Let f(x) be a real differentiable function such that f(0)=1 and f(x+y)=f(x)f′(y)+f′(x)f(y) for all x,y∈R. Then ∑n=1100logef(n) is equal to:
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Step-by-Step Solution
Step 1: Analyze the given functional equation
We are given a functional equation f(x+y)=f(x)f′(y)+f′(x)f(y) and the initial condition f(0)=1. To simplify the functional equation, we substitute x=0 into the equation. This allows us to relate f(y) and f'(y) using the given initial condition.
Step 2: Solve the differential equation
Rearranging the equation from the previous step, we get a first-order linear differential equation. Let k=1−f′(0). Then the equation becomes f′(y)=kf(y). This is a standard separable differential equation.
Step 3: Integrate to find f(y)
Integrating both sides of the differential equation f(y)f′(y)=k with respect to y, we get loge∣f(y)∣=ky+C. Exponentiating both sides gives f(y)=eky+C=eCeky. Let A=eC. So, f(y)=Aeky.
Step 4: Apply initial conditions to find constants
Using the initial condition f(0)=1, we find that A=1. So, f(y)=eky. Now we need to find the value of k. We know k=1−f′(0). Differentiating f(y)=eky gives f′(y)=keky. Substituting these back into the simplified functional equation f(y)=f′(y)+f′(0)f(y) leads to 1=k+f′(0), which is consistent with our definition of k. This means k can be any real constant, and f′(0) is determined by k. The problem statement implies a unique function, so we must re-examine the original functional equation.
Step 5: Re-evaluate using the original functional equation
Let's substitute f(x)=ekx and f′(x)=kekx into the original functional equation f(x+y)=f(x)f′(y)+f′(x)f(y). We find that ekxeky=2kekxeky. This implies 1=2k. Therefore, k=21.
Step 6: Determine f(n) and calculate the sum
With k=21, the function is f(x)=ex/2. Now we need to calculate logef(n), which simplifies to 2n. The sum ∑n=1100logef(n) becomes ∑n=11002n. This is an arithmetic series, and we can use the formula for the sum of the first N natural numbers, ∑n=1Nn=2N(N+1).