Continued differentiation

Calculus Level 4

a 2 n x n − 1 / 2 \large \frac{a}{2^n x^{n - 1/2}}

Suppose we differentiate the function f ( x ) = x f(x) = \sqrt x many times, prove that the n th n^\text{th} derivative of f ( x ) f(x) can be stated as the expression above.

Give the last three digits for the value for a a , when n = 20 n = 20 .

Details and Assumptions : As an explicit example, if a = − 123456 a = - 123456 , then submit your answer as 456; if a = 654321 a = 654321 , then submit your answer as 321.


The answer is 375.

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Roel Baars by Roel Baars , 33, Netherlands

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1 solution

d n d x n x = 1 2 ( − 1 2 ) ( − 3 2 ) ( − 5 2 ) ⋯ ( − 2 n − 3 2 ) 1 x n − 1 / 2 = ( − 1 ) n + 1 ( 2 n − 3 ) ! ! 2 n x n − 1 / 2 ⟹ a = ( − 1 ) n + 1 ( 2 n − 3 ) ! ! \begin{aligned} \frac {d^n}{dx^n} \sqrt x & = \frac 12\left(-\frac 12\right) \left(-\frac 32\right) \left(-\frac 52\right) \cdots \left(-\frac {2n-3}2\right) \frac 1{x^{n-1/2}} \\ & = \frac {(-1)^{n+1}(2n-3)!!}{2^nx^{n-1/2}} \\ \implies a & = (-1)^{n+1} (2n-3)!! \end{aligned}

For n = 20 n=20 , a = − 37 ! ! a = -37!! and we need to find 37 ! ! m o d 1000 37!! \bmod 1000 . We note that 37 ! ! m o d 5 3 = 0 37!! \bmod 5^3 = 0 and we need to find 37 ! ! m o d 2 3 37!! \bmod 2^3 .

37 ! ! ≡ 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 11 ⋅ 13 ⋅ 15 ⋅ 17 ⋅ 19 ⋅ 21 ⋅ 23 ⋅ 25 ⋅ 27 ⋅ 29 ⋅ 31 ⋅ 33 ⋅ 35 ⋅ 37 (mod 8) ≡ 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 1 ⋅ 3 ⋅ 5 (mod 8) ≡ 1 ⋅ 15 ⋅ 7 ⋅ 1 ⋅ 15 ⋅ 7 ⋅ 1 ⋅ 15 ⋅ 7 ⋅ 1 ⋅ 15 ⋅ 7 ⋅ 1 ⋅ 15 (mod 8) ≡ 1 ( − 1 ) ( − 1 ) ⏟ = 1 1 ( − 1 ) ( − 1 ) ⏟ = 1 1 ( − 1 ) ( − 1 ) ⏟ = 1 1 ( − 1 ) ( − 1 ) ⏟ = 1 1 ( − 1 ) (mod 8) ≡ − 1 (mod 8) \begin{aligned} 37!! & \equiv 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19 \cdot 21 \cdot 23 \cdot 25 \cdot 27 \cdot 29 \cdot 31 \cdot 33 \cdot 35 \cdot 37 \text{ (mod 8)} \\ & \equiv 1 \cdot {\color{#3D99F6}3 \cdot 5} \cdot {\color{#D61F06}7} \cdot 1 \cdot {\color{#3D99F6}3 \cdot 5} \cdot {\color{#D61F06}7} \cdot 1 \cdot {\color{#3D99F6}3 \cdot 5} \cdot {\color{#D61F06}7} \cdot 1 \cdot {\color{#3D99F6}3 \cdot 5} \cdot {\color{#D61F06}7} \cdot 1 \cdot {\color{#3D99F6}3 \cdot 5} \text{ (mod 8)} \\ & \equiv 1 \cdot {\color{#3D99F6}15} \cdot {\color{#D61F06}7} \cdot 1 \cdot {\color{#3D99F6}15} \cdot {\color{#D61F06}7} \cdot 1 \cdot {\color{#3D99F6}15} \cdot {\color{#D61F06}7} \cdot 1 \cdot {\color{#3D99F6}15} \cdot {\color{#D61F06}7} \cdot 1 \cdot {\color{#3D99F6}15} \text{ (mod 8)} \\ & \equiv \underbrace{1 {\color{#3D99F6}(-1)} {\color{#D61F06}(-1)}}_{=1} \underbrace{1 {\color{#3D99F6}(-1)} {\color{#D61F06}(-1)}}_{=1} \underbrace{1 {\color{#3D99F6}(-1)} {\color{#D61F06}(-1)}}_{=1} \underbrace{1 {\color{#3D99F6}(-1)} {\color{#D61F06}(-1)}}_{=1} 1 {\color{#3D99F6}(-1)} \text{ (mod 8)} \\ & \equiv - 1 \text{ (mod 8)} \end{aligned}

Therefore, 37 ! ! ≡ 8 n − 1 37!! \equiv 8 n - 1 , where n n is an integer. And

8 n − 1 ≡ 0 (mod 5 3 ) 8 n ≡ 1 (mod 125) ⟹ n ≡ 47 ⟹ 37 ! ! ≡ 8 ∗ 47 − 1 ≡ 375 (mod 1000) \begin{aligned} 8n - 1 & \equiv 0 \text{ (mod }5^3) \\ 8n & \equiv 1 \text{ (mod 125)} \\ \implies n & \equiv 47 \\ \implies 37!! & \equiv 8*47 - 1 \equiv \boxed{375} \text{ (mod 1000)} \end{aligned}

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Sir There is typo , it should be 375 (final step)

Naren Bhandari - 2 years, 8 months ago

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Thanks. I have amended it.

Chew-Seong Cheong - 2 years, 8 months ago

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