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

Question

Japoe25 2015-4-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/196477-how-can-i-make-my-function-accept-vectors-instead-of-scalars-i-m-really-new-to-matlab

评论： Julia 2015-4-7

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!

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Julia 2015-4-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/196477-how-can-i-make-my-function-accept-vectors-instead-of-scalars-i-m-really-new-to-matlab#answer_174261

在 MATLAB Online 中打开

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

3 个评论
显示 1更早的评论隐藏 1更早的评论

Japoe25 2015-4-7

在 MATLAB Online 中打开

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
0000    2.0000    3.0000    4.0000
5000    2.0000    4.5000    8.0000
1667    1.3333    4.5000   10.6667
0417    0.6667    3.3750   10.6667