Writing the Sum when Plotting Taylor Series

We want to plot f(x) and Taylor polynomials P0(x), P1(x),P2(x), P3(x) of f(x)=cot(x) ; about base point x=a=5*pi/8. step size is 0.01

The problem is we can't make our Taylor Series sum( T1) work. It gives this error:

Error using diff

Difference order N must be a positive integer scalar.

Error in matlab_hw1_q3 (line 11)

T1=sum( (diff(f(a),n(1,i))) ./ factorial(n(1,i)) .* ((x-a).^n(1,i)) , n(1,i))

Our code:

clc 
clear all
close all

f=@(x)  cot(x)
x= pi/8:0.01:7*pi/8 ;
a=5*pi/8 ;

n=[0 1 2 3]
for i=1:1:4
T1=sum( (diff(f(a),n(1,i))) ./ factorial(n(1,i)) .* ((x-a).^n(1,i)) , n(1,i))
end

Answer 1

[Walter Roberson]

There are two functions named diff that are quite different in what they do.

When you apply diff() to a symbolic function or symbolic expression, then diff() does calculus differentiation.

When you apply diff() to a numeric variable then the operation calculates successive differences, such as diff([3 7 -2]) would give [7-3, -2-7] = [4, -9]. In its simplest form, diff(v) where v is a vector, is equivalent to v(2:end)-v(1:end-1) . When diff() is applied to a numeric scalar, then because there are not at least two values to subtract, the result is empty.

When used with a second parameter, the parameter must be positive integer, and numeric diff() repeats the basic diff() operation that many times -- so diff([3 7 -2],2) would be diff(diff([3 7 -2])) which would be diff([4,-9]) which would be -13.

Now, you define the function handle f=@(x) cot(x) and you assign a=5*pi/8 so a is a numeric scalar. f(a) is going to calculate cot(5*pi/8) which is about -0.41 . Then you try diff(-0.41, n(1,1)) which is diff(-0.41, 0) which is an error because the order number is not a positive integer. If you had happened to start with n(1,1) being 1, then you would have had diff(-0.41, 1) which would have returned empty because there were not at least two numeric values to subtract.

If you are trying to take the 0'th derivative of cot(x), then the 1'st derivative, then 2'nd derivative, then you should not be working with function handles applied to numeric values: you should be working with symbolic expressions. Like

syms X
f = cot(X);
diff(f, X, n(1,i))

Once you have that abstract derivative in terms of the variable X, then you need to evaluate it at a :

   subs( diff(f, X, n(1,i)), X, a )

Answer 2

[Sena Ece Uzungil]

Thank you for your kind and very explaining answer . We tried implementing your solution but we are having an invalid index error right now. Our updated code is above ; we would be very happy if you can take another look at it.

clc 
clear all
close all
f=@(x)  cot(x)
X= pi/8:0.01:7*pi/8 ;
a=5*pi/8 ;
n=[0 1 2 3]

syms X
f = cot(X);
diff(f, X, n(1,i))

subs( diff(f,X,n(1,i), X, a))

for i=0:1:4
T1=sum( (diff(f(a),n(1,i))) ./ factorial(n(1,i)) .* ((x-a).^n(1,i)) , n(1,i))
end

And the error we get now is:

Index in position 2 is invalid. Array indices must be positive integers or

logical values.

Error in matlab_hw1_q3 (line 11)

diff(f, X, n(1,i))

[Walter Roberson]

At the point you first call diff, you have not assigned a value to i, so i will have its default value. The default value of the variables i and j are both sqrt(-1)

for i = 0:1:4
    the_diff = diff(f, X, n(1,i) );
    value_at_diff = subs(the_diff, X, a);
    T1 = sum( value_at_diff ./ .....)
end

I suggest you rethink this, though. The vector in your assignment to T1 is x, so when you sum() the expression, you are summing over all of the x values. You do not want to do that: you want to sum over all of i values.

T1 = 0;
for i=0:1:4
    T1 = T1 + expression_in_x;
end

[Sena Ece Uzungil]

Thank you for the answer!