Writing a sin function program dealing with factorials, help in editing and error?

I am trying to write a specific program involving the sine function and factorials.

The problem states that:

sin(x) = x - x^3 / 3! + x^5 / 5! - x^7 / 7! + x^9 / 9!

I am trying to get the error rate very low, basically having the number of terms it takes to get enough accuracy by 10^-5, so that:

abs[sin(x) - sk(x)] / abs[sin(x)] = 10^-5

I know the sk equation should be written as:

sk = x-x^3 / 3! + .......x^(2k-1) / (2k-1)!

I am starting out my program with:

for s = 1: maxs

nsteps(s) = (ones(s)

end

I am confused on how to add on to my code to create my desired result, I know intuitively that when plotting this, the #'s should increase as x increases.

[Merged information from duplicate question]

I am trying to write a specific program involving the sine function and factorials.

The problem states that:

sin(x) = x - x^3 / 3! + x^5 / 5! - x^7 / 7! + x^9 / 9!

I am trying to get the error rate very low, basically having the number of terms it takes to get enough accuracy by 10^-5, so that:

abs[sin(x) - sk(x)] / abs[sin(x)] = 10^-5

I know the sk equation should be written as:

sk = x-x^3 / 3! + .......x^(2k-1) / (2k-1)!

I am starting out my program with:

for s = 1: maxs

nsteps(s) = (ones(s)

end

I am confused on how to add on to my code to create my desired result, I know intuitively that when plotting this, the #'s should increase as x increases.

Here is my code:

function [s] = fact(t)

s = 1:max(s);

for s = 1:max(s)

nsteps(s) = ones(s);

end
for t = (1:2:2*k-1);
(-1)^(0:k-1) .* x^t ./ factorial(t);
end

I receive this error when performing it:

??? Undefined function or variable 's'.

??? Undefined function or variable "s".

Error in ==> fact at 2
s = 1:max(s);

HELP PLEASE!

[Walter Roberson]

Elige pointed out that "s" has not been defined in the new code.

Answer 1

[Walter Roberson]

t = (1:2:2*k-1);
(-1).^(0:k-1) .* x.^t ./ factorial(t)

Or something like that.

(Note: corrected -- ^ operators had to be .^ )

[Reelz]

here's my code:

function sin(t) = fact(t)

for t = (1:2:2*k-1);

(-1)^(0:k-1) .* x^t ./ factorial(t);

end

And here is the error I receive:

??? Error: File: fact.m Line: 1 Column: 17

The expression to the left of the equals sign is not a valid target for an

assignment.

?? it seems valid to me

[Walter Roberson]

function s = mysin(t)

[Reelz]

function s = mysin(t)

s = 1:max(s);

for s = 1:max(s)

nsteps(s) = ones(s);

end

for t = (1:2:2*k-1);

(-1)^(0:k-1) .* x^t ./ factorial(t);

end

??? Undefined function or variable "s".

Error in ==> mysin at 2

s = 1:max(s);

still undefined?

[Walter Roberson]

You have to define s before you use it.

You will also have to define k before you use that.

Your nsteps variable will fail if s somehow becomes greater than 1, as ones(2) would be a 2 x 2 array and that is not going to fit within the single numeric array location nsteps(2)

I do not know what you are trying to do with your for loop, or your nsteps variable.

[Walter Roberson]

Hint: (-1).^(0:k-1) .* x.^t ./ factorial(t) is an expression, not a command. You should be doing something with the expression, not just discarding the value of it (which is what a semi-colon means)

[Reelz]

I rewrote my code trying to define variables as is:

function s = mysin(t)

s = 0;

for s = 1:max(s)

nsteps(s) = ones(s);

k = 0:.1:pi/2;

x = 0:.1:2;

t = (1:2:2*k-1);

end

while t = (-1)^(0:k-1) .* x^t ./ factorial(t)

end

still receiving this error though:

??? Error: File: mysin.m Line: 9 Column: 9

The expression to the left of the equals sign is not a valid target for an

assignment.

[Walter Roberson]

"=" is not permitted in a "while". "while" has to be given a logical test.

Warning: You initialize s to 0, so max(s) is 0, and 1:0 is the empty vector, so your "for" loop will not do any iterations.

What is the purpose of your nsteps vector? Why are you trying to initialize it to a square matrix such as

1 1 1

1 1 1

1 1 1

(i.e., ones(3))

Answer 2

[Richard Brown]

I take it this is a programming exercise, not a straight mathematics question - you can work out the number of terms pretty easily using Taylor's theorem with remainder ...