matrix dimensions must agree.

Hi everyone, I hope you all are doing very well. I have been working on Matlab about taylor series recently, however, I usually get an error about matrix dimensions. Line 17 which starts with y1=y0 + ... . So can anybody help me out? thank you from now!

clc;
clear;
clear all;

x= -2:0.5:2;
dx=0.5;
yy=sin(x);
a=1;

b = diff(yy)/dx; 
b1= diff(diff(yy)/dx)/dx;
b2= diff(diff(diff(yy)/dx)/dx)/dx;
b3= diff(diff(diff(diff(yy)/dx)/dx)/dx)/dx;
b4= diff(diff(diff(diff(diff(yy)/dx)/dx)/dx)/dx)/dx;
b5= diff(diff(diff(diff(diff(diff(yy)/dx)/dx)/dx)/dx)/dx)/dx;
b6= diff(diff(diff(diff(diff(diff(diff(yy)/dx)/dx)/dx)/dx)/dx)/dx)/dx;
b7= diff(diff(diff(diff(diff(diff(diff(diff(yy)/dx)/dx)/dx)/dx)/dx)/dx)/dx)/dx;

y0= sin(a);
y1= y0 + b.*(x-a);
y2= y0 + b.*(x-a) + b1/factorial(2).*((x-a)^2);
y3= y0 + b.*(x-a) + b1/factorial(2).*((x-a)^2) + b2/factorial(3).*((x-a)^3);
y4= y0 + b.*(x-a) + b1/factorial(2).*((x-a)^2) + b2/factorial(3).*((x-a)^3) + b3/factorial(4).*((x-a)^4);
y5= y0 + b.*(x-a) + b1/factorial(2).*((x-a)^2) + b2/factorial(3).*((x-a)^3) + b3/factorial(4).*((x-a)^4) + b4/factorial(5).*((x-a)^5);
y6= y0 + b.*(x-a) + b1/factorial(2).*((x-a)^2) + b2/factorial(3).*((x-a)^3) + b3/factorial(4).*((x-a)^4) + b4/factorial(5).*((x-a)^5) + b5/factorial(6).*((x-a)^6);
y7= y0 + b.*(x-a) + b1/factorial(2).*((x-a)^2) + b2/factorial(3).*((x-a)^3) + b3/factorial(4).*((x-a)^4) + b4/factorial(5).*((x-a)^5) + b5/factorial(6).*((x-a)^6) + b6/factorial(7).*((x-a)^7);
y8= y0 + b.*(x-a) + b1/factorial(2).*((x-a)^2) + b2/factorial(3).*((x-a)^3) + b3/factorial(4).*((x-a)^4) + b4/factorial(5).*((x-a)^5) + b5/factorial(6).*((x-a)^6) + b6/factorial(7).*((x-a)^7) + b7/factorial(8).*((x-a)^8);

[Rik]

You are not taking a true derivative, but you are calculating the difference between each array element.

Also, you could make your code much more readable by doing something like this:

b0 = diff(yy)/dx; 
b1= diff(b)/dx;
b2= diff(b1)/dx;
b3= diff(b2)/dx;
b4= diff(b3)/dx;
b5= diff(b4)/dx;
b6= diff(b5)/dx;
b7= diff(b6)/dx;

By not using numbered variables, but using a cell array instead, you can further simplify the code:

b=cell(1,8);
b{1}=diff(yy)/dx;
y=cell(1,numel(b));
y{1}=sin(a);
y{2}=y{1}+b{1}*(x-a)
for n=2:numel(b)
    b{n}=diff(b{n-1})/dx;
    y{n+1}=y{n} + b{n}*((x-a)^n)/factorial(n);
end

Note that Matlab is one-indexed, meaning that y0 becomes y{1}.

If you use the symbolic toolbox to generate the variable x, this should work as intended. For a numerical fix, you will have to think about how you want to extend you vector after you apply diff.

[Onur Totos]

Thank you for your help! It really assisted me to learn about how to make it short and clear. If you were in my shoes, how would you write this example on the matlab?

[Rik]

I think I would take the mid-points of the x-vector as well when doing a diff for the b.

So then you would have

x_new=x_old(2:end)-dx/2;

That would enable you to continue the loop not to an arbitrary point, but to the point where you don't have a long enough x.

Also, you should change that first assingment to this:

dx=0.5;
x=-2:dx:2;

The other route takes you a bit away from numerical approximation: symbolic variables.

[Onur Totos]

Thanks for responding to me. I appreciate your help! The problem is still going on the program... Still I am having a problem about matrix dimensions and line 20.. Anyways, thanks again..!!

[Rik]

What is the code you are currently using?

[Onur Totos]

clc; clear ;clear all;

dx= 0.5;
x= -2:dx:2;
yy=sin(x);
a=1;

b = diff(yy)/dx; 
b1= diff(b)/dx;
b2= diff(b1)/dx;
b3= diff(b2)/dx;
b4= diff(b3)/dx;
b5= diff(b4)/dx;
b6= diff(b5)/dx;
b7= diff(b6)/dx;

y0= sin(a);
y1= y0 + b.*(x-a);
y2= y1 + b1/factorial(2).*((x-a)^2);
y3= y2 + b2/factorial(3).*((x-a)^3);
y4= y3 + b3/factorial(4).*((x-a)^4);
y5= y4 + b4/factorial(5).*((x-a)^5);
y6= y5 + b5/factorial(6).*((x-a)^6);
y7= y6 + b6/factorial(7).*((x-a)^7);
y8= y7 + b7/factorial(8).*((x-a)^8);

I am trying to use taylor series for sin(x)

[Jan]

@Onur Totos: Please use the code style to improve the readability of your code in the forum.

[Rik]

There are a few problems here:

1. you insist on numbered variables
2. you did not account for diff changing the size of your vector
3. you are still using clear all, instead of clear variables

Answer 1

[nanren888]

>> whos
  Name      Size            Bytes  Class     Attributes

  a         1x1                 8  double              
  b         1x8                64  double              
  b1        1x7                56  double              
  b2        1x6                48  double              
  b3        1x5                40  double              
  b4        1x4                32  double              
  b5        1x3                24  double              
  b6        1x2                16  double              
  b7        1x1                 8  double              
  dx        1x1                 8  double              
  x         1x9                72  double              
  yy        1x9                72  double              

The result of diff is shorter, being always the difference between elements.

sometimes;

  something = [0,diff(somethingElse)] 

can be useful.

[Onur Totos]

Thank you very much for your respons. It makes sense. However, I am tired of dealing with this problem since the morning. So what exactly do I have to do for fixing this problem? Regards