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mathmath333
 one year ago
Functions
mathmath333
 one year ago
Functions

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align}& f:\mathbb{R}\rightarrow \mathbb{R} ,\ \{x,y\} \in \mathbb{R} \hspace{1.5em}\\~\\ & f(x+f(y))=f(x)+y1, \ f(0)=1 , f(1)= ?\hspace{1.5em}\\~\\ \end{align}}\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if y is 0 then we have \[f(x+f(0))=f(x)+01 \\ f(x+1)=f(x)1 \text{ since } f(0)=1 \\ \text{ now enter in } x \text{ as 1 }\]

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.2I'm thinking that since functions have only one output for any given input, and we are dealing with a function f that can take an argument x and y, that f is a bijection. Therefore, for every input there is exactly one output and for every output there is exactly one input. So f(y) must equal to x: $$ f(x+f(y))=f(x)+y1\\ f(x+x)=f(x)+f(x)1\\ f(2x)=2f(x)1\\ f(x)=\cfrac{f(2x)+1}{2} $$ f(0)=1 Let's check $$ f(0)=1=\cfrac{f(2\cdot 0)+1}{2}=\cfrac{f(0)+1}{2}=1\\ $$ So $$ f(1)=\cfrac{f(2)+1}{2}\\ f(2)=\cfrac{f(4)+1}{2}\\ ... $$ Not sure if this is the right path, but we can probably find a solution to the recursive equation...

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.2@freckles path sounds more sane and much easier

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0so we just solve this type of questions by putting random numeric value

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ones that will help :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.2like we know f(0)=1 so wanted to try to use that and then we knew we wanted to find f(1) so I play with some 0's and 1's or what would the insides of f( ) 0 or 1 but I played with f(0) first since I knew its output
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