n th Fibonacci number with function

Hi guys

I'm playing around with the fibonacci row.

I've been trying the for loops so far, not too hard.

n = input('n:');

fibo = zeros(1,n);

for n = 1:n
    if n==1
        fibo(n)=0;
    else
        if n==2
            fibo(n)=1;
        else
    fibo(n)=fibo(n-1)+fibo(n-2);
        end
    end
end


disp(fibo);

But I want to have the whole thing in one function and only get the n-th fibonacci number.

any ideas

That's my idea, but it' s actually a very dumb.

(I know its not working)

function [c] = fibo (a, b, n)
n = input('n:');
a =  n - 2;
b = n - 1;
c = a + b;
disp (c)
end

Answer 1

[Steven Lord]

function [c] = fibo (a, b, n)
n = input('n:');

Every time your user calls your fibo function, you're going to ask them to pass a value into it for n (as the third input) then immediately turn around and prompt them to type in a replacement value for n.

IMO if part of your design requires your user to enter data manually, you should have (at least) two separate functions: the interface function that can call input and then just calls the processing function and the processing function that operates solely on data passed into it (no prompting of the user allowed.)

But that's a bit of an aside. You already have most of what you want. It's not too difficult to translate your first script into a function.

n = input('n:');

Turn this first statement into the function declaration line. "function theOutputs = fibonacci(theInputs)". I'll let you decide what to write in place of theOutputs and theInputs.

fibo = zeros(1,n);

for n = 1:n

Reusing the variable name technically works, but it can be confusing that the role of n kind of changes. I'd use a different variable name for the loop variable.

    if n==1
        fibo(n)=0;
    else
        if n==2
            fibo(n)=1;
        else
    fibo(n)=fibo(n-1)+fibo(n-2);
        end
    end
end


disp(fibo); 

Turn this last line into the line that defines whatever variables you need to return. It should assign into whatever variables appear in theOutputs above.

Answer 2

[Ameer Hamza]

As compared to using for-loop, an alternative approach is to use a closed-form expression of the Fibonacci sequence as given here: https://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression. For example

function y = Fib(n)
    phi = (1+sqrt(5))/2;
    y = round((phi^n-(-phi)^-n)/(2*phi-1));
end

or

function y = Fib(n)
    phi = (1+sqrt(5))/2;
    y = floor(phi^n/sqrt(5)+1/2);
end

Result

>> Fib(1)
ans =
     1
>> Fib(2)
ans =
     1
>> Fib(3)
ans =
     2
>> Fib(4)
ans =
     3
>> Fib(5)
ans =
     5
>> Fib(10)
ans =
    55
>> Fib(15)
ans =
   610

Answer 3

[Voss]

Notice that your function is calculating n-1 + n-2 and returning that as the nth term in the Fibonacci sequence. In actuality the nth term of the Fibonacci sequence is the (n-1)th term + the (n-2)th term, not just n-1 + n-2. Your loop is doing this correctly, so you just need to make your function do it that way. This is a good example of recursion.