f-x-y-f-x-f-y-xy-for-all-x-and-y-fromR-and-f-4-10-calculate-f-1319-

Question Number 145516 by mathmax by abdo last updated on 05/Jul/21

f (x + y) = f (x) + f (y) + xy for all x and y fromR

and f (4) = 10 calculate f (1319)

Answered by Olaf_Thorendsen last updated on 05/Jul/21

f (x + y) = f (x) + f (y) + x y

if x = y = 2

f (4) = 2 f (2) + 4 \Rightarrow f (2) = \frac{10 - 4}{2} = 3

if x = y = 1

f (2) = 2 f (1) + 1 \Rightarrow f (1) = \frac{3 - 1}{2} = 1

if x = y = 2^{n - 1}

f (2^{n}) = 2 f (2^{n - 1}) + 2^{2 n - 2}

f (2^{n}) = 2 (2 f (2^{n - 2}) + 2^{2 n - 4}) + 2^{2 n - 2}

f (2^{n}) = 2^{2} f (2^{n - 2}) + 2^{2 n - 2} + 2^{2 n - 3}

\dots

f (2^{n}) = 2^{n - 2} f (2^{2}) + \underset{k = 2}{\sum^{n - 1}} 2^{2 n - k}

f (2^{n}) = 10 . 2^{n - 2} + 2^{2 n} \underset{k = 2}{\sum^{n - 1}} {(\frac{1}{2})}^{k}

f (2^{n}) = 10 . 2^{n - 2} + 2^{2 n} \times \frac{1}{4} . \frac{1 - {(\frac{1}{2})}^{n - 2}}{1 - \frac{1}{2}}

f (2^{n}) = 10 . 2^{n - 2} + 2^{2 n} \frac{1 - {(\frac{1}{2})}^{n - 2}}{2}

f (2^{n}) = 10 . 2^{n - 2} + 2^{2 n - 1} - 2^{n + 1}

1319 = 1024 + 256 + 32 + 4 + 2 + 1

1319 = 2^{10} + 2^{8} + 2^{5} + 4 + 2 + 1

f (1024) = 10 . 2^{8} + 2^{19} - 2^{11} = 524800

f (256) = 10 . 2^{6} + 2^{15} - 2^{9} = 32896

f (32) = 10 . 2^{3} + 2^{9} - 2^{6} = 528

f (1024 + 256) = f (1280) = f (1024) + f (256) + 1024 \times 256

f (1280) = 819840

f (1280 + 32) = f (1312) = f (1280) + f (32) + 1280 \times 32

f (1312) = 861328

f (1312 + 4) = f (1316) = f (1312) + f (4) + 1312 \times 4

f (1316) = 866586

f (1316 + 2) = f (1318) = f (1316) + f (2) + 1316 \times 2

f (1318) = 869221

f (1318 + 1) = f (1319) = f (1318) + f (1) + 1318 \times 1

f (1319) = 870540

Commented by mathmax by abdo last updated on 05/Jul/21

thanks sir olaf

Answered by mathmax by abdo last updated on 05/Jul/21

we have f (2 + 2) = f (2) + f (2) + 4 \Rightarrow f (4) = 2 f (2) + 4 \Rightarrow 2 f (2) = 10 - 4 = 6

\Rightarrow f (2) = 3

f (2 + 1) = f (2) + f (1) + 2 = 5 + f (1) = f (3)

f (1 + 1) = 2 f (1) + 1 \Rightarrow 2 f (1) = f (2) - f (1) = 2 \Rightarrow f (1) = 1 \Rightarrow f (3) = 6

we have f (1) = 1, f (2) = 3 = \frac{2 (2 + 1)}{2}, f (3) = 6 = \frac{3 (3 + 1)}{2}

let suppose f (n) = \frac{n (n + 1)}{2} \Rightarrow f (n + 1) = f (n) + f (1) + n

= \frac{n (n + 1)}{2} + 1 + n = (n + 1) (\frac{n}{2} + 1) = \frac{(n + 1) (n + 2)}{2} \Rightarrow

\forall n \in N^{★} f (n) = \frac{n (n + 1)}{2} \Rightarrow

f (1319) = \frac{1}{2} (1319) \times (1320) = 660 \times 1319 = 66 \times 13190 = \dots