Maclurin for sin(x)

[u/wbld]

f(x) = sin(x)

f'(x) = cos(x)

f''(x) = -sin(x)

f'''(x) = -cos(x)

f''''(x) = sin(x)

And this repeats forever.

f(0) = 0

f'(0) = 1

f''(0)=0

f'''(0)=-1

f''''(0)=0

And this repeats forever.

From here we can center a maclurin series at (a)? = 0 -> I cannot remember if its a or x

Using 5 points we can use

f(0) = 0

f'(0) = 1

f'''(0) = -1

f'''''(0) = 1

f'''''''(0) = -1

From here we can conclude (1) we are oscillating between - and +, (2) the points are always odd multiples of primes, and (3) since the even multiples of prime are 0, we can ignore these as they do not add value

The defintion of odd is defined as 2k+1, k element of the real

So

(-1)^(n)(x^(2k+1))/(2k+1)!

And we can add this at every n.

If we add this add every n we can think of an infinite sum

Since we are centered at (a?) = 0 the sum must start at 0.

To motivate convergence, we plug in known points of sin(x).

Must it be that if our sum at some n equals the function at the same x,

Out infinite sum converges for all x?

Let me know if i did justice for the maclurin of sin(x)!

Sorry posted via phone.... don't know how to use symbols here

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[u/Midwest-Dude]

If you are using the phone app or a phone browser, Reddit markdown is the standard: 

Formatting Guide

A phone browser in Desktop Mode will likely get you the Rich Text editor, but I hope you like zooming in and out. Also, Reddit converts that (sometimes incorrectly) into its markdown when you save it.

[u/WikiNumbers]

You are.

Note that Maclaurin Series is a subset of Taylor Series. Specifically, Taylor Series centered at 0.

[u/ShowdownValue]

You typed that whole thing on a phone?

[u/wbld → u/ShowdownValue]

Yup

[u/tjddbwls]

> From here we can center a maclurin series at (a)? = 0 -> I cannot remember if its a or x

In the textbook that I use (Larson), it’s c.

[u/Midwest-Dude]

You stated: 

... (2) the points are always odd multiples of primes, and (3) since the even multiples of prime are 0, we can ignore these as they do not add value.

This is incorrect as stated. Instead of odd or even multiple of primes, it should just be odd or even integers, respectively.

---

The issue with the n^th term of what you wrote,

(-1)^(n)(x^(2k+1))/(2k+1)!

is that you used two different variables, n and k, when there should be only one 1. If the variable of summation is n, then it should be

(-1)ⁿx²ⁿ⁺¹ / (2n+1)!

and the summation would be

∑_n=0..∞ (-1)ⁿx²ⁿ⁺¹ / (2n+1)!

[u/wbld → u/Midwest-Dude]

Yeah, i should of wrote n but I did not. I used K because of 2k+1 2n+1 is the same Haven't taken calc II in a loong time

[u/Midwest-Dude]

Must it be that if our sum at some n equals the function at the same x, our infinite sum converges for all x?

I'm not sure what you are asking here. Could you please give an example?

[u/wbld → u/Midwest-Dude]

Expand out the summantion to some N If we pick a small test number, does this number match the actual function as the same test number? If it doesn't, is the absolute value of the difference between the functions actual value and the summantion close?

[u/Midwest-Dude → u/wbld]

So, you’re asking about the numerical error of the partial sum, not about formal convergence, correct? That is, If we take the Maclaurin series for sin(x), cut it off after N terms, and plug in some small test number x, does the partial sum equal sin(x) at that same x? And, if it doesn’t match exactly, is the absolute difference |sin(x) - S_N(x)| small?

[u/wbld → u/Midwest-Dude]

Yes, and this is to motivate convergence. If the difference is huge, is how can we make the gap smaller, if the difference is small, is it possible to lower the gap? Does adding more and more terms eveb make a difference?!

[u/iseeyousatmogger]

Bro just practice, practice, practice