# Thread: Stratergies for tricky Taylor Expansions

1. ## Stratergies for tricky Taylor Expansions

I am required to taylor expand ln(sec(x)) to obtain the first 3 non-zero terms. What I wanted to know was if anyone knows a quicker method than differentating 6 times (this gets very messy)?

2. Note that $\displaystyle \int{\tan{x}\,dx} = \ln{|\sec{x}|}$.

So find the Taylor series for $\displaystyle \tan{x}$, and integrate it term by term to find the Taylor series for $\displaystyle \ln{|\sec{x}|}$.

3. Ignore this. I didn't realise that only the first three terms were wanted.

4. Originally Posted by Resilient
Isn't that very hard to find?
Not particularly. It's usually given in a table or can be found quite easily on Wikipedia...

5. Originally Posted by Prove It
Note that $\displaystyle \int{\tan{x}\,dx} = \ln{|\sec{x}|}$.

So find the Taylor series for $\displaystyle \tan{x}$, and integrate it term by term to find the Taylor series for $\displaystyle \ln{|\sec{x}|}$.
Thanks, that makes sense. Another part of the question asked me to expand tan(x+pi/4) for the first three terms. I did this by the following method:

tan(x+pi/4) cos(x + pi/4) = sin(x + pi/4)
tan(x+pi/4) * 0.5 *root(2) (cos(x) - sin(x)) = 0.5 * root(2) (sin(x) + cos(x))
tan(x+pi/4) (cos(x) - sin(x)) = sin(x) + cos(x)

I then let tan(x+pi/4) = a + bx + cx^2 , subbed in the expansions for sin(x) and cos(x) and equated coefficients. Again, can you see an improvement upon this method? Time is very precious in the exam so I'm trying to optimise my working.

6. Why don't you just substitute $\displaystyle x + \frac{\pi}{4}$ into the series for $\displaystyle \tan{X}$?

7. Originally Posted by Prove It
Why don't you just substitute $\displaystyle x + \frac{\pi}{4}$ into the series for $\displaystyle \tan{X}$?
I get the following working trying to follow this method which seems to be much more difficult:

$\newline tan(x) = 1 + \frac{x^3}{3} + \frac{2x^5}{15} + ... \newline tan(x + \frac{\pi}{4}) = 1 + \frac{1}{3}(x + \frac{\pi}{4})^3 + \frac{2}{15}(x + \frac{\pi}{4})^5 +... \newline Considering \hspace{2} x^0 \hspace{2} terms: \newline 1 + \frac{1}{3} (\frac{\pi}{4})^3 + \frac{2}{15} (\frac{\pi}{4})^5 + ... = 1 + tan(\frac{\pi}{4}) = 2 \newline Considering \hspace{2} x^1 \hspace{2} terms: \newline \frac{1}{3} \binom{3}{1}(\frac{\pi}{4})^2 x + \frac{2}{15} \binom{5}{1}(\frac{\pi}{4})^4 x + ... =?$