extracting arrays from a taylor series and plotting

Max
2015 2 11
3 个回答
更新时间：2015 2 11
4 次查看（30 天）

0 个投票

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

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回答（3 个）

per isakson
per isakson 2015-2-11
编辑：per isakson 2015-2-11

1 个投票

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

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Max
Max 2015-2-11
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
per isakson 2015-2-11
编辑：per isakson 2015-2-11
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
Max 2015-2-11
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
per isakson 2015-2-11
编辑：per isakson 2015-2-11
Hint
>> whos ...
Name Size Bytes Class Attributes
do 1x720 5760 double
approx5 1x51 408 double
Length differs. Reconsider T = zeros(51);
Max
Max 2015-2-11
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
per isakson 2015-2-11
编辑：per isakson 2015-2-11
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
Max 2015-2-11
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
Max 2015-2-11
Thats a good idea Per Isakson, il ltry that out
per isakson
per isakson 2015-2-11
编辑：per isakson 2015-2-11
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
Max 2015-2-11
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
Max 2015-2-11
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
per isakson 2015-2-11
This is the starting point
Note summation from k=0
Max
Max 2015-2-11
I got that but matlab cannot start at 0 i though
Max
Max 2015-2-11
hence the reason why k was always definded as k = 1:n
per isakson
per isakson 2015-2-11
Yes, Matlab's indexing is "one-based", but in one way or other you have to include the first term in the series!
Max
Max 2015-2-11
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

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Roger Stafford
Roger Stafford 2015-2-11

1 个投票

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.)

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Max
Max 2015-2-11
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
Roger Stafford 2015-2-11
编辑：Roger Stafford 2015-2-11
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
Max 2015-2-11
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
Max 2015-2-11
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]...

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daniel
daniel 2015-2-11

0 个投票

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.

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Max
Max 2015-2-11
Thats an excellent point, I could just transpose the approx values but then it still wouldnt match size wise

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