Function whose derivative is itself, but

Calculus Level 4

Let there be a function f ( x ) f(x) such that

f ( x ) = f ( x 1 ) f'(x) = f(x-1)

If f ( π ) = [ W ( 1 ) ] π f(\pi) = [W(1)]^{-\pi} , where W ( x ) W(x) represents the Lambert's W function determine 1000 × f ( 1 Ω ) \lfloor 1000 \times f(\frac{1}{\Omega}) \rfloor , where Ω \Omega is the Omega constant .


The answer is 2718.

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1 solution

Efren Medallo
May 7, 2017

First we take that the solution to such functional differential equation is of the form

f ( x ) = c 1 e A x f(x) = c_1 e^{Ax}

for some unknown constant A A and some arbitrary constant c 1 c_1 .

Substituting that to the said differential equation,

c 1 A e A x = c 1 e A ( x 1 ) c_1 Ae^{Ax} = c_1 e^{A(x-1)}

c 1 A e A x = c 1 e A e A x c_1 Ae^{Ax} = c_1 e^{-A} e^{Ax}

A = e A A = e^{-A}

A e A = 1 Ae^A =1

A = W ( 1 ) = Ω 0.567143 A= W(1) = \Omega \approx 0.567143

Where Ω \Omega is the Omega Constant . This means our function is

f ( x ) = c 1 e W ( 1 ) x f(x) = c_1 e^{W(1) x}

Now, the Lambert's W function has the property such that

e W ( x ) = 1 W ( x ) e^{W(x)} = \frac{1}{W(x)}

So we can rephrase the function as

f ( x ) = c 1 [ W ( 1 ) ] x f(x) = c_1 [W(1)]^{-x}

From which we can see that since f ( π ) = [ W ( 1 ) ] π f(\pi) = [W(1)]^{-\pi} , so that makes c 1 = 1 c_1 =1 .

From here we can now determine f ( 1 Ω ) f(\frac{1}{\Omega}) .

f ( 1 Ω ) = e Ω × 1 Ω = e f(\frac{1}{\Omega}) = e^{\Omega \times \frac{1}{\Omega}} = e

And thus, 1000 × e = 2718 \lfloor 1000 \times e \rfloor = 2718 .

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