Functions

[zinc flume]

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[rose coral]

No idea

[pine leaf]

brother i swear this was in the senior maths challenge last year

[rose coral]

HUH

like

how senior

[pine leaf]

year 12-13

💀

[rose coral]

what age bracket attempts it?

[pine leaf]

16-18

[rose coral]

im 17 🥀

i should be able to

do it

i gues-

[pine leaf]

this is assuming $f : /mathbb{Z} /right \mathbb{Z}$ right?

[ashen raven]

curry supplier

[rose coral]

😭

[pine leaf]

holy fricken airball

[rose coral]

Z to Z ?

if that, then yes

[pine leaf]

$f\colon \bZ \to \bZ$

[ashen raven]

curry supplier

[pine leaf]

yeah

[rose coral]

yes

[pine leaf]

and f is a linear function/constant right

[rose coral]

how do you know its linear

i aint given

[pine leaf]

i remember doing this sorta question last year

[rose coral]

😭

[pine leaf]

it was a linear function

or a constant

[rose coral]

how do you KNOW THAT

[pine leaf]

huh

ok anyways back to solving ts

uhhhh

[rose coral]

i first put a =0

then b = 0

we get 2f(x) = f(2x)

f = 2x??

[last forum]

Well, if we assume its a polynomial, it'll maximally be of degree 1. Notice why.

[rose coral]

because LHS is applying function only once

and RHS is applying twice

so powers wont match on either side if degree is more than one

right?

[last forum]

Yes.

[last forum]

[rose coral]

i mean

[last forum]

Just let f=cx+d and compare coefficients.

[rose coral]

oh shitte

okay thats one way

but

what if the function is not a polynomial

[rose coral]

i got C = 2

and D equals to anything

[last forum]

Well, what other functions could exist?

You might want to invoke a proof by contradiction.

[pine leaf]

[north inlet]

Assume b as constant and do the derivative wrt a I always do that idk but it works well

Yep I got it

It's a linear fxn 2x+c

[north inlet]

Try this ping me if stuck

[rose coral]

i arrived at it too

let me see your method

[north inlet]

You cant just assume it is a linear fxn lmao

That's called guessing

[rose coral]

thats eactly what i did 😭

[north inlet]

I also learned how to do these types of questions recently (functional equation) I was taught to either do it by first principle of derivate or by what I told you

[rose coral]

oooh first principle of deivative

that sounds interesting to be used here

these type of questions always trouble me

[north inlet]

That would be harder here, works for simple functional equations

[rose coral]

i'd still like to learn both methods

currently studying chem, will reach back later

[slate comet]

this is your average FE, try random shit it'll work out in the end

can't have enough of em really

[rose coral]

yes it did easily

but im trying to accumalte

every single method

that exists

[slate comet]

f(2x)+2f(x)=f(f(2x))
what can you deduce from this alone?

[rose coral]

degree = 1

[slate comet]

no degrees are involved

this isn't a polynomial

[rose coral]

what else can i deduce 😭

[slate comet]

think

use your brain

I'm not trying to be rude or implying you aren't already doing that

but that's the only way to do these most of the time

[rose coral]

💀

let me

use it

more

[north inlet]

Just use the derivative 😭

Do the derivative wrt x and put x=y

[rose coral]

i currently

doing chem

[north inlet]

Only do this when it's given that fxn is differentiable*

[rose coral]

2f'(2a) = f'(f(a+b)).f'(a+b)

now?

[north inlet]

Put a = b

[rose coral]

why

just for the

sake of simplifying 💀

[north inlet]

Exactly

You can also put a = 0 or smth doesn't matter

Here a=b works

[rose coral]

f'(f(2a)) = 2

[north inlet]

*If f is not constant

You cant just cancel

[rose coral]

cancel what

[north inlet]

f'(2a)

[rose coral]

i did

[rose coral]

this is what you get after

[north inlet]

We are assuming it's not a constant fxn

[rose coral]

yes

[north inlet]

Yes so f'(x) = 2

f(x) = 2x + c

[rose coral]

f' (f(2a)) = 2 ??

[north inlet]

If you dont get that you should learn functions first

[rose coral]

dummy variable f(2a) =x

breh

yesok

got it

[north inlet]

I gave an adv mock test yesterday it had one question like this wanna try?

[rose coral]

yes

[north inlet]

Multiple correct

[rose coral]

A,C ??

[north inlet]

No it's B,C what fxn did you get?

You should have gotten
||f'(x) = 1 + lnx||

[rose coral]

waht

f(1) is zero right

[north inlet]

Yes

[rose coral]

f'x = (1+lnx) f(1) - f'(1)

did i mess up calculation

let me check

[north inlet]

Probably

[rose coral]

but the f(1) term will always come with the 1+lnx term

@north inlet

[north inlet]

f(1) Will be zero

f'(1) is 1 as given in the question

@rose coral

[rose coral]

uhh we havent done differentiation yet

real quick

derivative of f(x^x) is f' (x^x) or f'(x^x) * x^x (1+lnx) ??

latter one

got it

[north inlet]

we use chain rule

[north inlet]

Yes

[rose coral]

aah

okay

got it

thats where the mistake was

so the assumptions of y being constant and then y = x is just for the sake of our ease because we are free to use any inputs right

[north inlet]

Yeh

[rose coral]

okay got it

thankyou

[north inlet]

And I haven't learned integration yet so I just guessed which fxns differentiation will be that and got f(x) = x.ln(x)

[rose coral]

int lnx is just xlnx - x i guess

using by parts

[north inlet]

Oh I didn't know that I am yet to learn integration officially

[rose coral]

alr alr

thankyouuu

+close

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[north inlet]

Np

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