The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Compositions

Let ff be a N→N\mathbb{N} \rightarrow \mathbb{N} function defined by

Unparseable latex formula:

f(n) =\begin{Bmatrix}2n+1\ \ \ \ \ \mathrm{if}\ n>10, \\ 2n-2\ \ \ \ \ \ \mathrm{if} \ 0<n<10,



How would I go about solving f∘ff \circ f?

Any hints? I'm aware that this means f(f(n))f(f(n)) but I just don't know how to go about solving this.

(I made this question up btw, the actual question is different but if I can understand what goes on here I can probably do the actual question)
(edited 7 years ago)
Reply 1
Avatar for Notnek
Notnek
21
Original post by edothero
Let ff be a N→N\mathbb{N} \rightarrow \mathbb{N} function defined by

Unparseable latex formula:

f(n) =\begin{Bmatrix}2n+1\ \ \ \ \ \mathrm{if}\ n>10, \\ 2n-2\ \ \ \ \ \ \mathrm{if} \ 0<n<10,



How would I go about solving f∘ff \circ f?

Any hints? I'm aware that this means f(f(n))f(f(n)) but I just don't know how to go about solving this.

(I made this question up btw, the actual question is different but if I can understand what goes on here I can probably do the actual question)

The question you made up would need adusting since it's not a function (elements from the domain aren't mapped anywhere) and f∘ff \circ f wouldn't be defined for the full domain.

Try this function instead with domain Z\mathbb{Z}.

Unparseable latex formula:

f(x) = \begin{cases}[br]x+2, \ x<-1 \\[br]x-1, \ x\geq -1[br]\end{cases}



We'll start by trying different inputs e.g. a negative number like -10:

Inputting -10 into ff gives -8. Then input that into ff again and we get -6.

What's happening is we're getting f(f(x))=f(x+2)=(x+2)+2=x+4f(f(x)) = f(x+2) = (x+2) + 2 = x+4.


But what happens if we start with an input of -3. Putting that into ff gives -1 and putting the result into ff gives -2.

What's happening here is f(f(x))=f(x+2)=(x+2)−1=x+1f(f(x)) = f(x+2) = (x+2)-1 = x + 1.


Try to think about all the different cases. Post all your ideas/working if you get stuck.
Reply 2
Avatar for edothero
edothero
OP
15
Original post by notnek
The question you made up would need adusting since it's not a function (elements from the domain aren't mapped anywhere) and...


So I've tried inputting various values and I've ended up with 4 cases:

f(f(x))=f(x+2)=(x+2)+2=x+4f(f(x)) = f(x+2) = (x+2)+2 = x + 4

f(f(x))=f(x+2)=(x+2)−1=x+1f(f(x)) = f(x+2) = (x+2)-1 = x + 1

f(f(x))=f(x+2)=(x−1)+2=x+1f(f(x)) = f(x+2) = (x-1)+2 = x + 1

f(f(x))=f(x+2)=(x−1)−1=x−2f(f(x)) = f(x+2) = (x-1)-1 = x - 2

Though I'm not sure how the domain will be effected in each case;

For example:

f(−2)=−2+2=0f(-2) = -2+2 = 0
f(0)=0−1=−1f(0) = 0-1 = -1

The input: −2<−1-2<-1 but the output: −1≮−1 -1 \not< -1

Quite confused right now :laugh: (I study Computer Science, not Maths)
(edited 7 years ago)
Reply 3
Avatar for Zacken
Zacken
22
Original post by edothero
...


Essentially, you need to be worrying about the range of ff as well.

So, looking at the first part of the function (the x+2x+2 part) you need to know when that goes into the first part, and when that goes into the second part.

So for x+2<−1x+2 < -1 your f∘f=(x+2)+2=x+4f \circ f = (x+2) + 2 = x+4. This is your function. You only put x+2 into x+2 when both x+2<−1  ⟺  x<−3x+2 < -1 \iff x < -3 and when x<−1x < -1. So we fill in:

f=x+4f = x+4 when x<−3x<-3 (since that's when both conditions are fulfilled).

This is because, for the first choice of which part of function to use, we look at what our xx is, if it's less than −1-1, then we put it into x+2x+2. We get an output from this, the output is x+2x+2 (but remember that x<−1x < -1), we want to put this output into the function again. We need to look at what the output is. If the output is less than -1, then it goes into the first part of the function again. So if x+2<−1x+2 < -1 (which is the same as x<−3x<-3) then it goes into x+2x+2 so you get (x+2)+2(x+2) + 2 whenever x<−3x < -3.

Now you repeat this process, when x<−1x < -1, you get an output x+2x+2 that you want to put into the function again. This time, you want to put it into the second part of the function. This means your ouput x+2≥−1x+2 \geq -1 so x≥−3x \geq -3 (along with x < -1) means that if you start with x satisfying those constraints, you put it into the first part of the function the first time around, get an output that tells you to put it into the second part of the function. So whenever −3≤x<−1-3 \leq x <-1 your function is (x+2)−1=x+1(x+2) - 1 = x+1.

Now we look at the second part of your function, we only use it when x≥−1x \geq -1. When this happens, we get an output x−1x-1. We want to see whether to put this output into the first part of the function or the second.

Whenever x−1<−1  ⟺  x<0x -1 < -1 \iff x < 0 happens, we put it into the second part of the function. So we get f(x)=(x−1)+2=x+1f(x) = (x-1) + 2 = x+1 whenever x<0x < 0 (along with x≥−1x\geq -1) which combined together gives you −1≤x<0-1 \leq x < 0.

See if you can figure this bit out yourself:

Spoiler


(sorry for stealing your example notnek)

Putting all of that together, our function is:

Spoiler

Reply 4
Avatar for edothero
edothero
OP
15
Original post by Zacken
Essentially, you need to be worrying about the range of ff as well.

So, looking at the first part...


This is quite a good explanation! Thanks a ton. I was able to answer the original question; whether it's correct or not: time will tell :laugh:. Or you can be a bae and check it for me (its not assesed material).

How is Cambridge treating you btw?

Also thanks to notnek for the kick-start!
Reply 5
Avatar for Zacken
Zacken
22
Original post by edothero
This is quite a good explanation! Thanks a ton. I was able to answer the original question; whether it's correct or not: time will tell :laugh:. Or you can be a bae and check it for me (its not assesed material).

Also thanks to notnek for the kick-start!


Sure, I'm going to bed at 2:00, so if you get it up quickly, I'll have a look at it.
Reply 6
Avatar for edothero
edothero
OP
15
Original post by Zacken
Sure, I'm going to bed at 2:00, so if you get it up quickly, I'll have a look at it.


Can't seem to make the Latex work, if I end up posting it you can check it when you're free tomorrow.

Have a good sleep!
Reply 7
Avatar for Zacken
Zacken
22
Original post by edothero
Can't seem to make the Latex work, if I end up posting it you can check it when you're free tomorrow.

Have a good sleep!


Just take a picture of it on paper or something or even write it out without latex, doesn't matter

thanks, you too
Reply 8
Avatar for edothero
edothero
OP
15
Original post by Zacken
Just take a picture of it on paper or something or even write it out without latex, doesn't matter

thanks, you too


Lmaooo well, lets just say I forgot to post it, but if you're still interested:

Question:

Spoiler

Answer

Spoiler

Reply 9
Avatar for Zacken
Zacken
22
Original post by edothero
Lmaooo well, lets just say I forgot to post it, but if you're still interested:

Question:

Spoiler

Answer

Spoiler



A few comments:

1. Remember that you can collect the two middle functions together to get f(n)=nf(n) = n when
Unparseable latex formula:

4 < n \leqg 6



2. You can immediately tell that the last part of your function is wrong, since it's mapping −1-1 to something, but −1-1 is not even in your domain! The problem is that you're forgetting to put together the conditions on nn. n+1n+1 goes into n+1n+1 when TWO things are true. First, you need 0<n≤50 <n \leq 5 to hold so that you choose n+1n+1 in the first place, you then get an output of n+1n+1 (remembering that 0<n≤50 < n \leq 5) and then you put that back into n+1n+1 if the output 0<n+1≤50 < n+1 \leq 5 holds.

So f(n)=n+2f(n) = n+2 when you have (i) 0<n≤50 < n \leq 5 and (ii) −1<n≤4-1 < n \leq 4 which can be cleaned up as 0<n≤40 < n \leq 4.
Reply 10
Avatar for edothero
edothero
OP
15
Original post by Zacken
A few comments:


Thank you as always, I can imagine you're quite swamped with work at Cambridge so I appreciate taking time out to help :yy:
Reply 11
Avatar for Zacken
Zacken
22
Original post by edothero
Thank you as always, I can imagine you're quite swamped with work at Cambridge so I appreciate taking time out to help :yy:


No problemo! And this is a good way of procrastinating on my four example sheets due in. :wink:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you