Why are these zeros adding themselves to the array?

clear
clc

function result = expn(x,n)

result = 1;

    for i = 1:n
    result = (result + x.^i/factorial(i));

    end

end

x = [1,-2];
n = [1,2,4,6,10];
ex1 = zeros(1,5);
ex_2 = zeros(1,5);

for k = n
    t = expn(x(1),k);
    ex1(k) = t;
end

for k = n
    t = expn(x(2),k);
    ex_2(k) = t;
end

ex1

ex1 = 1×10

2.0000 2.5000 0 2.7083 0 2.7181 0 0 0 2.7183

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ex_2

ex_2 = 1×10

-1.0000 1.0000 0 0.3333 0 0.1556 0 0 0 0.1354

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Not sure why the zeros are appearing as elements in the arrays. If someone could help me out it would be much appreciated!

[Torsten]

The zeros are at the positions that are not covered by n = [1,2,4,6,10];. Would you prefer a number different from 0 to be set there ?

[Andrew]

Oh that makes sense. No I'd rather there be no zeros, and have the array consist only of the nonzero numbers.

Answer 1

[Voss]

k goes 1,2,4,6,10

n = [1,2,4,6,10];
for k = n
    fprintf('k = %d\n',k);
end

k = 1
k = 2
k = 4
k = 6
k = 10

so inside your loop, e.g., ex1(k) = t; sets ex1(1), ex1(2), ex1(4), ex1(6), ex1(10)

When you set an element of an array that's outside the current size of the array, the array is expanded as necessary with elements containing zeros. So that's where the zeros are coming from.

Example:

vec = zeros(1,5)

vec = 1×5

0 0 0 0 0

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vec(4) = 40 % not expanded

vec = 1×5

0 0 0 40 0

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vec(10) = 100 % vec gets expanded to length 10 with zeros

vec = 1×10

0 0 0 40 0 0 0 0 0 100

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You probably meant for ex1 and ex_2 to be length-5 vectors for their entire lifetimes, in which case you'd do something like this:

x = [1,-2];

n = [1,2,4,6,10];

ex1 = zeros(1,5);

ex_2 = zeros(1,5);

m = numel(n);

for k = 1:m

t = expn(x(1),n(k));

ex1(k) = t;

end

for k = 1:m

t = expn(x(2),n(k));

ex_2(k) = t;

end

[Andrew]

Yes this is exactly what I meant! Thank you very much!

[Voss]

You're welcome!

Answer 2

[Walter Roberson]

function result = expn(x,n)

result = 1;

    for i = 1:n
    result = (result + x.^i/factorial(i));

    end

end

x = [1,-2];
n = [1,2,4,6,10];
ex1 = dictionary();
ex_2 = dictionary;

for k = n
    t = expn(x(1),k);
    ex1(k) = t;
end

for k = n
    t = expn(x(2),k);
    ex_2(k) = t;
end

ex1

ex1 =

  dictionary (double --> double) with 5 entries:

    1  --> 2
    2  --> 2.5000
    4  --> 2.7083
    6  --> 2.7181
    10 --> 2.7183

ex_2

ex_2 =

  dictionary (double --> double) with 5 entries:

    1  --> -1
    2  --> 1
    4  --> 0.3333
    6  --> 0.1556
    10 --> 0.1354

Answer 3

[Catalytic]

ex1=[2.0000    2.5000         0    2.7083         0    2.7181         0         0         0    2.7183]

ex1 = 1×10

2.0000 2.5000 0 2.7083 0 2.7181 0 0 0 2.7183

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ex1=nonzeros(ex1)'

ex1 = 1×5

2.0000 2.5000 2.7083 2.7181 2.7183

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