Question

f (x + 2) + f(x = √2 f(x+1), find the period of f(x)... please solve it ​

Answer 1

Answer:

Period of f(x) = 8.

Step-by-step explanation:

⇒ f(x+2) + f(x) = √2 f(x+1) ...(1)

Now all we have to do is play with the equation by replacing the value of x with appropriate value of x in the question which will help us eliminating the other terms and leaving us  with f(x)  and another extra term containing our period of f(x).

Replacing x by (x+2) we get :

⇒ f(x+4) + f(x+2) = √2 f(x+3)...(2)

Add (1) & (2) :

⇒ f(x) + 2f(x+2) + f(x+4) = √2 (f(x+1) + f(x+ 3)...(3)

We can write f(x=1) + f(x+3) as √2 f(x+2) using equation (1).

Put this above creatively obtained value in equation (3)

⇒ f(x) + 2 f(x+2) + f(x+4) = √2(√2 f(x+2))

⇒ f(x) + 2 f(x+2) + f(x+4) = 2 f(x+2)

⇒ f(x) + f(x+4) = 0...(eqn 4) [∵ using bhagwan kee dain..aankhein xD]

Now replace x by x+4

⇒ f(x+4) + f(x+8) = 0...(5)

Now let us subtract (4) and (5)

⇒ f(x) - f(x+8) = 0

⇒ f(x) = f(x+8)

So period of f(x) = 8.

So the above method can be used by Gangadhar,but i love character Shaktiman more,so let me allow you to introduce the SHAKTIMAN method xD but this method might feel difficult if you have no knowledge about complex number but that should not be the case because you seem to have started calculus...the very last portion of math. So let me move ahead.

What can be done here is that we can create a quadratic equation representing our functional equation.  

So the above functional equation can  be written as (in quadratic form) :

⇒ t² + 1 = √2t

Let us solve it and reach to final result first.

⇒ t² - √2t + 1 = 0

⇒ t =$\frac{-(-\sqrt{2})\pm\sqrt{4-2} }{2}$

⇒ t = $\frac{\sqrt{2}\pm\sqrt{-2} }{2}$

⇒ t = $\frac{1\pm i}{\sqrt{2} }$

⇒ t = e^(iπ/4)   OR  t = e^(-iπ/4)

(sorry for not writing this line in Latex,but I did write it in LATEX format earlier but it didn't render properly..this LaTeX never likes seeing me happy xD)

Now we need to think of the number to which t must be raised so as to obtain the answer to be 1.

⇒t^8 = e^[(i±π)/4]^8

Taking the positive and negative sign separately and solving,we find that t raised to 8 did the necessary work,so period = 8.

(I didn't solve it complete because I expect you to be better than me at calculation stuff... I don't support spoonfeeding because meko chamach se khana pasand nahi xD moreover,isse jaayda LaTeX mei likhne kee aukad nahi h humari, so aage wale 2 steps udaa diya xD if you don't get the solution..toh kuch nahi kr sakte hain,fees wapis maang lena 11th mei jaha bhi coaching gye the unse xD)