extracting arrays from a taylor series and plotting

I have been working on a script that calculates a Taylor series without using the built-in function.

I am having a lot of trouble extracting the proper array of numbers for the three iteration values I need to plot together with the error (I need to plot N = 2 5 & 50 with the exact function of 5sin(3x)).

This is what I have:

clear;clc

n =[2 5 50];
do=linspace(-2*pi,2*pi,720);
T = zeros(51);

for i =1:720

      for k=1:1:50;
          ns=2*k+1;
          T(i)=T(i)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);
      end
  end

approx2 = T(:,2); 
approx5 = T(:,5);
aprrox50 = T(:,50);
exact = 5*sin(3*do);
plot(do,exact, '-r')
hold on
ez1=ezplot('approx2');
ez2=ezplot('approx5');
ez3=ezplot('approx50');

legend('5sin(3x)','T2','T5','T50')

set(ez1, 'color', [0 1 0])
set(ez2, 'color', [0 0 1])
set(ez3, 'color', [1 0 1])

title('Graph of 5sin(3x) and taylor expansions T2, T5 and T50')

I cannot get it to graph properly though......

I have looked through the matlab help file and searched through google.

I am thinking it might have to do with not defning the Taylor iterations properly.

This is a homework assignment so I AM NOT LOOKING FOR SOMEONE TO DO THIS FOR ME.

Any hints or tips to put me in the right direction will be greatly appreciated!

Thank You

Answer 1

[per isakson]

Hints:

• aprrox50 &nbsp is a typo
• read the doc on ezplot, note that approx.. are double vectors
• use plot to plot approx..
• I find it easier to work with functions than with scripts when it comes to debugging - it used to be anyhow.
• try &nbsp figure, imagesc(T), colorbar
• ALWAYSONTOP, by Elmar Tarajan makes it possible to see the effect on the diagram of each line of code; step through the code and watch - or dock the figure

[Max]

The problem states that i have to write a script file.

Zeros wise, I have to define the T function before the loop correct?

So i tried just T = 0, bu then I receive an error saying "index out of bounds because numel(T)=1

thats why I defined it as the 51x51 zero matrix

[per isakson]

I've edited my answer

• "have to write a script file" &nbsp better not argue with that
• "I have to define the T function before the loop correct" &nbsp not exactly have to, but it critical for performance. zeros(51,51) communicates intent better; I was fooled by T(ii), which you use in the assignment

[Max]

clear;clc

n =[2 5 50];
do=linspace(-2*pi,2*pi,720);
T = zeros(51);

for i =1:720
   for k=1:1:50;
        ns=2*k+1;
        T(i)=T(i)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

     end
   end

for i =1:720
   for k=1:1:5;
        ns=2*k+1;
        T(i)=T(i)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

     end
   end
     for i =1:720
     for k=1:1:2;
          ns=2*k+1;
          T(i)=T(i)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

     end
   end

approx2 = T(2,:); 
approx5 = T(5,:);
aprrox6 = T(50,:);
exact = 5*sin(3*do);
plot(do,exact, '-r')
hold on
ez1=plot(do,approx2, '-m');
ez2=plot(do,approx5, '-g');
ez3=plot(do,approx6, '-b');

legend('5sin(3x)','T2','T5','T50')

title('Graph of 5sin(3x) and taylor expansions T2, T5 and T50')

When I have the 3 separate loops it gives me better values,but still having trouble plotting them all together.

[per isakson]

Hint

    >> whos ...
      Name      Size             Bytes  Class     Attributes
      do        1x720             5760  double              
      approx5   1x51               408  double

Length differs. Reconsider T = zeros(51);

[Max]

T = zeros(720) correct?

I do that and it makes them all the same dimension.

Problem I am having though is the approx2, approx5, and approx6 arrays are just 1 value followed by a bunch of zeros......

hmm do i need to store each iteration at the end of the loop before the end?

[per isakson]

    T = zeros(720);

is not compatible with

    approx2 = T(2,:); 
    approx5 = T(5,:);
    aprrox6 = T(50,:);
    ...
    T(i)=T(i) ...

T is not needed, it just make the code more complicated. Allocating 720 rows and using 3. Why not

    approx2 = zeros(size(do));
    ect.

and use approx2 in the loop.

[Max]

So ive gotten this far:

clear;clc

n =[2 5 50];
do=linspace(-2*pi,2*pi,720);
T = zeros(720);

for i =1:720
   for k=1:1:50;
        ns=2*k+1;
        T(i)=T(i)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

     end
   end

for i =1:720
   for k=1:1:5;
        ns=2*k+1;
        T(i)=T(i)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

     end
   end
   for i =1:720
       for k=1:1:2;
          ns=2*k+1;
          T(i)=T(i)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

       end
   end

approx2 = T(2,:); 
approx5 = T(5,:);
approx6 = T(50,:);
exact = 5*sin(3*do);
plot(do,exact, '-r')
hold on
plot(do,approx2, '-m');
plot(do,approx5, '-g');
plot(do,approx6, '-b');

legend('5sin(3x)','T2','T5','T50')
title('Graph of 5sin(3x) and taylor expansions T2, T5 and T50')

Its almost right except it is figuring the approx2 - approx6 values as points, so it just ends up graphing a line........

even stranger is it graphs the exact value as being equal to (-5,5)

[Max]

Thats a good idea Per Isakson, il ltry that out

[per isakson]

I would say: Take three steps away from the keyboard and write done some pseudo code on a piece of paper.

And

    do=linspace(-2*pi,2*pi,720);

You chosen the name do. I would have chosen x.

[Max]

I tried this:

clear;clc

n =[2 5 50];
do=linspace(-2*pi,2*pi,100);
approx2 = zeros(size(do));
approx5 = zeros(size(do));
approx6 = zeros(size(do));
for i =1:100
   for k=1:1:50;
        ns=2*k+1;
        approx6 = approx6 +5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

     end
   end

for i =1:100
   for k=1:1:5;
        ns=2*k+1;
        approx5 = approx5 +5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

     end
   end
   for i =1:100
       for k=1:1:2;
          ns=2*k+1;
          approx2 = approx2 +5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

       end
   end

exact = 5*sin(3*do);
plot(do,exact, '-r')
hold on
plot(do,approx2, '-m');
plot(do,approx5, '-g');
plot(do,approx6, '-b');

legend('5sin(3x)','T2','T5','T50')
title('Graph of 5sin(3x) and taylor expansions T2, T5 and T50')

It doesnt work because I need to define in the taylor series function that it is added to the approx before it not approx6 = approx6 + 5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);

im trying to think of a way that i could define it to have the first term be the preceding factor....

[Max]

Im getting closer:

clear;clc

n =[2 5 50];
x=linspace(-2*pi,2*pi,100);
approx2 = zeros(size(x));
approx5 = zeros(size(x));
approx6 = zeros(size(x));
for i =1:100
   for k=1:1:50;
        ns=2*k+1;
        approx6(i) = approx6(i) +5*(((-1)^k)*(3*x(i))^(ns))/factorial(ns);

     end
   end

for i =1:100
   for k=1:1:5;
        ns=2*k+1;
        approx5(i) = approx5(i) +5*(((-1)^k)*(3*x(i))^(ns))/factorial(ns);

     end
   end
   for i =1:100
       for k=1:1:2;
          ns=2*k+1;
          approx2(i) = approx2(i) +5*(((-1)^k)*(3*x(i))^(ns))/factorial(ns);

       end
   end

exact = 5*sin(3*x);
plot(x,exact, '-r')
hold on
plot(x,approx2, '-m');
plot(x,approx5, '-g');
plot(x,approx6, '-b');

legend('5sin(3x)','T2','T5','T50')
title('Graph of 5sin(3x) and taylor expansions T2, T5 and T50')

The script still gets weird with the graphing portion though

[per isakson]

This is the starting point

Note summation from k=0

[Max]

I got that but matlab cannot start at 0 i though

[Max]

hence the reason why k was always definded as k = 1:n

[per isakson]

Yes, Matlab's indexing is "one-based", but in one way or other you have to include the first term in the series!

[Max]

well the first term is always 15x

so would defining the series as

(for example)

approx2(i) = 15*x + 5*(((-1)^k)*(3*x(i))^(ns))/factorial(ns);

i want to say that takes care of the 0 term

Answer 2

[Roger Stafford]

It looks as though your T is incorrectly computed. For each of three different counts, n = 2, 5, and 50, you need to compute n terms of the Taylor series for each of 720 different values of the angle. That would imply three, not two, nested for-loops or their equivalent, and the resulting T should be two-dimensional, not one-dimensional.

   for n = [2,5,50]
     for i = 1:720
       for k = 1:n
         etc.

(Of course, with the appropriate use of matlab's 'sum' function you could eliminate at least one of these loops.)

[Max]

I am thinking that multiple loops is the answer, unfortunately the problem I am working on specifically states you cannot use any of the built in sum features....

[Roger Stafford]

I really should have written the loops in my suggestion this way:

   T = zeros(720,3);
   n = [2,5,50];
   for j = 1:3
     for i = 1:720
       for k = 1:n(j)
         T(i,j) = T(i,j) + ....
         .......

   approx2 = T(:,1); 
   approx5 = T(:,2);
   approx50 = T(:,3);

Does that help you any?

[Max]

Yes this does help, but for the three loops would i define the j values as

j = 1:2

j = 1:5

J = 1:50

I know they would go at the begging of each loop for that specific number of iterations.

[Max]

I tried this:

T = zeros(720,3);
n = [2 5 50];
do=linspace(-2*pi,2*pi,720);
for j = 1:2
    for i = 1:720
        for k = 1:n(j)
            ns=2*k+1;
            T(i,j)=T(i,j)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);
        end
    end
end
for j = 1:5
    for i = 1:720
        for k = 1:n(j)
            ns=2*k+1;
            T(i,j)=T(i,j)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);
        end
    end
end
for j = 1:50
    for i = 1:720
        for k = 1:n(j)
            ns=2*k+1;
            T(i,j)=T(i,j)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);
        end
    end
end

approx2 = T(:,1);
approx5 = T(:,5);
approx50 = T(:,50);

But matlab complains about n(4) being out of bonds because numel(n)=3

which makes no sense to me because its defined at the begging of the script that n = [2 5 50]...

Answer 3

[daniel]

The first problem I see is that "exact" is a 1x720 vector and your approximations are 51x1 vectors; my first suggestion would be to get them all in the same domain. Once they all span the same domain and are of the same size, your plot should look better.

[Max]

Thats an excellent point, I could just transpose the approx values but then it still wouldnt match size wise