If f(x+y) + f(x-y) =2f(x)f(y) and f(0)=k then find f(x)?

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In summary, the given equation is f(x+y) + f(x-y) = 2f(x)f(y) and the initial condition is f(0) = k. The general solution for f(x) is f(x) = c*cos(x) + d*sin(x), where c and d are constants. To find the specific solution for f(x) with the given initial condition, we can substitute c = k and d = f'(0) into the general solution. This equation and initial condition can have infinitely many solutions due to the infinite possible values for c and d.
  • #1
4
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if
f(x+y) + f(x-y) =2f(x)f(y)
and f(0)=k
then find f(x)?

What should be approach followed to solve this question .
please ans it soon :blushing:
 
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  • #2
You have an expression with f(x) and f(y) and you only want f(x). So you have to get rid of y.

One more hint: that identity holds for all values of y.
 

1. What is the given equation and initial condition?

The given equation is f(x+y) + f(x-y) = 2f(x)f(y) and the initial condition is f(0) = k.

2. What is the general solution for f(x)?

The general solution for f(x) is f(x) = c*cos(x) + d*sin(x), where c and d are constants.

3. How do you use the initial condition to find the values of c and d?

Substituting x = 0 and f(0) = k into the general solution, we get k = c*cos(0) + d*sin(0) = c. This means that c = k. Substituting this into the general solution, we get f(x) = k*cos(x) + d*sin(x). To find d, we can differentiate both sides of the equation with respect to x and substitute x = 0 to get d = f'(0).

4. How do you find the specific solution for f(x) given the initial condition?

Substituting c = k and d = f'(0) into the general solution, we get f(x) = k*cos(x) + f'(0)*sin(x). This is the specific solution for f(x) with the given initial condition.

5. Can this equation and initial condition have multiple solutions?

Yes, this equation and initial condition can have infinitely many solutions because there are infinitely many possible values for c and d. Each pair of values for c and d will result in a different solution for f(x).

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