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Back to derivative basics (Posted on 2024-01-22) Difficulty: 3 of 5
Find all functions f:R+->R satisfying:

f(1)=1,
f has second derivative and
f'(1/x)=-(1/2)f(x).

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 2 of 2 |
The line "f has second derivative" is a clue.  So start by taking the derivative of the given functional equation to get:
-f"(1/x)/x^2 = -f'(x)/2

Then make the substitution x->1/x and simplify to get:
2x^2*f"(x) = f'(1/x)

Now plug this back into the original equation to get:
2x^2*f"(x) = -(1/2)f(x)

Now just a bit of cleanup to get an equation for a regular second-order differential equation:
4x^2*f"(x) + f(x) = 0

At this point I let Wolfram Alpha solve 4x^2*y" + y = 0 and got the general solution of:
f(x) = b*sqrt(x) + c*sqrt(x)*ln(x)
Also, then f'(x) = b*[(1/2)/sqrt(x)] + c*[(1/2)*ln(x)/sqrt(x) + 1/sqrt(x)]

We are given f(x)=1 Substitute x=1 into the original equation to get f'(1)=-1/2.
Then 1 = f(1) = b*sqrt(1) + c*sqrt(1)*ln(1)
and -1/2 = f'(1) = b*[(1/2)/sqrt(1)] + c*[(1/2)*ln(1)/sqrt(1) + 1/sqrt(1)]

These reduce to: 1 = f(1) = b
and: -1/2 = f'(1) = b*(1/2) + c
So then b=1 and c=-1

Then f(x) = sqrt(x) - sqrt(x)*ln(x)

Checking this 
f'(x) =  (1/2)/sqrt(x) - (1/2)*ln(x)/sqrt(x) - 1/sqrt(x)
Then f'(1/x) = (-1/2)*sqrt(x) - (1/2)*ln(x)*sqrt(x)
And (-1/2)*f(x) = (-1/2)*sqrt(x) - (1/2)*sqrt(x)*ln(x)
These match, so solution confirmed.

  Posted by Brian Smith on 2024-01-23 12:01:50
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