Functions Need help in it

bigmindedone

For number 6 and 7 can you explain what to do i dont need the answer just how to approach it

(edited 9 years ago)

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Reply 1

ghostwalker

Original post by bigmindedone

...

For the first one, you may already have come across the questions/proofs that

If, g: A ----> B, and h: B----> C

and hog is a bijection, then one of them is a surjection and the other an injection.

If so, just adapt that.

Reply 2

bigmindedone

i still dont understand your method can you explain some more

Reply 3

james22

Original post by bigmindedone

i still dont understand your method can you explain some more

Show that it is injective and surjective. For injectivity start with asusming f(f(x))=f(f(y)) and try to show that x=y.

For surjective pick a y, and try to find an x such that f(f(x))=y.

Reply 4

bigmindedone

Original post by james22

Show that it is injective and surjective. For injectivity start with asusming f(f(x))=f(f(y)) and try to show that x=y.

For surjective pick a y, and try to find an x such that f(f(x))=y.

Thanks ill try that but do u have any idea about 7

Reply 5

Smaug123

Original post by bigmindedone

Can someone give me a guideline of how to do this question please i have no idea how to start it it seems obvious that f(x) = g(x) but can you tell me what this question wants from me

Hmm. The answer isn't immediately obvious to me. I can prove that

f^2(3) = 3

very fast…

Why do you say it's obvious that f=g? Certainly it's possible for f=g, but I don't see why it's the only choice.

ETA: Ah, got it. I just didn't go far enough. It's not trivial, though - there are a couple of steps to do.

(edited 9 years ago)

Reply 6

bigmindedone

Original post by Smaug123

Hmm. The answer isn't immediately obvious to me. I can prove that

f^2(3) = 3

very fast…

can you explain what to do

Reply 7

Smaug123

Original post by bigmindedone

can you explain what to do

When you've explained why "it seems obvious"…

Reply 8

james22

This was a fun one, and it is far from obvious what to do.

I started by picking an x such that f(x)=3 (you can dio this since f and g are bijections as they have inverses).

Plugging this into the idendtities given you get f(g(x))=g(3)=x and 3g(x)=x^2

Now try to calculate g(x) in 2 different ways, using each identity and then make them equal and solve (remember everything is positive so you can square root, and everything in 1-1 so if g(x)=g(y) you can say that x=y).

EDIT: Hold on, I made an error but I think this still works, give me a minutes.

(edited 9 years ago)

Reply 9

bigmindedone

Original post by Smaug123

When you've explained why "it seems obvious"…

i meant its obvious to see that f(x) = g(x) since their multiples is a square number but i dont have any idea what to do from there

Reply 10

james22

Original post by Smaug123

I can prove that

f^2(3) = 3

very fast…

How did you prove this? I can only get this result by first doing the question.

Reply 11

bigmindedone

Original post by james22

How did you prove this? I can only get this result by first doing the question.

How do you know they are bijections

Reply 12

james22

Original post by bigmindedone

How do you know they are bijections

They have inverses.

Reply 13

bigmindedone

Original post by james22

They have inverses.

Is this question hard or am i just bad because im still lost

Reply 14

james22

Original post by bigmindedone

Is this question hard or am i just bad because im still lost

It's hard.

Do you see why they are bijections?

Reply 15

bigmindedone

Original post by james22

It's hard.

Do you see why they are bijections?

You said because they have inverses but i dont see that too lol
how did you see that offhand

(edited 9 years ago)

Reply 16

Smaug123

Original post by james22

How did you prove this? I can only get this result by first doing the question.

My answer was wrong :frown:

but I've found a better one. Completely different method.

Stupidly, I directly proved bijectiveness, rather than using inverseness :P

Spoiler

I then examine the behaviour of

f_i(x) = f(f_{i-1}(x)), f_1(x) = f(x)

the iteration of f. I derive the formula

\frac{f_{i+1}(x)}{f_i(x)} = \frac{f_i(x)}{f_{i-1}(x)}

, from which the result follows because we have that

f

is bounded both above and below, but if

f_i(x) > f_{i-1}(x)

then

f_i

inductively increases geometrically, and in particular goes to infinity. Likewise it decreases geometrically if

f_i(x) < f_{i-1}(x)

, so we must have

f_i(x) = f_{i-1}(x)

Reply 17

Smaug123

Original post by bigmindedone

You said because they have inverses but i dont see that too lol
how did you see that offhand

It's precisely property 1.

Reply 18

bigmindedone

Original post by Smaug123

My answer was wrong :frown:

but I've found a better one. Completely different method.

Stupidly, I directly proved bijectiveness, rather than using inverseness :P

Spoiler

I then examine the behaviour of

f_i(x) = f(f_{i-1}(x)), f_1(x) = f(x)

the iteration of f. I derive the formula

\frac{f_{i+1}(x)}{f_i(x)} = \frac{f_i(x)}{f_{i-1}(x)}

, from which the result follows because we have that

f

is bounded both above and below, but if

f_i(x) > f_{i-1}(x)

then

f_i

inductively increases geometrically, and in particular goes to infinity. Likewise it decreases geometrically if

f_i(x) < f_{i-1}(x)

, so we must have

f_i(x) = f_{i-1}(x)

Where did you derive that formula from

Reply 19

james22

Original post by Smaug123

My answer was wrong :frown:

but I've found a better one. Completely different method.

Stupidly, I directly proved bijectiveness, rather than using inverseness :P

Spoiler

I then examine the behaviour of

f_i(x) = f(f_{i-1}(x)), f_1(x) = f(x)

the iteration of f. I derive the formula

\frac{f_{i+1}(x)}{f_i(x)} = \frac{f_i(x)}{f_{i-1}(x)}

, from which the result follows because we have that

f

is bounded both above and below, but if

f_i(x) > f_{i-1}(x)

then

f_i

inductively increases geometrically, and in particular goes to infinity. Likewise it decreases geometrically if

f_i(x) < f_{i-1}(x)

, so we must have

f_i(x) = f_{i-1}(x)

So you've basically showed that the only solution to the functional equation is the identity?

I spend ages trying to find something special about the number 3. Good solution.