f ( x ) = e x − e − x 2 ( 1 + ∫ 1 x f ( t ) d t )
Suppose a function f defined on x > 0 satisfy the equation above, find the value of ⌊ 1 0 0 ⋅ a → ∞ lim ∫ a 1 a f ( t ) d t ⌋ .
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Good problem! My solution takes the derivative of both sides of the original functional equation where I obtained the ODE:
f'(x) + [[cosh(x) -1]/sinh(x)]*f(x) = 0; f(1) = csch(1)
which yields the function:
f(x) = [coth(1/2)]*[tanh(x/2) / sinh(x)].
Using this function in the following "floored" limit still gives me 216 for a final result!
Same as i DID....(+1)
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