determine Maclaurin development of order 3 with residual term with o notation e^sinx

**Riazy** · February 23rd 2011, 05:50 AM

determine Maclaurin development of order 3 with residual term with o notation e^sinx

My main problem is that i don't understand the rules of O notation

1. e^sinx ... u= sinx

we get e^u

2. Develeopment of e^x = 1 + x + x^2/2 + (O(x^4))
Develeopment of sinx = x - 1/6 x^3 + O(x^5)

3. e^sinx = e^x-x^3/6 * e^(O(x^5))
= e^x * e^-x^3/6 * e(O(x^5))

I could need some help finish this off
secondly I don't understand why we put an O notation of O(x^5)) here out of a sudden, we had an ordo of 4 as well didnt we?

Would be grateful if someone could explain all the Ordo (O notation) rules so that i can start working on problem self without using hints like the one above. All that we have is a booklet handout with problems + a very bad book that does not cover the topic well.

Thanks alot

**chisigma** · February 23rd 2011, 06:57 AM

Setting $f(x)=e^{\sin x}$ You can observe that is...

$\displaystyle \frac{d}{dx} \ln f(x)= \frac{f^{'}(x)}{f(x)}= \cos x$ (1)

Now if is...

$\displaystyle f(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n}$ (2)

... the (1) becomes...

$\displaystyle a_{1} + 2\ a_{2}\ x + 3\ a_{3}\ x^{2} + 4\ a_{4}\ x^{3} + ... =$

$\displaystyle = (a_{0} + a_{1}\ x + a_{2}\ x^{2} + a_{3}\ x^{3} + ...)\ (1-\frac{x^{2}}{2} + ...)=$

$\displaystyle = a_{0} + a_{1}\ x + (a_{2}-\frac{1}{2})\ x^{2} - \frac{a_{1}}{2}\ x^{3}+ ...$ (3)

... and from (3) You derive...

$\displaystyle a_{0}=1$

$\displaystyle a_{1}=a_{0}=1$

$\displaystyle a_{2}= \frac{a_{1}}{2}= \frac{1}{2}$

$\displaystyle a_{3}= \frac{1}{3}\ (a_{2}-\frac{1}{2}) = 0$

$\displaystyle a_{4}= -\frac{a_{1}}{12}= -\frac{1}{12}$

... so that is...

$\displaystyle e^{\sin x} = 1 + x + \frac{x^{2}}{2} + O(x^{4})$ (4)

Kind regards

$\chi$ $\sigma$

**Prove It** · February 23rd 2011, 07:53 AM

Originally Posted by chisigma

Setting $f(x)=e^{\sin x}$ You can observe that is...

$\displaystyle \frac{d}{dx} \ln f(x)= \frac{f^{'}(x)}{f(x)}= \cos x$ (1)

Now if is...

$\displaystyle f(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n}$ (2)

... the (1) becomes...

$\displaystyle a_{1} + 2\ a_{2}\ x + 3\ a_{3}\ x^{2} + 4\ a_{4}\ x^{3} + ... =$

$\displaystyle = (a_{0} + a_{1}\ x + a_{2}\ x^{2} + a_{3}\ x^{3} + ...)\ (1-\frac{x^{2}}{2} + ...)=$

$\displaystyle = a_{0} + a_{1}\ x + (a_{2}-\frac{1}{2})\ x^{2} - \frac{a_{1}}{2}\ x^{3}+ ...$ (3)

... and from (3) You derive...

$\displaystyle a_{0}=1$

$\displaystyle a_{1}=a_{0}=1$

$\displaystyle a_{2}= \frac{a_{1}}{2}= \frac{1}{2}$

$\displaystyle a_{3}= \frac{1}{3}\ (a_{2}-\frac{1}{2}) = \frac{1}{6}$

$\displaystyle a_{4}= -\frac{a_{1}}{12}= -\frac{1}{12}$

... so that is...

$\displaystyle e^{\sin x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + O(x^{4})$ (4)

Kind regards

$\chi$ $\sigma$

Incase the OP was wondering the MEANING of the "Big Oh" notation, to use what $\displaystyle \chi\sigma$ so kindly posted...

$\displaystyle e^{\sin{x}} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O\left(x^4\right)$

The $\displaystyle O\left(x^4\right)$ in this example means that when you are evaluating this function for values close to $\displaystyle x=0$ (as you would with a MacLaurin Series - if not, you would use a Taylor Series centred around a point closer to the point you are evaluating the function at), the absolute error (remainder) is no greater than some constant multiple of $\displaystyle |x^4|$ .

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