\[f _{n}\] denotes the nth Fibonacci number. Prove \[\sum_{k=0}^{n} f _{k}^{2} =f _{n} f _{n+1}\]
if you would be kind enough, please show me the steps as u solve it
Have you tried induction?
i don't know how to do induction ..im very bad at discrete math
i think i just had a really bad teacher
Have you learnt induction in your course?
yes i have but it just went over my head..i don't even know where to start or what the steps are...but if you are willing to teach me how to do it step by step then that would be good..we got a take home final and i must pass it to have the slightest chance of passing the class..so u can show me how to do it and explain it good or just give me the answer and show all the steps...
The reason I ask is induction is the way to prove this result. When you learn induction, it's probably one of the most involved mathematical procedures you will have learnt up to that time. Let me see if I can find a good video or web site that can walk you through it and you can review at your pace.
ok thank you
Here is a good example of proof by induction. http://www.khanacademy.org/video/proof-by-induction?playlist=Algebra Watch this, make sure you really understand it. If that means you watch it 2 or 3 times, nothing wrong with that. And then we can talk about your problem again.
ok will do..thanks again :)
Once you've digested the Khan Academy video, watch this example so you see one more example which is slightly more complicated and definitely faster. http://www.youtube.com/watch?v=3qTJrWCgVwI&feature=related Your problem is a step up in complexity again.
(This second video also cracks me up because you feel like he wouldn't hesitate to beat you with that black rod if you made an error.)
ok so i watched both videos and understand what the authors were doing...but i dont know where to take the numbers from my problem and plug it in to the formula \[s(n) = n(n+1) / 2\]
lol
that formula for the sum of integers has nothing to do with your problem.
hey brb i have to use the rest room...but keep typing i'll bb in a min..
The general method for using induction is the following 1. Define carefully the statement you want to prove by induction P(n), as per the second video and define the domain of n (usually n are natural numbers) 2. Show that P(1) is true. (or whatever the lowest value of n is in the domain of P(n)) 3. Show that for arbitrary k in the domain of P(n), that P(k) ==> P(k+1) i.e., the result P(k) logically implies P(k+1) 4. Then conclude by the Principle of Mathematical Induction that P(n) is true for all n
So ... step 1: For your problem, what is the statement P(n) you are trying to prove?
still there?
back
ok sooo how do i start to do my problem?
\[\sum_{k=0}^{n} f _{k}^{2} =f _{n} f _{n+1}\]
Right and what is the domain of n here?
im not sure to be honest
n is non-negative integers; n = 01,2,3,4,... Now. Before we go any further, what is the definition of the Fibonacci numbers?
i don't know that either lol =/
im telling u ..this class has been soo hard for me man i have no clue what i am doing
i take notes and everything but i have no idea what i am writing down
Do you have a textbook? If so, look up Fibonacci numbers there. If not, go and google it. Then tell me what you find.
oh yeah it is when u add the current number with the previous number to get the next number
but it starts' with 0 and 1
so it's 0,1, then 1+0= 1, then 1+1=2, 2+1=3, 3+2=5, 5+3=8.....right?
Ok, so in terms of the \[ f_n \] write down the definition
\[ f_0 = 0, f_1 = 1 \] and \[ f_{n+2} = f_n + f_{n+1} \]
ok. So returning now to the four steps: 2. Show that P(1) is true. (or whatever the lowest value of n is in the domain of P(n)) Hence, show that P(0) is true.
see i dont understand that when it's put like that it has to be dummied down for me hehe
given those values of f0 and f1, the formula is saying f2 = f0 + f1 = 0 + 1 = 1. So what is f3?
f3 = f1 + f2 = ...
f3 = f0 + f1+ f2= 0 + 1+ 2 = 3?
No, f_n+2 = f_n+1 + f_n. Hence f3 = f2 + f1 = 1 + 1 = 2 What is f4?
f0 = 0, f1 = 1, f2 = 1, f3 = 2 What is f4?
im so confused now
hold on
f0 = 0, f1 = 1, f2 = f0 + f1 = 0 + 1 = 1 f3 = f1 + f2 = 1 + 1 = 2 f4 = f2 + f3 = ...
let me ask something so how would you find f(64) or some large number like that?
so f4 = f2, which is 1 + f3, which is 2, so 2+1 = 3
so to find f(64) do u have to find all the f from 0-63 ?
Yes, f4 = f2 + f3 = 1 + 2 = 3 and f5? The short answer to your question is as we currently define this sequence, there's no quick way to find f(64) or f(640), which isn't to say there ins't a quick way; you just don't know it yet.
so f5 is (and only because we found f 0-4 right?) = f3 + f4, which is 2+3 respectively, which = 5 so f5=5?
yes.
Ok. So follow now the procedure I've outlined for you. Step 2: Show P(0) is true. Then steps 3 and 4.
one question...where are u getting 0 from? is it because k=0 under the sum sign?
So P(0) says that \[ \sum_{k=0}^0 f_k^2 = f_0f_1 \] This is true because f0 = 0 and hence the LHS (left hand side) of that expression is \[ f_0^2 = 0^2 \] and the RHS is \[ f_0f_1 = 0 . 1 = 0 \] and are therefore equal. Hence the statement P(0) is true.
Step 3: show that P(k) implies P(k+1) for arbitrary k. (And yes I know that n can equal zero from the way the question is posed.)
If P(k) is true then \[ \sum_{j=0}^k f_j^2 = f_kf_{k+1} \] We've changed the variable of summation to j to avoid having k mean two things. Show that this implied P(k+1) i.e., that \[ \sum_{j=0}^{k+1} f_j^2 = f_{k+1}f_{k+2} \]
i have no idea wht u are saying man
then go back and watch that second video again. I am following exactly the procedure he is using there, but applying it to your problem.
are u solving it or are some other step?
that's what i need help with ...what am i taking from my problem and what am i plugging in?
I've completely outlined it for you. You need to read a few more proofs by induction to understand what's going on here. I'm sure your text book has a few. Go and find them. Copy them out. Then back a blank piece of paper and replicate one or two of them without looking. When you can do that, then you'll be ready to tackle this problem.
Very patient, JJ.
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