Function won't accept vector

function result = g(x)
result= (2*x*exp(cos(6*x))*exp(-x)) + 15

" Error using * Inner matrix dimensions must agree."

I tried inputting g([0 1 2]), but I get this error. Can someone help me understand why? I would like the function to accept scalars and vectors.

I know there's such a thing as ".*", but when I created this function: (13/400)*(x.^3 - 30.876*x.^2 + 32.454*x + 99.89), it accepted [0 1 2] just fine.

Answer 1

[KSSV]

g = @(x) (2*x.*exp(cos(6*x)).*exp(-x)) + 15 ;
result = g([0 1 2])

When you input a vector, you need to do element by element multiplication. This is done by using .*. Please read about matlab element by element operations. https://in.mathworks.com/help/fixedpoint/ref/times.html

[Rachel Dawn]

Hello. Thank you for your response. I would like to know why is it that in my 2nd function, I use ".^" and in the first function I use ".*" ? Why does my 2nd function accept vectors without the use of ".*"?

[James Tursa]

@Rachel: scalar*array gives the same result as scalar.*array, so in this particular case the * and the .* do the same thing. So in your 2nd function:

(13/400)*(x.^3 - 30.876*x.^2 + 32.454*x + 99.89)

The only multiply operations that you have involve a scalar (the (13/400) and the 30.876 and the 32.454 are all scalars). Hence there was no need in this particular case to use the .* operator.

In your 1st function, x and exp(cos(6*x)) and exp(-x) are all vectors that you are multiplying by each other so you must use the .* operator to get the element-wise multiplication that you want.

[Rachel Dawn]

Thank you, James!

Answer 2

[Shrirang]

I just tried it..Your function behaves perfectly fine with ".*" operator Here is an example.

function result = Test(x) 
result= (2 .* x .* exp(cos(6 .* x)) .* exp(- x)) + 15;
and answer was ==> 15.0000   16.9219   16.2588

May be there was typo mistake in your trial In your second example you have properly handled the dot operator before ^ so there was no error.

Let me know if it works for you.

[Rachel Dawn]

Hello. Thank you for your response. I would like to know why is it that in my 2nd function, I use ".^" and in the first function I use ".*" ? Why does my 2nd function accept vectors without the use of ".*"?