Why can't CS fa/g/s into maths?

Why can't CS fa/g/s into maths?

They act all superior but then act like maths isn't useful at all, because they can't do it.

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Why are mathfags so smug?

Not smug, just annoyed when csfags talk about maths thinking they're right when they're blatantly wrong.

Why can't you do your own homework?

Math minor here.

Who even cares apart from it looking good on a resume?

how do you do this without a computer?

An answer is 89, I've already done it.
>minor
dropped
Look for patterns in the Fibonacci sequence, you don't need to compute many terms and it can be done easily by hand

Yo math niggers, do this problem with your so called fancy pants equations and shit while I'll have the answer ready before you can get to fib(10)

>brainlet who thinks the power of his computer contributes to his own ""intelligence""

F(n) | F(kn) where k and n are positive integers.

Therefore, as 2013 = 11*183, then F(11) | F(2013).

F(11) = 89 which is prime, therefore 89 is a factor of F(2013)

Good luck computing F(2013) even on a powerful computer

what's the big deal about fibonacci sequence?

I can write a program to calculate the index of a fibonacci number in under 5 mins. Wtf?

Why study algorithms if you're not going to implement them

That's not what the question was asking
It's showing an understanding of the properties that certain sequences/operations have. Without paying attention to such things, we wouldn't have finite fields and consequently wouldn't have public key cryptography

I'm not saying I'm intelligent, I'm most certainly not, but I'm pragmatic

I want to get shit done, not stroke my own ego

>Why can't CS fa/g/s into maths?
what makes you think that ? just because you go to a shitty universtiy ? math classes are part of every decent CS curriculum.

Are you retarded
Do you not understand how fast the native addition operation is
A modern computer can literally do trillions of additions per second

Easily done. Fibonacci might often be used as an example of recursive algorithms, but that's not an efficient way to do it.

Store fibonacci values in an array and go up from F(2) to F(2013) and it becomes linear.

The problem with the Fibonacci sequence is that there is no real solution to the recursive sequence F(n) = F(n-1) + F(n-2). You can't guess the growth rate of it (seems to be between N log N and N^2). You also can't use the formula ( a of n = (alpha)(r of 1)^n + (beta)(r of 2)^n) where alpha and beta are constants and r of 1 and r of 2 are non identical roots found in r^2 + (c of 1)r + (c of 2), where c of 1 and c of 2 are the coefficients of the recursion sequence.

Computer engineering student here.
>The Fibonacci numbers are a function
>Babby's first math tier questions
You are a fucking brainlet, OP.

Fair enough, find an odd prime factor of F(279262929216353) (hint, I know the answer)
>desktop thread
>rice thread
>buyer's remorse
You aren't intended to find the growth rate of it, and lim n->infinity F(n) = 1.61803399...*F(n-1)

A sequence is a function with the domain being the set of natural numbers. Therefore the fibonacci sequence is a function.
The question wasn't intended to be hard, I'm not even that good at maths and I did it.

1

>1
>prime number
confirmed for cs student

If you wish to compute a Fibonacci number at any location, such as at F(2013), you need a solution to the recurrence relation.

Look at the Jacobthal numbers. {0, 1, 1, 3, 5, 11, 21 .. } = W(n) = W(n-1) + 2W(n-2). These have a recurrence relation of ((-1)^n-1 + (2)^n) / 3.

Recurrence solution*

discrete math undergrad detected

>you need a solution to the recurrence relation
No you don't.
And evaluating F(279262929216353) isn't going to happen, (it's a fairly big number)

You're fucking retarded, using addition to calculate the Fibonacci numbers is always slow
that's still not a good way to do it
The fast doubling and matrix exponentiation are the best way to calculate yhem because they have Θ(logn) running time nayuki.io/page/fast-fibonacci-algorithms

The growth rate of the recursive algorithm is Θ(Φ^n) for time and Θ(n) for space. You have no clue what you are talking about

Fair enough, but I still managed F(2013) fast enough not to perceive a delay after hitting the key and getting the answer with the following:

static BigInteger fibo(int n) {
Vector vals = new Vector();

vals.ensureCapacity(n);
vals.setSize(n);

vals.add(1, BigInteger.valueOf(1));
vals.add(2, BigInteger.valueOf(1));

if(n

what does | do?

a|b means that the the integer a divides the integer b

x ∣ y is equivalent to y = x * k , k ∈ ℕ

that's so weird

that's induction. math style.

Ur gay

nice symbols dude

This guy is right.

Yesterday I was writing a RESTful API for a PostgreeSQL database. Suddenly I was wondering: "what if F(2013) was odd?"
I was thinking about this the whole day, unable to work: Odd? Even? Odd? Even?

I think they should really teach such necessary skills.

Yes but exponentiation is amortized logarithmic time at best, plus floating point numbers have limited precision and easily break down for large exponents of non-integers (in the case of fibonacci's recurrence relation).

I love math :)

How do you know that F(n) always divides F(kn)? It's not intuitively obvious.

meant for

What does the arrow mean in "N --> N"?

Name 1 thing learning such math is useful for.

It means the function F takes domain from the natural numbers and outputs to the range of natural numbers.

If OP can't prove this himself, then he actually is shit at math and this thread is invalidated

>tfw on the school team for the Putnam this year

>Laplace transform and other complex calculus is required for any circuit work
>BUT YOU CAN'T MATH!
Sorry what? In b4 you went to some shit college where CS is exclusively theoretical.

You used the closed form expression instead.