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:)

|dw:1436502484982:dw|

-1* mod 7

so thats the answer for you dan?

no just thinking

You will get it dan lol

hm

1

is the answer 1

Well nope

i think my arithmetic is wrong maybe, i got a repetition at f(9) ill put what i got

|dw:1436504022956:dw|

k ya i saw myin writing same thing

lol nope @myininaya

I didn't give an answer

Oh thought you did lol sorry

i have a feeling we will run into a repeating pattern in mod 7 like every 9th maybe

Lets conjecture
\[7\mid f(3+9k)\]

:P 4022 / 9 = 446.8 pretty close hehe

based on @myininaya 'w work..

WOw great work

Great job :D

brute force cheattoorr

lol dan got right after geerky lol

yep

ask another one now!

k let start typing it :D

the proof might be interesting

by induction it can be proven

I would love to see a proof on this but I think that is above me.

once you get a full repettiion it must be true

but there could be a neater proof this is a lame way to prove it

that is true, once the remainders start repeating, we're technically done!

good night myin

see ya @myininaya D:

f(1+x)=9x
f(1-x)=x?

x*f(1-x)= x^2+ax+b
f(1+x)=cx+d

oh and since u applied x after there can be no constant term in both

therefore
f(1-x)=x+b
and
f(1+x)=ax
they must of this tform

f(1+x) cannot have a constant term

and f(1-x) cannot have a coefficient on the x

so we are restrited with the unknowns here

restricted*

hm

hm lol waiting lol falling asleep lol

f(-7)=-7+b
f(-7)=-7a
a+b=9
f(-7)=-7+ 9-a
=2-a
f(-7)=-7*(a)
2-a=-7a
a=1/3

lol what is going on

I don't know but I'll guess 13 and then I think I'll try solving it to make sure.

\[\large a+b=217\]

howd u do it

but your close

simply plugin \(x=\pm 8\), you will get two linear eqn's which you can solve

\[\large a+b=229\]
?

Correcto :D
Finally now i can sleep lol

dont know who to give the stars D:

ah i liked it

It looks like you also get the answer to what f(9) is for free as well haha

I subed 1+x=u and 1-x=v and tried to get a differential eqn like yesterday but got stuck

Ahh okay I see that should make it easy..

do you think this assumption is right
f(1+x)=ux
f(1-x)=x+v

Hmmm I don't know?

if we ignore the fourier series and other orthogonal representations to polynomial functions

yeah at least we will get all linear functions, if not all

f(1-x)= x + u
f(1+x)=v*x
u+v=9

oo

Looks f(x) cannot be a polynomial

i think so, now u can rewrite the right side as functions of 1-x ?

wow yeah 1-x = u should do

Nice, that wasn't so bad after all haha

So if we look at f(x) as being a generating function, what's the sequence it generates? :O