Whoa! Extreme!

Calculus Level 3

f ( x ) = x 1 0 100 + x 1 0 100 − 1 + x 1 0 100 − 2 + x 1 0 100 − 3 . . . + x 1 ( π e π e π e π e π ) − 1 ( 1 0 100 ) ! f(x) = \frac{x^{10^{100}} + x^{10^{100} - 1} + x^{10^{100} - 2} + x^{10^{100} - 3} ... + x^1}{(\pi e \pi e \pi e \pi e \pi)^{-1} (10^{100})!}

What's the googol'th derivative of f ( x ) f(x) ?

Details and assumptions:

  • g o o g o l = 1 0 100 googol = 10^{100}
  • Round to the nearest tenth.


The answer is 16708.1.

Brock Brown by Brock Brown , 25, USA

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1 solution

Brock Brown
Sep 29, 2015

If you take the n n th derivative of x A x^{A} for any positive natural A A , then...

  • if n = A n = A then the n n th derivative is A ! A! .
  • if n > A n > A then the n n th derivative is 0.

Because we're taking the googol'th derivative, we can actually get rid of every single instance of x x here. Everything in the sum evaluates to zero at the googol'th derivative, except for x 1 0 100 x^{10^{100}} , where we're left with a googol'th derivative of 1 0 100 ! 10^{100}! .

So, let's rephrase our function...

f ( x ) = ( π e π e π e π e π ) 1 1 0 100 ! [ x 1 0 100 + x 1 0 100 − 1 + x 1 0 100 − 2 + x 1 0 100 − 3 . . . + x 1 ] ⟹ f(x) = (\pi e \pi e \pi e \pi e \pi)\frac{1}{10^{100}!}[{x^{10^{100}} + x^{10^{100} - 1} + x^{10^{100} - 2}+ x^{10^{100}-3} ... + x^1}] \implies

f 1 0 100 ( x ) = π 5 e 4 1 0 100 ! × 1 0 100 ! = π 5 e 4 ≈ 16708.1 f^{10^{100}}(x) = \frac{\pi^{5} e^{4}}{10^{100}!} \times 10^{100}! = \pi^{5} e^{4} \approx \boxed{16708.1}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

You can sum up your bullets as:

d n d x n x A = A ! ( A − n ) ! x A − n \frac{d^n}{dx^n} x^A = \frac{A!}{(A-n)!}x^{A-n}

if we follow the convention ( − k ) ! = (-k)! = " ∞ \infty " (note the inverted apostrophes - ( − k ) ! (-k)! is actually undefined) for positive integral k k . That covers extra ground, too, like what happens when n < A n < A .

Jake Lai - 5 years, 8 months ago

WOAHHHHH!!! I DIDN'T THOUGHT OF THIS!

Pi Han Goh - 5 years, 8 months ago

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