Christmas Streak 28/88: Similar But Different

Calculus Level 5

Define P \mathbb{P} as the set of positive reals, and R \mathbb{R} as the set of reals.

Two functions f − : P → R f^-:\mathbb{P}\to\mathbb{R} and f + : P → R f^+:\mathbb{P}\to\mathbb{R} both have their second derivatives well-defined in their domains.

{ f − ( x ) = ( x − 1 ) e x − 1 + ∫ 1 x 2 t − f − ( t ) t d t ; f + ( x ) = ( x − 1 ) e x − 1 + ∫ 1 x 2 t + f + ( t ) t d t . \cases{\displaystyle f^-(x)=(x-1)e^{x-1}+\int_{1}^{x}\frac{2t-f^-(t)}{t}dt; \\\\ \displaystyle f^+(x)=(x-1)e^{x-1}+\int_{1}^{x} \frac{2t+f^+(t)}{t}dt.}

f + ( 2 ) + f − ( 2 ) = a ⋅ e + ln ⁡ b + c , f^+(2)+f^-(2)=a\cdot e+\ln b+c, for some integers a , b a,~b and c . c. Find the value of a 3 + b 3 + c 3 . a^3+b^3+c^3.

This problem is a part of <Christmas Streak 2017> series .


The answer is 4122.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Oct 26, 2017

Let f − = g f^-=g and f + = h f^+=h for clarity.

g ( x ) = ( x − 1 ) e x − 1 + ∫ 1 x 2 t − g ( t ) t d t \displaystyle g(x)=(x-1)e^{x-1}+\int_{1}^{x} \frac{2t-g(t)}{t}dt

From this we see that g ( 1 ) = 0. g(1)=0. Differentiate both sides with respect to x . x.

g ′ ( x ) = x e x − 1 + 2 x − g ( x ) x g'(x)=xe^{x-1}+\dfrac{2x-g(x)}{x}

x g ′ ( x ) = x 2 e x − 1 + 2 x − g ( x ) xg'(x)=x^2e^{x-1}+2x-g(x)

x g ′ ( x ) + g ( x ) = x 2 e x − 1 + 2 x xg'(x)+g(x)=x^2e^{x-1}+2x

Integrate both sides, using the fact that ( x g ( x ) ) ′ = x g ′ ( x ) + g ( x ) . (xg(x))'=xg'(x)+g(x).

x g ( x ) = ( x 2 − 2 x + 2 ) e x − 1 + x 2 + C 1 xg(x)=(x^2-2x+2)e^{x-1}+x^2+C_1

Now substitute x = 1 x=1 and use that g ( 1 ) = 0. ⇒ C 1 = − 2. g(1)=0.~\Rightarrow~C_1=-2.

Therefore x g ( x ) = ( x 2 − 2 x + 2 ) e x − 1 + x 2 − 2. xg(x)=(x^2-2x+2)e^{x-1}+x^2-2.

∴ g ( x ) = x 2 − 2 x + 2 x e x − 1 + x − 2 x . \therefore~g(x)=\dfrac{x^2-2x+2}{x}e^{x-1}+x-\dfrac{2}{x}.

h ( x ) = ( x − 1 ) e x − 1 + ∫ 1 x 2 t + h ( t ) t d t \displaystyle h(x)=(x-1)e^{x-1}+\int_{1}^{x} \frac{2t+h(t)}{t}dt

From this we see that h ( 1 ) = 0. h(1)=0. Differentiate both sides with respect to x . x.

h ′ ( x ) = x e x − 1 + 2 x + h ( x ) x h'(x)=xe^{x-1}+\dfrac{2x+h(x)}{x}

x h ′ ( x ) = x 2 e x − 1 + 2 x + h ( x ) xh'(x)=x^2e^{x-1}+2x+h(x)

What's different here is that we'll differentiate both sides this time.

Before doing that, substitute x = 1 x=1 and notice that h ′ ( 1 ) = 3 + h ( 1 ) = 3. h'(1)=3+h(1)=3.

x h ′ ′ ( x ) + h ′ ( x ) = ( x 2 + 2 x ) e x − 1 + 2 + h ′ ( x ) xh''(x)+h'(x)=(x^2+2x)e^{x-1}+2+h'(x)

h ′ ′ ( x ) = ( x + 2 ) e x − 1 + 2 x h''(x)=(x+2)e^{x-1}+\dfrac{2}{x}

Integrate both sides.

h ′ ( x ) = ( x + 1 ) e x − 1 + 2 ln ⁡ x + C 2 h'(x)=(x+1)e^{x-1}+2\ln x + C_2

Substitute x = 1 x=1 and use that h ′ ( 1 ) = 3. ⇒ C 2 = 1. h'(1)=3.~\Rightarrow~C_2=1.

Integrate again to get h ( x ) = x e x − 1 + 2 x ( ln ⁡ x − 1 ) + x + C 3 . h(x)=xe^{x-1}+2x(\ln x -1)+x+C_3.

Substitute x = 1 x=1 and use that h ( 1 ) = 0. ⇒ C 3 = 0. h(1)=0.~\Rightarrow~C_3=0.

∴ h ( x ) = x e x − 1 + 2 x ln ⁡ x − x . \therefore~h(x)=xe^{x-1}+2x\ln x - x.

From above, we see that g ( 2 ) + h ( 2 ) = ( e + 1 ) + ( 2 e + 4 ln ⁡ 2 − 2 ) = 3 e + ln ⁡ 16 − 1. g(2)+h(2)=(e+1)+(2e+4\ln 2-2)=3e+\ln 16 - 1.

∴ a 3 + b 3 + c 3 = 3 3 + 1 6 3 + ( − 1 ) 3 = 4122 . \therefore~a^3+b^3+c^3=3^3+16^3+(-1)^3=\boxed{4122}.

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