# Thread: Finding Maclaurin Series By Division

1. ## Finding Maclaurin Series By Division

Find the first four nonzero terms of the Maclaurin series for the function by dividing appropriate Maclaurin series

secx

I know secx=1/cosx and the Maclaurin series of cosx=1-x^2/2!+x^4/4!-x^6/6!............

I do not know how one divide those terms

Thank you

2. Originally Posted by soleilion
Find the first four nonzero terms of the Maclaurin series for the function by dividing appropriate Maclaurin series

secx

I know secx=1/cosx and the Maclaurin series of cosx=1-x^2/2!+x^4/4!-x^6/6!............

I do not know how one divide those terms

Thank you
$\displaystyle \sec(x)=\frac{1}{\cos(x)}=\frac{1}{1-x^2/2!+x^4/4!-x^6/6!+...} $$\displaystyle =\frac{1}{\cos(x)}=\frac{1}{1- (x^2/2!-x^4/4!+x^6/6!-...) } ............. \displaystyle =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2 ................... \displaystyle +(x^2/2!-x^4/4!+x^6/6!-...)^3+.. The first four non-zero terms will be the \displaystyle 0,\ 2,\ 4,\ 6 th powers of \displaystyle x So: \displaystyle \sec(x) = 1 +(x^2/2!-x^4/4!+x^6/6!)+[x^4/(2!)^2-2x^6/(2!4!)]+[x^6/(2!)^3]+O(x^8) RonL 3. Originally Posted by CaptainBlack \displaystyle \sec(x)=\frac{1}{\cos(x)}=\frac{1}{1-x^2/2!+x^4/4!-x^6/6!+...}$$\displaystyle =\frac{1}{\cos(x)}=\frac{1}{1- (x^2/2!-x^4/4!+x^6/6!-...) }$

............. $\displaystyle =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2$
................... $\displaystyle +(x^2/2!-x^4/4!+x^6/6!-...)^3+..$

The first four non-zero terms will be the $\displaystyle 0,\ 2,\ 4,\ 6$ th powers of $\displaystyle x$

So:

$\displaystyle \sec(x) = 1 +(x^2/2!-x^4/4!+x^6/6!)+[x^4/(2!)^2-2x^6/(2!4!)]+[x^6/(2!)^3]+O(x^8)$

RonL

I misunderstood
How the series have 1power, 2power and 3power????

4. Originally Posted by soleilion
I misunderstood
How the series have 1power, 2power and 3power????

Those terms don't appear, that is they have zero coefficients.

RonL

5. Originally Posted by CaptainBlack
$\displaystyle \sec(x)=\frac{1}{\cos(x)}=\frac{1}{1-x^2/2!+x^4/4!-x^6/6!+...}$$\displaystyle =\frac{1}{\cos(x)}=\frac{1}{1- (x^2/2!-x^4/4!+x^6/6!-...) }$

............. $\displaystyle =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2$
................... $\displaystyle +(x^2/2!-x^4/4!+x^6/6!-...)^3+..$

The first four non-zero terms will be the $\displaystyle 0,\ 2,\ 4,\ 6$ th powers of $\displaystyle x$

So:

$\displaystyle \sec(x) = 1 +(x^2/2!-x^4/4!+x^6/6!)+[x^4/(2!)^2-2x^6/(2!4!)]+[x^6/(2!)^3]+O(x^8)$

RonL
............. $\displaystyle =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2$
................... $\displaystyle +(x^2/2!-x^4/4!+x^6/6!-...)^3+..$

How does this come from 1/cosx

what method did you use?
Do you have any similar example?

6. Originally Posted by soleilion
............. $\displaystyle =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2$
................... $\displaystyle +(x^2/2!-x^4/4!+x^6/6!-...)^3+..$

How does this come from 1/cosx

what method did you use?
Do you have any similar example?
For $\displaystyle |u|<1$:

$\displaystyle \frac{1}{1-u}=1+u+u^2+...$

In our case, for small $\displaystyle x$:

$\displaystyle u=x^2/2!-x^4/4!+x^6/6!-...$

RonL

7. Originally Posted by CaptainBlack
For $\displaystyle |u|<1$:

$\displaystyle \frac{1}{1-u}=1+u+u^2+...$

In our case, for small $\displaystyle x$:

$\displaystyle u=x^2/2!-x^4/4!+x^6/6!-...$

RonL
Thank you very much
I got it

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