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mathmath333

  • one year ago

Functions

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& f:\mathbb{R}\rightarrow \mathbb{R} ,\ \{x,y\} \in \mathbb{R} \hspace{1.5em}\\~\\ & f(x+f(y))=f(x)+y-1, \ f(0)=1 , f(-1)= ?\hspace{1.5em}\\~\\ \end{align}}\)

  2. freckles
    • one year ago
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    if y is 0 then we have \[f(x+f(0))=f(x)+0-1 \\ f(x+1)=f(x)-1 \text{ since } f(0)=1 \\ \text{ now enter in } x \text{ as -1 }\]

  3. ybarrap
    • one year ago
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    I'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: $$ f(x+f(y))=f(x)+y-1\\ f(x+x)=f(x)+f(x)-1\\ f(2x)=2f(x)-1\\ f(x)=\cfrac{f(2x)+1}{2} $$ 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...

  4. ybarrap
    • one year ago
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    @freckles path sounds more sane and much easier

  5. mathmath333
    • one year ago
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    wow , so quick

  6. mathmath333
    • one year ago
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    so we just solve this type of questions by putting random numeric value

  7. freckles
    • one year ago
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    ones that will help :p

  8. freckles
    • one year ago
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    like 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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