A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

zmudz

  • one year ago

The function \(f(n)\)takes the integers to the real numbers such that \(f(m + n) + f(m - n) = 2f(m) + 2f(n)\) for all integers \(m\) and \(n\) and \(f(4) = 16\). Find \(f(n)\).

  • This Question is Closed
  1. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    so far I have come up with a value for f(0) and I came up with the function f is even: how I did this: ... m+n=4 m-n=4 ------ 2m=8 m=4 when n=0 so \[f(4+0)+f(4-0)=2f(4)+2f(0) \\ f(4)+f(4)=2f(4)+2 f(0) \\ 0=2 f(0) \\ f(0)=0 \\ \text{ now I set } m=0 \\ \text{ and so we have } \\ f(0+n)+f(0-n)=2f(0)+2f(n) \\ f(n)+f(-n)=0+2f(n) \\ f(-n)=f(n) \text{ which says } f \text{ is even }\] so maybe we can use these two facts somehow or maybe not

  2. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    me too, i got what you got but didn't see the link of f(4) =16 to the problem :(

  3. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well we can come with f(8) using that fact with the already found stuff

  4. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[f(4+4)+f(4-4)=2f(4)+2f(4) \\ f(8)+f(0)=4 f(4) \\ f(8)+0=4(16) \\ f(8)=64\]

  5. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    It seems the function is f(x) = x^2

  6. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    lol let's test \[(m+n)^2+(m-n)^2=2m^2+2n^2 \\ m^2+2nm+n^2+m^2-2nm+n^2=2m^2+2n^2 \\ \] the equation holds for f(x)=x^2

  7. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You are well organized. I just look at f(0) =0 , f(4) =16, f(8) =64 to conclude that f(x) =x^2

  8. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh, we are done. hahaha. it asked us to find f(n) and we get f(x) = x^2 hence f(n) = n^2 right?

  9. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah!! I think so, with your argument!!! we prove that f(n) = n^2 . And by your testing, we have the expression hold for any m, n in Z

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.