jagr2713
  • jagr2713
Fun question :D #Bored
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jagr2713
  • 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
  • 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
  • jagr2713
:)

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More answers

dan815
  • dan815
|dw:1436502484982:dw|
dan815
  • dan815
-1* mod 7
jagr2713
  • jagr2713
so thats the answer for you dan?
dan815
  • dan815
no just thinking
jagr2713
  • jagr2713
You will get it dan lol
jagr2713
  • jagr2713
hm
dan815
  • dan815
1
dan815
  • dan815
is the answer 1
jagr2713
  • jagr2713
Well nope
dan815
  • dan815
i think my arithmetic is wrong maybe, i got a repetition at f(9) ill put what i got
jagr2713
  • jagr2713
Lol @dan815
myininaya
  • 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
  • dan815
|dw:1436504022956:dw|
dan815
  • dan815
k ya i saw myin writing same thing
jagr2713
  • jagr2713
lol nope @myininaya
myininaya
  • myininaya
I didn't give an answer
jagr2713
  • jagr2713
Oh thought you did lol sorry
myininaya
  • 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
  • dan815
i have a feeling we will run into a repeating pattern in mod 7 like every 9th maybe
geerky42
  • 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
  • ganeshie8
Lets conjecture \[7\mid f(3+9k)\]
dan815
  • dan815
:P 4022 / 9 = 446.8 pretty close hehe
ganeshie8
  • ganeshie8
based on @myininaya 'w work..
jagr2713
  • jagr2713
WOw great work
jagr2713
  • jagr2713
Great job :D
dan815
  • dan815
brute force cheattoorr
jagr2713
  • jagr2713
lol dan got right after geerky lol
myininaya
  • 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
  • dan815
yep
dan815
  • dan815
ask another one now!
jagr2713
  • jagr2713
k let start typing it :D
ganeshie8
  • ganeshie8
the proof might be interesting
dan815
  • dan815
by induction it can be proven
myininaya
  • myininaya
I would love to see a proof on this but I think that is above me.
dan815
  • dan815
once you get a full repettiion it must be true
dan815
  • dan815
but there could be a neater proof this is a lame way to prove it
ganeshie8
  • ganeshie8
that is true, once the remainders start repeating, we're technically done!
myininaya
  • myininaya
sadly i can't stay for the fun @jagr2713 will post it is time for me to retire :(
dan815
  • dan815
good night myin
jagr2713
  • 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
  • jagr2713
see ya @myininaya D:
dan815
  • dan815
f(1+x)=9x f(1-x)=x?
dan815
  • dan815
x*f(1-x)= x^2+ax+b f(1+x)=cx+d
dan815
  • dan815
oh and since u applied x after there can be no constant term in both
dan815
  • dan815
therefore f(1-x)=x+b and f(1+x)=ax they must of this tform
dan815
  • dan815
f(1+x) cannot have a constant term
dan815
  • dan815
and f(1-x) cannot have a coefficient on the x
dan815
  • dan815
so we are restrited with the unknowns here
dan815
  • dan815
restricted*
jagr2713
  • jagr2713
hm
dan815
  • 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
  • jagr2713
hm lol waiting lol falling asleep lol
dan815
  • 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
  • dan815
lol what is going on
Empty
  • Empty
I don't know but I'll guess 13 and then I think I'll try solving it to make sure.
ganeshie8
  • ganeshie8
\[\large a+b=217\]
dan815
  • dan815
howd u do it
jagr2713
  • jagr2713
Nope ganesh D: is that your answer @dan815
jagr2713
  • jagr2713
but your close
ganeshie8
  • ganeshie8
simply plugin \(x=\pm 8\), you will get two linear eqn's which you can solve
ganeshie8
  • ganeshie8
\[\large a+b=229\] ?
jagr2713
  • jagr2713
Correcto :D Finally now i can sleep lol
jagr2713
  • jagr2713
dont know who to give the stars D:
Empty
  • Empty
Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.
Empty
  • Empty
Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.
dan815
  • dan815
ah i liked it
Empty
  • Empty
It looks like you also get the answer to what f(9) is for free as well haha
ganeshie8
  • ganeshie8
I subed 1+x=u and 1-x=v and tried to get a differential eqn like yesterday but got stuck
Empty
  • 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
  • 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
  • ganeshie8
Ahh okay I see that should make it easy..
Empty
  • 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
  • dan815
do you think this assumption is right f(1+x)=ux f(1-x)=x+v
Empty
  • Empty
Hmmm I don't know?
dan815
  • dan815
if we ignore the fourier series and other orthogonal representations to polynomial functions
dan815
  • 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
  • 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
  • ganeshie8
yeah at least we will get all linear functions, if not all
dan815
  • 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
  • dan815
f(1-x)= x + u f(1+x)=v*x u+v=9
Empty
  • 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
  • dan815
oo
ganeshie8
  • ganeshie8
Looks f(x) cannot be a polynomial
ganeshie8
  • 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
  • dan815
i think so, now u can rewrite the right side as functions of 1-x ?
ganeshie8
  • ganeshie8
wow yeah 1-x = u should do
Empty
  • Empty
Nice, that wasn't so bad after all haha
Empty
  • Empty
So if we look at f(x) as being a generating function, what's the sequence it generates? :O

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