Fun question :D #Bored

- jagr2713

Fun question :D #Bored

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- jagr2713

Function F(n) defined over all positive integers n satisfies the following:
\[f(1)=1,f(2)=2, and\]
\[f(n+2)=f(n+1)+(f(n))^{2}+4022 \ for \ n \ge1\]
How many of the 4022 integers f(1), f(2),....f(4022) are multiple of 7?

- myininaya

well I notice f(3) is a multiple of 7
but checking all of those will be crazy
so there has to be a slick way of doing this

- jagr2713

:)

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## More answers

- dan815

|dw:1436502484982:dw|

- dan815

-1* mod 7

- jagr2713

so thats the answer for you dan?

- dan815

no just thinking

- jagr2713

You will get it dan lol

- jagr2713

hm

- dan815

1

- dan815

is the answer 1

- jagr2713

Well nope

- dan815

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

- jagr2713

Lol @dan815

- myininaya

well going to the whole mod thing I guess we can really just look at:
\[f(n+2)=f(n+1)+(f(n))^2+4 \text{ thinking in mod } 7\]
these will make the outputs easier to look at
but that is still a lot of f( ) thingys to check:
\[f(1)=1,f(2)=2,f(3)=7=0,f(4)=15=1, f(5)=68=5\]
hmmm...
\[f(6)=f(5)+(f(4))^2+4 \\ f(6)=5+1^2+4=10=3 \\ f(7)=f(6)+(f(5))^2+4 \\ f(7)=3+(5)^2+4=25=4 \\ f(8)=f(7)+(f(6))^2+4 \\ f(8)=4+(3)^2+4 \\ f(8)=1+9=10=3 \\ f(9)=f(8)+(f(7))^2+4 \\ f(9)=3+4^2+4=4^2=2 \\ f(10)=f(9)+(f(8))^2+4 \\ f(10)+2+3^2+4=6+2=8=1 \\ f(11)=f(10)+(f(9))^2+4 \\ f(11)=1+2^2+4=9=2\]
\[f(12)=f(11)+(f(10))^2+4 \\ f(12)=2+1^2+4=7=0\]
I think f(3) and f(12) so far are multiples of 7

- dan815

|dw:1436504022956:dw|

- dan815

k ya i saw myin writing same thing

- jagr2713

lol nope @myininaya

- myininaya

I didn't give an answer

- jagr2713

Oh thought you did lol sorry

- myininaya

I'm showing my work so far hoping it can help anyone else trigger more thoughts
but my thought actually came from dan's work above

- dan815

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

- geerky42

I solved it by programming and I asked @jagr2713 for confirmation of my answer but he insisted me to post my answer here lol...
` numbers = [1, 2]
total = 0
for x in range(3, 4023):
NewNumber = (numbers[1] + (numbers[0])**2 + 4022)%7
if NewNumber%7 == 0:
total +=1
numbers = [numbers[1], NewNumber]
print(total) `
Output is 447

- ganeshie8

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

- dan815

:P 4022 / 9 = 446.8 pretty close hehe

- ganeshie8

based on @myininaya 'w work..

- jagr2713

WOw great work

- jagr2713

Great job :D

- dan815

brute force cheattoorr

- jagr2713

lol dan got right after geerky lol

- myininaya

I see pattern
1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,... I wonder if this is the pattern
we have a 0
at 3rd term,12th term, 21th term,
oh and I see where @ganeshie8 get's his conjecture now

- dan815

yep

- dan815

ask another one now!

- jagr2713

k let start typing it :D

- ganeshie8

the proof might be interesting

- dan815

by induction it can be proven

- myininaya

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

- dan815

once you get a full repettiion it must be true

- dan815

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

- ganeshie8

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

- myininaya

sadly i can't stay for the fun @jagr2713 will post
it is time for me to retire :(

- dan815

good night myin

- jagr2713

For all real numbers x, the function f(x) satisfies
\[f(1+x)+xf(1-x)=x ^{2}+9x\]
If the values of f(-7) can be expressed as
\[\frac{ a }{ b }\]
Where a and b are coprime positive integers/ What is a+b
Here you go @dan815

- jagr2713

see ya @myininaya D:

- dan815

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

- dan815

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

- dan815

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

- dan815

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

- dan815

f(1+x) cannot have a constant term

- dan815

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

- dan815

so we are restrited with the unknowns here

- dan815

restricted*

- jagr2713

hm

- dan815

f(1-x)=x+b and
f(1+x)=ax
f(-7)=-7+b
f(-7)=-7a
and
f(-7)= c/d, where c and d are co prime int
hmm i am wondering if the first lines are really okay to assume

- jagr2713

hm lol waiting lol falling asleep lol

- dan815

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

- dan815

lol what is going on

- Empty

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

- ganeshie8

\[\large a+b=217\]

- dan815

howd u do it

- jagr2713

Nope ganesh D: is that your answer @dan815

- jagr2713

but your close

- ganeshie8

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

- ganeshie8

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

- jagr2713

Correcto :D
Finally now i can sleep lol

- jagr2713

dont know who to give the stars D:

- Empty

Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.

- Empty

- dan815

ah i liked it

- Empty

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

- ganeshie8

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

- Empty

@ganeshie8 What you could try instead is plug in 1-x=y+h and 1-x=y and that'll turn it into a differential equation.

- ganeshie8

this could be an interesting question
For all real numbers x, the function f(x) satisfies
\[f(1+x)+xf(1-x)=x ^{2}+9x\]
find \(f(x)\)
Here you go @dan815

- ganeshie8

Ahh okay I see that should make it easy..

- Empty

Unfortunately (I might have done it wrong) the differential equation we get is not a very pretty one to solve but we can do it with an integrating factor it seems, I just think it ends up having the erf(x) function in there as part of the answer #_#

- dan815

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

- Empty

Hmmm I don't know?

- dan815

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

- dan815

f(1+x)+x*f(1-x) = x^2+-9x
then can we say
f(1+x) must be 1 degree lower than x*f(1-x)
so the x^2 term must come from that term

- Empty

Ahhh I see what you're looking at now and why you would say that but I'm sure if I know if we can do that or not. Interesting observation though.
It looks like we should be able to solve for f(x)=ax+b by plugging in I think.

- ganeshie8

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

- dan815

yea because if this is really true u get restrictions all over the place
since then x*f(1-x) cannot have a constant term that means f(1-x) must have form ax+b
and then f(1+x) cannot have a constnat term as it all equalys x^2+9x

- dan815

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

- Empty

I was also considering if we get anything by splitting it up into an even and odd function: f(x)=g(x)+h(x)
g(x)=g(-x) and h(x)=-h(-x) since we can always do this, then make a pair of equations this way that will be easy to solve hopefully.

- dan815

oo

- ganeshie8

Looks f(x) cannot be a polynomial

- ganeshie8

does this work
f(1+x) + xf(1-x) = x^2+9x
f(1-x) -xf(1+x) = x^2-9x
-----------------------------
f(1-x)(1+x^2) = x^3+10x^2-9x
f(1-x) = (x^3+10x^2-9x)/(1+x^2)

- dan815

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

- ganeshie8

wow yeah 1-x = u should do

- Empty

Nice, that wasn't so bad after all haha

- Empty

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

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