Are we allowed, by the definition of induction, to assume that $f_{1}(x)$ still has the same formula, even...
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I'm pretty sure i'm supposed to demonstate this by induction but i think there is a problem with the way the proposition is stated that impossibilitates that, I need to prove that
For all natural number $n$, $f_{0}(x) = frac{x}{x+1} & f_{n+1}(x)=(f_{0}(f_{n}(x))) implies f_{n}(x) = frac{x}{(n+1)x + 1}$
So if we first verify it for n=0 then the explicit formula for $f_{n}$ is given. But then, if we try to verify it for any other $n$, let's say $n=1$, we would have the formula for $f_{0}(x)$ as always, and we would also have that $f_{2}(x) = f_{0}(f_{1}(x))$ but how is that suppose to lead us to conclude the explicit formula for $f_{1}$ if we don't know what $f_{2}(x)$ looks like?
Did I get induction wrong?
abstract-algebra algebra-precalculus proof-writing induction
asked Nov 18 at 10:49
ArielK
195
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show 1 more comment
up vote
0
down vote
favorite
I'm pretty sure i'm supposed to demonstate this by induction but i think there is a problem with the way the proposition is stated that impossibilitates that, I need to prove that
For all natural number $n$, $f_{0}(x) = frac{x}{x+1} & f_{n+1}(x)=(f_{0}(f_{n}(x))) implies f_{n}(x) = frac{x}{(n+1)x + 1}$
So if we first verify it for n=0 then the explicit formula for $f_{n}$ is given. But then, if we try to verify it for any other $n$, let's say $n=1$, we would have the formula for $f_{0}(x)$ as always, and we would also have that $f_{2}(x) = f_{0}(f_{1}(x))$ but how is that suppose to lead us to conclude the explicit formula for $f_{1}$ if we don't know what $f_{2}(x)$ looks like?
Did I get induction wrong?
abstract-algebra algebra-precalculus proof-writing induction
asked Nov 18 at 10:49
ArielK
195
1
Note that $f_1(x) = f_0(f_0(x))$.
– Fabio Somenzi
Nov 18 at 11:14
Yes but only when n = 0 right? How can that still hold if we consider other values of n?
– ArielK
Nov 18 at 11:15
1
Yes, but that's not a problem. You can verify that $f_1(x) = frac{x}{2x+1}$. That's not the inductive proof, but the inductive step is not much different.
– Fabio Somenzi
Nov 18 at 11:19
1
The induction step is like this: let $f_{n+1}(x) = f_0(f_n(x))$ and suppose $f_n(x)=frac{x}{(n+1)x+1}$. Prove that $f_{n+1}(x) = frac{x}{(n+2)x+1}$. It's a matter of plugging the hypothesis for $f_n(x)$ into the definition of $f_{n+1}(x)$ and simplifying.
– Fabio Somenzi
Nov 18 at 11:27
1
The main problem with this problem is that your $f_0$ should be called $f_1$ to begin with. In this way $f_0$ is the identity, and $f_{m+n}=f^{circ m}circ f^{circ n}$, etcetera.
– Christian Blatter
Nov 18 at 11:55
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm pretty sure i'm supposed to demonstate this by induction but i think there is a problem with the way the proposition is stated that impossibilitates that, I need to prove that
For all natural number $n$, $f_{0}(x) = frac{x}{x+1} & f_{n+1}(x)=(f_{0}(f_{n}(x))) implies f_{n}(x) = frac{x}{(n+1)x + 1}$
So if we first verify it for n=0 then the explicit formula for $f_{n}$ is given. But then, if we try to verify it for any other $n$, let's say $n=1$, we would have the formula for $f_{0}(x)$ as always, and we would also have that $f_{2}(x) = f_{0}(f_{1}(x))$ but how is that suppose to lead us to conclude the explicit formula for $f_{1}$ if we don't know what $f_{2}(x)$ looks like?
Did I get induction wrong?
abstract-algebra algebra-precalculus proof-writing induction
asked Nov 18 at 10:49
ArielK
195
I'm pretty sure i'm supposed to demonstate this by induction but i think there is a problem with the way the proposition is stated that impossibilitates that, I need to prove that
For all natural number $n$, $f_{0}(x) = frac{x}{x+1} & f_{n+1}(x)=(f_{0}(f_{n}(x))) implies f_{n}(x) = frac{x}{(n+1)x + 1}$
So if we first verify it for n=0 then the explicit formula for $f_{n}$ is given. But then, if we try to verify it for any other $n$, let's say $n=1$, we would have the formula for $f_{0}(x)$ as always, and we would also have that $f_{2}(x) = f_{0}(f_{1}(x))$ but how is that suppose to lead us to conclude the explicit formula for $f_{1}$ if we don't know what $f_{2}(x)$ looks like?
Did I get induction wrong?
abstract-algebra algebra-precalculus proof-writing induction
abstract-algebra algebra-precalculus proof-writing induction
asked Nov 18 at 10:49
ArielK
195
asked Nov 18 at 10:49
ArielK
195
edited Nov 18 at 11:08
asked Nov 18 at 10:49
ArielK
195
asked Nov 18 at 10:49
ArielK
195
asked Nov 18 at 10:49
ArielK
195
195
1
Note that $f_1(x) = f_0(f_0(x))$.
– Fabio Somenzi
Nov 18 at 11:14
Yes but only when n = 0 right? How can that still hold if we consider other values of n?
– ArielK
Nov 18 at 11:15
1
Yes, but that's not a problem. You can verify that $f_1(x) = frac{x}{2x+1}$. That's not the inductive proof, but the inductive step is not much different.
– Fabio Somenzi
Nov 18 at 11:19
1
The induction step is like this: let $f_{n+1}(x) = f_0(f_n(x))$ and suppose $f_n(x)=frac{x}{(n+1)x+1}$. Prove that $f_{n+1}(x) = frac{x}{(n+2)x+1}$. It's a matter of plugging the hypothesis for $f_n(x)$ into the definition of $f_{n+1}(x)$ and simplifying.
– Fabio Somenzi
Nov 18 at 11:27
1
The main problem with this problem is that your $f_0$ should be called $f_1$ to begin with. In this way $f_0$ is the identity, and $f_{m+n}=f^{circ m}circ f^{circ n}$, etcetera.
– Christian Blatter
Nov 18 at 11:55
|
show 1 more comment
1
Note that $f_1(x) = f_0(f_0(x))$.
– Fabio Somenzi
Nov 18 at 11:14
Yes but only when n = 0 right? How can that still hold if we consider other values of n?
– ArielK
Nov 18 at 11:15
1
Yes, but that's not a problem. You can verify that $f_1(x) = frac{x}{2x+1}$. That's not the inductive proof, but the inductive step is not much different.
– Fabio Somenzi
Nov 18 at 11:19
1
The induction step is like this: let $f_{n+1}(x) = f_0(f_n(x))$ and suppose $f_n(x)=frac{x}{(n+1)x+1}$. Prove that $f_{n+1}(x) = frac{x}{(n+2)x+1}$. It's a matter of plugging the hypothesis for $f_n(x)$ into the definition of $f_{n+1}(x)$ and simplifying.
– Fabio Somenzi
Nov 18 at 11:27
1
The main problem with this problem is that your $f_0$ should be called $f_1$ to begin with. In this way $f_0$ is the identity, and $f_{m+n}=f^{circ m}circ f^{circ n}$, etcetera.
– Christian Blatter
Nov 18 at 11:55
1
1
Note that $f_1(x) = f_0(f_0(x))$.
– Fabio Somenzi
Nov 18 at 11:14
Note that $f_1(x) = f_0(f_0(x))$.
– Fabio Somenzi
Nov 18 at 11:14
Yes but only when n = 0 right? How can that still hold if we consider other values of n?
– ArielK
Nov 18 at 11:15
Yes but only when n = 0 right? How can that still hold if we consider other values of n?
– ArielK
Nov 18 at 11:15
1
1
Yes, but that's not a problem. You can verify that $f_1(x) = frac{x}{2x+1}$. That's not the inductive proof, but the inductive step is not much different.
– Fabio Somenzi
Nov 18 at 11:19
Yes, but that's not a problem. You can verify that $f_1(x) = frac{x}{2x+1}$. That's not the inductive proof, but the inductive step is not much different.
– Fabio Somenzi
Nov 18 at 11:19
1
1
The induction step is like this: let $f_{n+1}(x) = f_0(f_n(x))$ and suppose $f_n(x)=frac{x}{(n+1)x+1}$. Prove that $f_{n+1}(x) = frac{x}{(n+2)x+1}$. It's a matter of plugging the hypothesis for $f_n(x)$ into the definition of $f_{n+1}(x)$ and simplifying.
– Fabio Somenzi
Nov 18 at 11:27
The induction step is like this: let $f_{n+1}(x) = f_0(f_n(x))$ and suppose $f_n(x)=frac{x}{(n+1)x+1}$. Prove that $f_{n+1}(x) = frac{x}{(n+2)x+1}$. It's a matter of plugging the hypothesis for $f_n(x)$ into the definition of $f_{n+1}(x)$ and simplifying.
– Fabio Somenzi
Nov 18 at 11:27
1
1
The main problem with this problem is that your $f_0$ should be called $f_1$ to begin with. In this way $f_0$ is the identity, and $f_{m+n}=f^{circ m}circ f^{circ n}$, etcetera.
– Christian Blatter
Nov 18 at 11:55
The main problem with this problem is that your $f_0$ should be called $f_1$ to begin with. In this way $f_0$ is the identity, and $f_{m+n}=f^{circ m}circ f^{circ n}$, etcetera.
– Christian Blatter
Nov 18 at 11:55
|
show 1 more comment
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1
Note that $f_1(x) = f_0(f_0(x))$.
– Fabio Somenzi
Nov 18 at 11:14
Yes but only when n = 0 right? How can that still hold if we consider other values of n?
– ArielK
Nov 18 at 11:15
1
Yes, but that's not a problem. You can verify that $f_1(x) = frac{x}{2x+1}$. That's not the inductive proof, but the inductive step is not much different.
– Fabio Somenzi
Nov 18 at 11:19
1
The induction step is like this: let $f_{n+1}(x) = f_0(f_n(x))$ and suppose $f_n(x)=frac{x}{(n+1)x+1}$. Prove that $f_{n+1}(x) = frac{x}{(n+2)x+1}$. It's a matter of plugging the hypothesis for $f_n(x)$ into the definition of $f_{n+1}(x)$ and simplifying.
– Fabio Somenzi
Nov 18 at 11:27
1
The main problem with this problem is that your $f_0$ should be called $f_1$ to begin with. In this way $f_0$ is the identity, and $f_{m+n}=f^{circ m}circ f^{circ n}$, etcetera.
– Christian Blatter
Nov 18 at 11:55