Let's analyze the given relations:
f(x+1x)=2f(x): This relation suggests a halving property.
If we substitute x=2n+12, we get: f(2n+12+12n+12)=2f(2n+12)
Simplifying the fraction inside the argument gives: f(2n+32)=2f(2n+12)
f(1−x)=1−f(x): This relation suggests a complementary property.
If we substitute x=21, we get: f(1−21)=1−f(21) Simplifying gives: f(21)=21
Now, let's consider the infinite series: S=f(32)+f(52)+f(72)+⋯+f(2n+12)+⋯
Using the halving property, we can rewrite this series as: S=2[f(52)+f(72)+⋯+f(2n+12)+⋯]
Notice that the terms inside the brackets are almost the same as the original series, except for the first term. So, we can write:
S=2(S−f(32))
Solving for S, we get: S=2f(32)
To find f(32), we can use the complementary property: f(1−32)=1−f(32) f(31)=1−f(32)
Now, using the halving property on f(31): f(31+131)=2f(31) f(41)=21−f(32)
We can continue this process to find a pattern: f(51)=21−21−f(32) f(61)=21−21−21−f(32)
We can see that this pattern converges to 21 as we keep applying the halving property. Therefore, f(31)=21.
Substituting this back into the equation for f(32), we get: 21=1−f(32) f(32)=21
Finally, substituting this value into the expression for S, we get: S=2⋅21=1
So, the value of the given infinite series is 1.
