series

**Enita** · January 30th 2007, 12:56 PM

hi everyone!

needed some help on two questions.

1. Find the first 3 non-zero terms in the series for $secx$

2. Assuming the series for $e^x$ and $tanx$ , determine the series for $e^xtanx$ up to and including the term in $x^4$

thank you.

**CaptainBlack** · January 30th 2007, 02:15 PM

Originally Posted by Enita

1. Find the first 3 non-zero terms in the series for $secx$

Well we need the Taylor series about 0.

So lets look at the sequence of derivatives and find the first three which are non-zero.

First the zero-th drrivative is non-zeros at $x=0$ and equal to $\sec(0)=1$

The first derivative:

$\frac{d \sec(x)}{dx}=\frac{\sin(x)}{\cos^2(x)}$

which is zero at $x=0$

Second derivative:

$\frac{d^2 \sec(x)}{dx^2}=\frac{\sin^2(x)+1}{\cos^3(x)}$

which is equal to $1$ at $x=0$

Third derivative:

$\frac{d^3 \sec(x)}{dx^3}=\frac{2 \sin(x)}{\cos^2(x)}+\frac{3 \sin^3(x)+3\sin(x)}{\cos^4(x)}$

which is zero at $x=0$

Fourth derivative:

$\frac{d^4 \sec(x)}{dx^4}=\frac{11 \sin^2(x)+5}{\cos^3(x)}+\frac{12 \sin^4(x)+12\sin^2(x)}{\cos^5(x)}$

which is $5$ when $x=0$ .

So the series expansion of $\sec(x)$ to the first three non zero terms is:

$\sec(x)=\sec(0)+ \left. \frac{d^2 \sec(x)}{dx^2}\right|_{x=0}x^2/2 + \left. \frac{d^4 \sec(x)}{dx^4}\right|_{x=0}x^4/24+...$

so (if I have done this right):

$\sec(x)=\sec(0)+ x^2/2+(5/24)x^4+...$

RonL

**Enita** · January 31st 2007, 12:41 PM

Originally Posted by CaptainBlack

Third derivative:

$\frac{d^3 \sec(x)}{dx^3}=\frac{2 \sin(x)}{\cos^2(x)}+\frac{3 \sin^3(x)+3\sin(x)}{\cos^4(x)}$

which is zero at $x=0$

hey thank you.

yes the answer is right checked the back of the book.

but what do you do for the line i quoted?

**CaptainBlack** · January 31st 2007, 01:04 PM

Originally Posted by Enita

hey thank you.

yes the answer is right checked the back of the book.

but what do you do for the line i quoted?

It does not contribute to the series as it is zero at x=0, but we need it
to construct the next higher order derivative.

RonL

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