 1 − x ) k = ( 1 − ( 1 − x ) ) n = x n \begin{aligned} g(x) & = f(x) + f'(x) + \frac {f''(x)}{2!} + \frac {f'''(x)}{3!} + \cdots + \frac {f^{(n)}(x)}{n!} \\ & = \frac {(1-x)^n}{0!} + \frac {-n(1-x)^{n-1}}{1!} + \frac {n(n-1)(1-x)^{n-2}}{2!} + \frac {-n(n-1)(n-2)(1-x)^{n-3}}{3!} + \cdots + \frac {(-1)^nn(n-1)\cdots 3 \cdot 2 \cdot 1(1-x)^0}{n!} \\ & = \sum_{k=0}^n \frac {(-1)^k n!}{k!(n-k)!} (1-x)^{n-k} = \sum_{k=0}^n (-1)^n {n \choose k} (1-x)^{n-k} = \sum_{k=0}^n (-1)^n {n \choose k} (1-x)^k \\ & = (1-(1-x))^n = x^n \end{aligned}

⟹ g ( 0 ) = 0 n = 0 \implies g(0) = 0^n = \boxed{0}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

It should be ∑ k = 0 n ( − 1 ) k ( n k ) ( 1 − x ) n − k = ∑ k = 0 n ( − 1 ) n − k ( n k ) ( 1 − x ) k = ( − 1 + 1 − x ) n = ( − x ) n \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} (1 - x)^{n - k} = \sum_{k=0}^{n} (-1)^{n - k} \binom{n}{k} (1 - x)^{k} = (-1 + 1 - x)^{n} = (-x)^{n} . Hence, g ( x ) = ( − x ) n g(x) = (-x)^{n} . This does not affect the answer though.

Krutarth Patel - 4 months, 1 week ago
Aravind V
Oct 10, 2017

We can see that the expression that we want is 1-nC0+nC1-nC2+...(-1)^(n-1) x nCn which is the binomial expansion of (1-x)^n when x is 1. So since the expression and f(x) are equal at x=1, by putting x=1 in f(x) gives the answer as 0

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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