Maclaurin series and Landau's big "Oh" problem.

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September 23rd 2007, 02:22 PM
Rombie

Maclaurin series and Landau's big "Oh" problem.

I have found the first 5 Taylor polynomials about x=0 for cosh(x). Being p0=1, p1=1, p2=1+ $\frac{1}{2}x^2$ , p3=1+ $\frac{1}{2}x^2$ , p4=1+ $\frac{1}{2}x^2$ + $\frac{1}{24}x^4$ .

How do I Find the Maclaurin series for cosh(x)?

I can see it should be something like
$f(x) \approx \sum_{n = 1}\frac{x^{2n}}{2n!}$
but how can I show a methodic way to get there from the polynomial approximations?

Also, how do I use Landau's big O notation to find

$\lim_{x \to 0}\frac{2cosh^2(x)-2}{1-cosh(3x)}$

I simplify this with identity $cosh^2(x)=\frac{1}{2}(1+cosh(2x))$ to

$\lim_{x \to 0}\frac{cosh(2x)-1}{1-cosh(3x)}$

then change them to Maclaurin approximations with big O error.

$\lim_{x \to 0}\frac{2x^2+O(x^3)}{\frac{-9}{2}x^2-O(x^3)}$

How do I simplify that expression to get the limit?

Thank you for any help.
September 23rd 2007, 06:29 PM
ThePerfectHacker

I have no idea what you are doing.

Just compute $\frac{f^{(n)}(0)}{n!}$ .
Where $f(x) = \cosh x$

And $f^{(\mbox{even})} = \cosh x$ and $f^{(\mbox{odd})} = \sinh x$ .
September 23rd 2007, 07:52 PM
Rombie

I know how to do the big O problem now.. it was trivial. divide through by x^2 and let x tend to 0. yielding the limit -4/9

I know how to get the polynomials, I've worked them out. I need a way to find the series from the polynomials. I don't want to just write down. This is the series I found for cosh(x) through google which makes sense.

$f(x) \approx \sum_{n = 0}\frac{x^{2n}}{2n!} $

(there should be an infinity symbol above the sum)

I want to get to that series from the Maclaurin polynomials which I have calculated if possible.
September 23rd 2007, 08:17 PM
CaptainBlack

Quote:

Originally Posted by Rombie

I have found the first 5 Taylor polynomials about x=0 for cosh(x). Being p0=1, p1=1, p2=1+ $\frac{1}{2}x^2$ , p3=1+ $\frac{1}{2}x^2$ , p4=1+ $\frac{1}{2}x^2$ + $\frac{1}{24}x^4$ .

How do I Find the Maclaurin series for cosh(x)?

I can see it should be something like
$f(x) \approx \sum_{n = 1}\frac{x^{2n}}{2n!}$
but how can I show a methodic way to get there from the polynomial approximations?

You don't. You use the definition of the Maclaurin series, you will
see that it involves $f^{(n)}(x)$ the $n$ -th derivative of your function $f(x)=\cosh(x)$ .

RonL
September 23rd 2007, 08:24 PM
CaptainBlack

Quote:

Originally Posted by Rombie

Also, how do I use Landau's big O notation to find

$\lim_{x \to 0}\frac{2cosh^2(x)-2}{1-cosh(3x)}$

I simplify this with identity $cosh^2(x)=\frac{1}{2}(1+cosh(2x))$ to

$\lim_{x \to 0}\frac{cosh(2x)-1}{1-cosh(3x)}$

then change them to Maclaurin approximations with big O error.

$\lim_{x \to 0}\frac{2x^2+O(x^3)}{\frac{-9}{2}x^2-O(x^3)}$

How do I simplify that expression to get the limit?

Thank you for any help.

(Assuming that what you have is correct) Divide top and bottom through by $x^2$ to get:

$\frac{2x^2+O(x^3)}{\frac{-9}{2}x^2-O(x^3)}=\frac{2+O(x)}{\frac{-9}{2}-O(x^x)}$

Hence:

$\lim_{x \to 0}\frac{2x^2+O(x^3)}{\frac{-9}{2}x^2-O(x^3)}$

.......... $=\frac{\lim_{x \to 0}[2+O(x)]}{\lim_{x \to 0}[\frac{-9}{2}-O(x^x)]}=\frac{2}{-(9/2)}=- ~\frac{4}{9}$
September 23rd 2007, 09:58 PM
Rombie

Quote:

Originally Posted by CaptainBlack

You don't. You use the definition of the Maclaurin series, you will
see that it involves $f^{(n)}(x)$ the $n$ -th derivative of your function $f(x)=\cosh(x)$ .

RonL

Would finding the series for the nth polynomial help? I'm confused by the fact that every second step yields a 0 term and how that translates to the Maclaurin series. I know the formula I'm just having trouble applying it. The definition has the nth derivative in it but then in the examples it's always an infinite series that looks similar to the series for the nth polynomial.
September 23rd 2007, 10:45 PM
CaptainBlack

Quote:

Originally Posted by Rombie

Would finding the series for the nth polynomial help? I'm confused by the fact that every second step yields a 0 term and how that translates to the Maclaurin series. I know the formula I'm just having trouble applying it. The definition has the nth derivative in it but then in the examples it's always an infinite series that looks similar to the series for the nth polynomial.

Every second step yields 0 because $\cosh(x)$ is an even function and only has even powers of $x$ in its Mclaurin series.

Forget about the polynomials for now they are not directly relevant to finding the general term of the series.

$ \cosh(x)=\frac{e^x+e^{-x}}{2} $

so:

$ D^{(n)} [\cosh(x)]=\frac{D^{(n)} [e^x]+D^{(n)} [e^{-x}]}{2}=\frac{e^x+(-1)^n e^{-x}}{2} $

Hence:

$ \left. D^{(n)} [\cosh(x)]\right| _{x=0}=1\ \ n \mbox{ even } $

and:

$ \left. D^{(n)} [\cosh(x)]\right| _{x=0}=0\ \ n \mbox{ odd } $

RonL
September 24th 2007, 12:59 PM
Rombie

So from that can I say?
$\sum_{n = 0}\frac{1}{n!}x^{n} $ (n even)

Since the nth even derivative will always be 1.
September 24th 2007, 07:35 PM
CaptainBlack

Quote:

Originally Posted by Rombie

So from that can I say?
$\sum_{n = 0}\frac{1}{n!}x^{n}$ $ $ (n even)

Since the nth even derivative will always be 1.

That is what the book says.

RonL
September 24th 2007, 08:46 PM
Rombie

Thanks for your time.