How do we write $f(x+2)$ in terms of $f(x)$?
$$f : R^+ rightarrow R^+$$
$$f(x) = dfrac{x}{x+1} $$
How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.
Regards
functions
asked Dec 10 '18 at 5:42
Hamilton
1798
add a comment |
$$f : R^+ rightarrow R^+$$
$$f(x) = dfrac{x}{x+1} $$
How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.
Regards
functions
asked Dec 10 '18 at 5:42
Hamilton
1798
Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
– Lucas Henrique
Dec 10 '18 at 5:45
3
One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
– platty
Dec 10 '18 at 5:45
If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
– Lucas Henrique
Dec 10 '18 at 5:47
Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
– achille hui
Dec 10 '18 at 5:50
add a comment |
$$f : R^+ rightarrow R^+$$
$$f(x) = dfrac{x}{x+1} $$
How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.
Regards
functions
asked Dec 10 '18 at 5:42
Hamilton
1798
$$f : R^+ rightarrow R^+$$
$$f(x) = dfrac{x}{x+1} $$
How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.
Regards
functions
functions
asked Dec 10 '18 at 5:42
Hamilton
1798
asked Dec 10 '18 at 5:42
Hamilton
1798
asked Dec 10 '18 at 5:42
Hamilton
1798
asked Dec 10 '18 at 5:42
Hamilton
1798
asked Dec 10 '18 at 5:42
Hamilton
1798
1798
Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
– Lucas Henrique
Dec 10 '18 at 5:45
3
One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
– platty
Dec 10 '18 at 5:45
If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
– Lucas Henrique
Dec 10 '18 at 5:47
Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
– achille hui
Dec 10 '18 at 5:50
add a comment |
Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
– Lucas Henrique
Dec 10 '18 at 5:45
3
One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
– platty
Dec 10 '18 at 5:45
If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
– Lucas Henrique
Dec 10 '18 at 5:47
Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
– achille hui
Dec 10 '18 at 5:50
Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
– Lucas Henrique
Dec 10 '18 at 5:45
Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
– Lucas Henrique
Dec 10 '18 at 5:45
3
3
One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
– platty
Dec 10 '18 at 5:45
One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
– platty
Dec 10 '18 at 5:45
If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
– Lucas Henrique
Dec 10 '18 at 5:47
If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
– Lucas Henrique
Dec 10 '18 at 5:47
Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
– achille hui
Dec 10 '18 at 5:50
Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
– achille hui
Dec 10 '18 at 5:50
add a comment |
3 Answers
3
active
oldest
votes
I can't quite tell, but I think you're looking for a function $g$ so that
$$g(f(x))=f(x+2).$$
To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then
$$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$
So, if $f(x)=y$, then
$$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$
So, your answer is
$$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$
answered Dec 10 '18 at 5:51
Carl Schildkraut
11.2k11441
Oh...I was too slow typing the answer :)
– tonychow0929
Dec 10 '18 at 5:53
add a comment |
A more general method:
$$f(x) = frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$
Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.
answered Dec 10 '18 at 5:53
tonychow0929
29825
add a comment |
If I get your question correctly, then you can try this:
$$f(x+2) = frac{x+2} {x+3}$$
$$f(x) = frac{x} {x+1}$$
$$ x = x.f(x) + f(x)$$
$$implies x(1-f(x)) = f(x)$$
$$implies x = frac {f(x)}{1-f(x)}$$
Now substitute this $x$ in $f(x+2)$ and get the desired answer.
answered Dec 10 '18 at 5:53
PradyumanDixit
847214
2
Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
– tonychow0929
Dec 10 '18 at 5:56
1
@tonychow0929 I am sorry, thank you for that :)
– PradyumanDixit
Dec 10 '18 at 5:56
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
I can't quite tell, but I think you're looking for a function $g$ so that
$$g(f(x))=f(x+2).$$
To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then
$$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$
So, if $f(x)=y$, then
$$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$
So, your answer is
$$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$
answered Dec 10 '18 at 5:51
Carl Schildkraut
11.2k11441
Oh...I was too slow typing the answer :)
– tonychow0929
Dec 10 '18 at 5:53
add a comment |
I can't quite tell, but I think you're looking for a function $g$ so that
$$g(f(x))=f(x+2).$$
To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then
$$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$
So, if $f(x)=y$, then
$$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$
So, your answer is
$$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$
answered Dec 10 '18 at 5:51
Carl Schildkraut
11.2k11441
Oh...I was too slow typing the answer :)
– tonychow0929
Dec 10 '18 at 5:53
add a comment |
I can't quite tell, but I think you're looking for a function $g$ so that
$$g(f(x))=f(x+2).$$
To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then
$$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$
So, if $f(x)=y$, then
$$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$
So, your answer is
$$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$
answered Dec 10 '18 at 5:51
Carl Schildkraut
11.2k11441
I can't quite tell, but I think you're looking for a function $g$ so that
$$g(f(x))=f(x+2).$$
To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then
$$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$
So, if $f(x)=y$, then
$$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$
So, your answer is
$$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$
answered Dec 10 '18 at 5:51
Carl Schildkraut
11.2k11441
answered Dec 10 '18 at 5:51
Carl Schildkraut
11.2k11441
answered Dec 10 '18 at 5:51
Carl Schildkraut
11.2k11441
answered Dec 10 '18 at 5:51
Carl Schildkraut
11.2k11441
11.2k11441
Oh...I was too slow typing the answer :)
– tonychow0929
Dec 10 '18 at 5:53
add a comment |
Oh...I was too slow typing the answer :)
– tonychow0929
Dec 10 '18 at 5:53
Oh...I was too slow typing the answer :)
– tonychow0929
Dec 10 '18 at 5:53
Oh...I was too slow typing the answer :)
– tonychow0929
Dec 10 '18 at 5:53
add a comment |
A more general method:
$$f(x) = frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$
Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.
answered Dec 10 '18 at 5:53
tonychow0929
29825
add a comment |
A more general method:
$$f(x) = frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$
Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.
answered Dec 10 '18 at 5:53
tonychow0929
29825
add a comment |
A more general method:
$$f(x) = frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$
Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.
answered Dec 10 '18 at 5:53
tonychow0929
29825
A more general method:
$$f(x) = frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$
Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.
answered Dec 10 '18 at 5:53
tonychow0929
29825
answered Dec 10 '18 at 5:53
tonychow0929
29825
answered Dec 10 '18 at 5:53
tonychow0929
29825
answered Dec 10 '18 at 5:53
tonychow0929
29825
29825
add a comment |
add a comment |
If I get your question correctly, then you can try this:
$$f(x+2) = frac{x+2} {x+3}$$
$$f(x) = frac{x} {x+1}$$
$$ x = x.f(x) + f(x)$$
$$implies x(1-f(x)) = f(x)$$
$$implies x = frac {f(x)}{1-f(x)}$$
Now substitute this $x$ in $f(x+2)$ and get the desired answer.
answered Dec 10 '18 at 5:53
PradyumanDixit
847214
2
Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
– tonychow0929
Dec 10 '18 at 5:56
1
@tonychow0929 I am sorry, thank you for that :)
– PradyumanDixit
Dec 10 '18 at 5:56
add a comment |
If I get your question correctly, then you can try this:
$$f(x+2) = frac{x+2} {x+3}$$
$$f(x) = frac{x} {x+1}$$
$$ x = x.f(x) + f(x)$$
$$implies x(1-f(x)) = f(x)$$
$$implies x = frac {f(x)}{1-f(x)}$$
Now substitute this $x$ in $f(x+2)$ and get the desired answer.
answered Dec 10 '18 at 5:53
PradyumanDixit
847214
2
Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
– tonychow0929
Dec 10 '18 at 5:56
1
@tonychow0929 I am sorry, thank you for that :)
– PradyumanDixit
Dec 10 '18 at 5:56
add a comment |
If I get your question correctly, then you can try this:
$$f(x+2) = frac{x+2} {x+3}$$
$$f(x) = frac{x} {x+1}$$
$$ x = x.f(x) + f(x)$$
$$implies x(1-f(x)) = f(x)$$
$$implies x = frac {f(x)}{1-f(x)}$$
Now substitute this $x$ in $f(x+2)$ and get the desired answer.
answered Dec 10 '18 at 5:53
PradyumanDixit
847214
If I get your question correctly, then you can try this:
$$f(x+2) = frac{x+2} {x+3}$$
$$f(x) = frac{x} {x+1}$$
$$ x = x.f(x) + f(x)$$
$$implies x(1-f(x)) = f(x)$$
$$implies x = frac {f(x)}{1-f(x)}$$
Now substitute this $x$ in $f(x+2)$ and get the desired answer.
answered Dec 10 '18 at 5:53
PradyumanDixit
847214
edited Dec 10 '18 at 5:57
answered Dec 10 '18 at 5:53
PradyumanDixit
847214
answered Dec 10 '18 at 5:53
PradyumanDixit
847214
answered Dec 10 '18 at 5:53
PradyumanDixit
847214
847214
2
Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
– tonychow0929
Dec 10 '18 at 5:56
1
@tonychow0929 I am sorry, thank you for that :)
– PradyumanDixit
Dec 10 '18 at 5:56
add a comment |
2
Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
– tonychow0929
Dec 10 '18 at 5:56
1
@tonychow0929 I am sorry, thank you for that :)
– PradyumanDixit
Dec 10 '18 at 5:56
2
2
Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
– tonychow0929
Dec 10 '18 at 5:56
Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
– tonychow0929
Dec 10 '18 at 5:56
1
1
@tonychow0929 I am sorry, thank you for that :)
– PradyumanDixit
Dec 10 '18 at 5:56
@tonychow0929 I am sorry, thank you for that :)
– PradyumanDixit
Dec 10 '18 at 5:56
add a comment |
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Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
– Lucas Henrique
Dec 10 '18 at 5:45
3
One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
– platty
Dec 10 '18 at 5:45
If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
– Lucas Henrique
Dec 10 '18 at 5:47
Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
– achille hui
Dec 10 '18 at 5:50