How can I make my function accept vectors instead of scalars(I'm Really New to Matlab)

I'm trying to make a taylor series expression for e^x = sum(x^n/n!) for n=0:50 and for x^n/n!>0.01 So far I have this:

function [ xs ] = myExpfuntion(x)

x=input('Enter x: ');

xl=zeros(51,1);                          %preallocation of the storage
for n= 1:50
    if ((x^n)/factorial(n))>0.01
        xl(n+1) = ((x^n)/factorial(n)); %save every iteration result
    end
end
xs=1+sum(xl) ;                            %calculate the sum of the iteration results

end

I want to delete the line "x=input('Enter x: ');" and just place my value for x in the function i.e. myExpfunction(x) and also make the function work with x being a row vector i.e x=[1 2 3]

My current function wont allow me to do so and I don't get what I'm doing wrong. Please help!

Answer 1

[Julia]

Hi,

I modified you code so that vectors can be used as input.

However, I am not sure, if I did it in the right way. It should help you though to get what you want:

function [ xs ] = myExpfuntion(x)

xs=zeros(1,length(x));
xl=zeros(51,length(x));                          %preallocation of the storage
for n= 1:50
    if ((x.^n)/factorial(n))>0.01
        xl(n+1,:) = ((x.^n)/factorial(n)); %save every iteration result
    end
end
for i = 1:length(x)
    xs(i)=1+sum(xl(:,i)) ;  %calculate the sum of the iteration results
end

end

This code results in:

>> xs=myExpfuntion([1,2,3,4])

xs =

      2.7083    7.0000   16.3750   34.3333

[Japoe25]

Hello Julia,

Thank you for taking some time in answering my question! Unfortunately i fear there is something wrong though since the answer at x=4 should be around 54.5940. I'll have a look at the code now :)

[Japoe25]

Hello again Julia,

If it helps, I've had a look at the iteration results and saw that the iterations stop when they reach the length of x instead of at >0.01

xl =

         0         0         0         0
    1.0000    2.0000    3.0000    4.0000
    0.5000    2.0000    4.5000    8.0000
    0.1667    1.3333    4.5000   10.6667
    0.0417    0.6667    3.3750   10.6667

[Julia]

Hello Japoe,

1/(5!) is smaller than 0.01. So for n greater than 4 the if statement is never true again. Perhaps you have to modify the if condition.