Comparing Maclaurin Series of cos x and sin x

[jackson_sun]

I have been asked to differentiate cos x and six...the maclaurin series versions...

I have done the general and specific terms as shown below.

Im not sure if these are correct?
thanks

General Terms
cos x = ∑ (-1)n (x^(2n) / (2n)!)

COS x = ∑ (-1)n (x^(2n+1)/ (2n +1) / (2n)!)


sin x = ∑ (-1)n (x^(2n+1) / (2n+1)!)

‘sin x = ∑ (-1)n (((2n+1) x^(2n))/ (2n+1)!)

Specific Terms
sin x = x – (x^3 / 3!) + (x^5 / 5!) – (x^7 / 7!) + (x^9 / 9!) - …
‘sin x = 1 – (3x^2 / 3!) + (5x^4 / 5!) - (7x^6 / 7!) + (9x^8 / 9!) - …

cos x = 1 – (x^2 / 2!) + (x^4 / 4!) – (x^6 / 6!) + (x^8 / 8!) - …
COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!) - …
thanks

 

[Gib Z]

You already posted the same thing else where, learn what the factorial actually means! Then you can simplify yours derivatives and realize what your not realizing right now!

 

[HallsofIvy]

I have been asked to differentiate cos x and six...the maclaurin series versions...

I have done the general and specific terms as shown below.

Im not sure if these are correct?
thanks

General Terms
cos x = ∑ (-1)n (x^(2n) / (2n)!)

COS x = ∑ (-1)n (x^(2n+1)/ (2n +1) / (2n)!)

You know to use "^" to indicate exponent of the "x"- use for the exponent on (-1) also: (-1)^n. Obviously "cos x" can't be equal to both of these. Do you mean cos' x? Also A/B/C, at best, ambiguous. Do you mean
((x^(2n+1)/(2n+1))/(2n)! or (x^(2n+1))/((2n+1)/(2n)!)?

Finally, if you are differentiating as you say, you are "going the wrong way"! The derivative of x^n is nx^(n-1), not (1/(n+1)) x^(n+1).

sin x = ∑ (-1)n (x^(2n+1) / (2n+1)!)

‘sin x = ∑ (-1)n (((2n+1) x^(2n))/ (2n+1)!)

Specific Terms
sin x = x – (x^3 / 3!) + (x^5 / 5!) – (x^7 / 7!) + (x^9 / 9!) - …
‘sin x = 1 – (3x^2 / 3!) + (5x^4 / 5!) - (7x^6 / 7!) + (9x^8 / 9!) - …

cos x = 1 – (x^2 / 2!) + (x^4 / 4!) – (x^6 / 6!) + (x^8 / 8!) - …
COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!) - …
thanks

 

[Gib Z]

It looks to me the cos x in capitals is the integral of cos x in normal...

 

[jackson_sun]

yeh COS x is the integral of cos x...i meant integral of cos x and derivative of sin x...sorry for the confusion.

I have realized how to simplify 'sin x so that it equals cos x...but i am not sure how to simplify the integral of cos x to equal sin x.

COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!)
= ?

any idea? and on the subject...what is the notation for the integral of cos x...I have used COS x but this is obviously incorrect.
thanks...

 

[Gib Z]

Take your second term...if you don't see the arithmetic, don't try this level of math yet, \frac{\frac{x^3}{3}}{2!} =\frac{x^3}{3\cdot2!} = \frac{x^3}{3!}.

I didn't take into account the sign of the integral, but you get what i mean.

To learn how to type the math graphics I am typing, click on the images I make appear. That will tell you what i typed to get that. Or at least say, "Integral of.."...

 

[jackson_sun]

I understand exactly what you have done.
How would it be done for the general terms is my bigger problem

 

[Gib Z]

I understand exactly what you have done.

Good

How would it be done for the general terms is my bigger problem

\cos x = \sum_{n=0}^{\infty} -1^n \frac{x^{2n}}{2n!}

Integrate every term in the sum, with respect to x. In every individual term, n is constant.

So

\int\cos x dx= \sum_{n=0}^{\infty} -1^n \frac{\frac{x^{2n+1}}{2n+1}}{2n!}

(2n+1)2n!=(2n+1)!,

therefore it simplifies to


\int\cos x dx= \sum_{n=0}^{\infty} -1^n \frac{x^{2n+1}}{(2n+1)!}, which is exactly your series for sin x.

 

[HallsofIvy]

Just one comment: it would be far better to write
\int\cos x dx= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}
with parentheses around the -1. I would be inclined to interpret -1ⁿ as -(1ⁿ) which, of course, would be just -1!

 

[Gib Z]

O yes i guess i would too, if someone else wrote that..thanks

 

[jackson_sun]

Thanks gents

 

[Gib Z]

No Problem, and welcome to PF