I try to apply Taylor's series for sin(x) + cos(x)

f(x) = sin(x) + cos(x) = (-1)^k[ x^2k+1 / (2k+1!) + x^2k / (2k!)]

let

Tn(x) = (-1)^k[ x^2k+1 / (2k+1!) + x^2k / (2k!)]

1) plot f (x) = sin(x) + cos(x), T1(x), T2(x), T5(x) for x ∈ [-π,π]

I come up with

n = input(prompt)

x = linspace(-pi,pi);

SumTerms = (-1)*(x.^(1)./factorial(1)) + (-1)*(x.^(0)./factorial(0)); %initail

for i = 1:n

SumTerms = SumTerms + (-1)^factorial(i)*(x.^(2*i+1)./factorial(2*i+1)) + (-1)^factorial(i)*(x.^(2*i)./factorial(2*i));

end

f = SumTerms

figure,

plot(x,f);

title('tayler series of sinx + cosx')

xlabel('xi')

ylabel('y')

2) Plot Tn(3π) for n = 1, 2, . . . , 20. Note: In this case, your x-axis should be n.

n = input(prompt)

x = 3*pi;

SumTerms = (-1)*(x.^(1)./factorial(1)) + (-1)*(x.^(0)./factorial(0)); %initail

for i = 1:n

SumTerms = SumTerms + (-1)^factorial(i)*(x.^(2*i+1)./factorial(2*i+1)) + (-1)^factorial(i)*(x.^(2*i)./factorial(2*i));

end

f = SumTerms

figure,

plot(n,f);

title('tayler series of sinx + cosx')

xlabel('x*pi')

ylabel('sin(x) + cos(x)')

I try to plot x as 3*pi and y as taylor series of sin(x) + cos(x) but its show nothing on the graph

[Jon]

What is your question? Are you getting errors? If so please copy and paste them here. Are you unsure how to do a certain step, if so please explain what you are trying to do and where you are stuck

[John D'Errico]

To add to the good points made by Jon, I need to add some points, but now from John. Sadly, I guess we are a dime a dozen. :)

First, note that the sum of sin(x) + cos(x) is still jjust a sine wave. The amplitude and phase angle change.

syms x
simplify((sin(x) + cos(x)) - sqrt(2)*sin(x+pi/4))

ans = 

0

So you should still see essentially a sine wave. If you don't, then what are you seeing? Why do you think there is a problem?

Next, you need to consider if this series is convergent. Well, theoretically, it is globally convergent, if you compute with an infinite number of digits. But the reason this assignment was given to you was surely to see if you can sum a series, but also so you should see that a series need not be convergent without taking a huge number of terms, and when you do that, you end up seeing massive subtractive cancellation. Effectively, don't expect that series to converge at all, ever in double precision for x at all large in magnitude. How large must x be to see that problem? I'd not be surprsed if things start going to hell for abs(x) as large as pi. At least there you will start to see some numerical junk happening. At x=3*pi? You really are going to see junk happen there.

Next, you might notice that at x=3*pi, cos(x) is technically -1. But what hapens is you have numerical problems that prevent this from happening. For example, at x = pi/3, so a relatively small number. we have this in the cosine series:

format long g
x = pi/3;
N = 16;
n = (0:2:N)';
T = (-1).^(n/2).*x.^n./factorial(n);
[T,cumsum(T)]

ans = 9×2

1 1 -0.548311355616075 0.451688644383925 0.0501075571162564 0.501796201500181 -0.0018316361712683 0.499964565328913 3.58681040022699e-05 0.500000433432915 -4.37041971752511e-07 0.499999996390943 3.63083448473851e-09 0.500000000021778 -2.18772283334568e-11 0.4999999999999 9.99627727053343e-14 0.5

What happens is you are starting to get large numbers, there, but of opposite signs. Now form a cumulative sum, as we see it is converging to 1/2, as expected, since

cos(pi/3)

ans = 
                       0.5

is 1/2.

So all is good there. Now see what happens at x = 3*pi. I'll even go out a few more terms.

x = 3*pi;
N = 32;
n = (0:2:N)';
T = (-1).^(n/2).*x.^n./factorial(n);
[T,cumsum(T)]

ans = 17×2

1.0e+00 * 1 1 -44.4132198049021 -43.4132198049021 328.755682239758 285.342462434856 -973.406558494995 -688.064096060139 1544.0042657847 855.940169724559 -1523.87112968894 -667.930959964381 1025.45490056393 357.523940599549 -500.480812074284 -142.956871474735 185.233035956592 42.2761644818571 -53.7699054971861 -11.493741015329

sum(T)

ans = 
        -0.999995780428758

cos(3*pi)

ans = 
    -1

That is actually a little better than I thought would happen, but we are starting to see contamination from floating point errors and massive subtractive cancellation.

Similar stuff happens for the sine series for large arguments. And 10 is starting to grow large in this context.

Answer 1

[Nadia Shaik]

Hi,

From my understanding you can plot the figure when x is in the range of -pi to pi. But you are not able to plot the figure when x=3*pi.

The graph is being plotted, but as it is a single point it is not visible. When you try to plot a marker and colour it will be visible

The following changes have been done to the code

prompt='enter a value  '
n = input(prompt)
x = 3*pi;
SumTerms = (-1)*(x.^(1)./factorial(1)) + (-1)*(x.^(0)./factorial(0)); %initail

for i = 1:n
SumTerms = SumTerms + (-1)^factorial(i)*(x.^(2*i+1)./factorial(2*i+1)) + (-1)^factorial(i)*(x.^(2*i)./factorial(2*i));
end

f = SumTerms

figure,
plot(x,f,"ko");
title('tayler series of sinx + cosx')
xlabel('xi')
ylabel('y')

Results for the graph

Hope it helps!

[Phuri Nittiwat]

I Think for the case ploting x = 3*pi with the series, by themself is a sinusoidal pattern.

As a funtion its self is a vector so that I create 3*pi as a vector quantity that this case I think is the solution for this problem

clear

clc

clear all

%get input from user

prompt = 'initail value start from k=1: ';

k = 1 %start from k = 1

prompt = 'Enter last value: ';

n = input(prompt) %last number of the summation

x = linspace(0,3*pi,n); % X is 3*pi which in range of (0,3*pi) circle which n is the point we consider

t = SumT(k,n,x)

r = 0:1:n-1 %counting number

%ploting n in x axis vs y of taylor's series of 3*pi

figure,

plot(r,t);

title('tayler series of sinx + cosx')

xlabel('n')

ylabel('Summation of 3*pi')