Let f: R→R be a continuous function satisfying f(0) = 1 and f(2x) - f(x) = x for all x ∈R. If limn→∞[f(x)−f(x/2n)]=G(x) then ∑r=110G(r2) is equal to:
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Step-by-Step Solution
Step 1: Express f(x) in terms of f(x/ 2n)
We are given the functional equation f(2x)−f(x)=x. We can rewrite this as f(y)−f(y/2)=y/2 by substituting y=2x. Using this, we can express the difference f(x)−f(x/2n) as a telescoping sum.
Step 2: Substitute the functional equation into the sum
For each term in the sum, we apply the relation f(y)−f(y/2)=y/2. Here, y=x/2k−1, so y/2=x/2k. Thus, each term f(x/2k−1)−f(x/2k) becomes x/2k.
Step 3: Evaluate the sum
The sum ∑k=1n(1/2)k is a geometric series with first term a=1/2 and common ratio r=1/2. The sum of the first n terms of a geometric series is given by a(1−rn)/(1−r).
Step 4: Find G(x) by taking the limit
As n→∞, the term (1/2)n approaches 0. Therefore, G(x) simplifies to x. The continuity of f and f(0)=1 are consistent with this result, as f(x/2n)→f(0)=1 as n→∞.
Step 5: Calculate the final sum
Since G(x)=x, we have G(r2)=r2. We need to calculate the sum of the first 10 squares. The formula for the sum of the first N squares is N(N+1)(2N+1)/6.
Step 6: Apply the sum of squares formula
Substitute N=10 into the formula for the sum of the first N squares to get the final result.