A community for students.
Here's the question you clicked on:
 0 viewing
jagr2713
 one year ago
Fun question :D #Bored
jagr2713
 one year ago
Fun question :D #Bored

This Question is Closed

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3Function 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
 one year ago
Best ResponseYou've already chosen the best response.3well 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
 one year ago
Best ResponseYou've already chosen the best response.3so thats the answer for you dan?

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3You will get it dan lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i think my arithmetic is wrong maybe, i got a repetition at f(9) ill put what i got

myininaya
 one year ago
Best ResponseYou've already chosen the best response.3well 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
 one year ago
Best ResponseYou've already chosen the best response.2k ya i saw myin writing same thing

myininaya
 one year ago
Best ResponseYou've already chosen the best response.3I didn't give an answer

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3Oh thought you did lol sorry

myininaya
 one year ago
Best ResponseYou've already chosen the best response.3I'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
 one year ago
Best ResponseYou've already chosen the best response.2i have a feeling we will run into a repeating pattern in mod 7 like every 9th maybe

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.4Lets conjecture \[7\mid f(3+9k)\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2:P 4022 / 9 = 446.8 pretty close hehe

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4based on @myininaya 'w work..

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3lol dan got right after geerky lol

myininaya
 one year ago
Best ResponseYou've already chosen the best response.3I 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

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3k let start typing it :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4the proof might be interesting

dan815
 one year ago
Best ResponseYou've already chosen the best response.2by induction it can be proven

myininaya
 one year ago
Best ResponseYou've already chosen the best response.3I would love to see a proof on this but I think that is above me.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2once you get a full repettiion it must be true

dan815
 one year ago
Best ResponseYou've already chosen the best response.2but there could be a neater proof this is a lame way to prove it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4that is true, once the remainders start repeating, we're technically done!

myininaya
 one year ago
Best ResponseYou've already chosen the best response.3sadly i can't stay for the fun @jagr2713 will post it is time for me to retire :(

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3For all real numbers x, the function f(x) satisfies \[f(1+x)+xf(1x)=x ^{2}+9x\] 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2x*f(1x)= x^2+ax+b f(1+x)=cx+d

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh and since u applied x after there can be no constant term in both

dan815
 one year ago
Best ResponseYou've already chosen the best response.2therefore f(1x)=x+b and f(1+x)=ax they must of this tform

dan815
 one year ago
Best ResponseYou've already chosen the best response.2f(1+x) cannot have a constant term

dan815
 one year ago
Best ResponseYou've already chosen the best response.2and f(1x) cannot have a coefficient on the x

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so we are restrited with the unknowns here

dan815
 one year ago
Best ResponseYou've already chosen the best response.2f(1x)=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
 one year ago
Best ResponseYou've already chosen the best response.3hm lol waiting lol falling asleep lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.2f(7)=7+b f(7)=7a a+b=9 f(7)=7+ 9a =2a f(7)=7*(a) 2a=7a a=1/3

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I don't know but I'll guess 13 and then I think I'll try solving it to make sure.

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3Nope ganesh D: is that your answer @dan815

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4simply plugin \(x=\pm 8\), you will get two linear eqn's which you can solve

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\large a+b=229\] ?

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3Correcto :D Finally now i can sleep lol

jagr2713
 one year ago
Best ResponseYou've already chosen the best response.3dont know who to give the stars D:

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1It looks like you also get the answer to what f(9) is for free as well haha

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I subed 1+x=u and 1x=v and tried to get a differential eqn like yesterday but got stuck

Empty
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 What you could try instead is plug in 1x=y+h and 1x=y and that'll turn it into a differential equation.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4this could be an interesting question For all real numbers x, the function f(x) satisfies \[f(1+x)+xf(1x)=x ^{2}+9x\] find \(f(x)\) Here you go @dan815

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ahh okay I see that should make it easy..

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Unfortunately (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
 one year ago
Best ResponseYou've already chosen the best response.2do you think this assumption is right f(1+x)=ux f(1x)=x+v

dan815
 one year ago
Best ResponseYou've already chosen the best response.2if we ignore the fourier series and other orthogonal representations to polynomial functions

dan815
 one year ago
Best ResponseYou've already chosen the best response.2f(1+x)+x*f(1x) = x^2+9x then can we say f(1+x) must be 1 degree lower than x*f(1x) so the x^2 term must come from that term

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Ahhh 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
 one year ago
Best ResponseYou've already chosen the best response.4yeah at least we will get all linear functions, if not all

dan815
 one year ago
Best ResponseYou've already chosen the best response.2yea because if this is really true u get restrictions all over the place since then x*f(1x) cannot have a constant term that means f(1x) must have form ax+b and then f(1+x) cannot have a constnat term as it all equalys x^2+9x

dan815
 one year ago
Best ResponseYou've already chosen the best response.2f(1x)= x + u f(1+x)=v*x u+v=9

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Looks f(x) cannot be a polynomial

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4does this work f(1+x) + xf(1x) = x^2+9x f(1x) xf(1+x) = x^29x  f(1x)(1+x^2) = x^3+10x^29x f(1x) = (x^3+10x^29x)/(1+x^2)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i think so, now u can rewrite the right side as functions of 1x ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4wow yeah 1x = u should do

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Nice, that wasn't so bad after all haha

Empty
 one year ago
Best ResponseYou've already chosen the best response.1So if we look at f(x) as being a generating function, what's the sequence it generates? :O
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.