What are functions that aren't constant and have properties 1),2)?

1)f(m+n)+f(m-n)=2(f(m)+f(n))

2)f(1)+f(2)+---+f(15)<=1987

f defined on domain N and with range N

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Considering `m = n = 0` and replacing 0 for m and n in the first condition provided by the problem, yields:

`f(0+0)+f(0-0)=2(f(0)+f(0)) => 2f(0) = 2f(0) = 0`

Considering `f(1) = a ` and the function `g(m) = f(m) - f(m-1)` yields:

`n = 1 => f(m+1) + f(m-1) = 2f(m) + 2*f(1)`

`f(m+1) + f(m-1) = 2f(m) + 2*a => f(m+1) + f(m-1) = f(m) + f(m) + 2*a`

`f(m+1) - f(m) = f(m) - f(m-1) + 2a`

Replacing `g(m+1)` for `f(m+1) - f(m)` and `g(m)` for `f(m) - f(m-1)` , yields:

`g(m+1) = g(m) + 2a`

The relation `g(m+1) = g(m) + 2a` proves that the function g(m) is an arithmetic progression whose common difference is `d = 2a` .

Using the formula that gives the general term of an arithmetic progression, yields:

`g(m) = g(1) + (m-1)*2a`

Since `g(1) = f(1) - f(0) = f(1) - 0 = a` yields:

`g(m) = a + (m-1)*2a `

Factoring out a yields:

`g(m) = a(1 + 2m - 2) =>g(m) = a(2m - 1)`

Using the relation `f(k) - f(k-1) = (2k - 1)*a` and giving values to `k` from `1` to `m` , yields:

`k = 1 => f(1) - f(0) = a`

`k = 2 => f(2) - f(1) = (4 - 1)*a => f(2) - a = 3a => f(2) = 4a = 2^2a`

`k = 3 => f(3) - f(2) = (6 - 1)*a => f(3) - 4a = 5a => f(3) = 9a = 3^2a`

.................

`k = m => f(m) = m^2*a`

You need to use the property 2 of the function, such that:

`f(1) + ... + f(15) <= 1987`

Using the fact that `f(m) = m^2*a` yields:

`a + 2^2a + 3^2a + ... + 15^2*a < = 1987`

`a(1 + 2^2 + ... + 15^2) <= 1987`

`a*(15(15+1)(2*15+1))/6 <= 1987`

`a*15*16*31/6 <= 1987`

`a*15*16*31 <= 6*1987 => 7440*a<= 11992`

If `a = 1` , the inequality `7440*1<= 11992` holds

For `a = 2,` the inequality `7440*2 = 14880 <= 11992` becomes invalid.

Hence, `a = 1` and the function becomes `f(n) = n^2` .

**Hence, evaluating the required function, under the given conditions, yields `f(n) = n^2` .**

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