Is There a Trick to Simplify Taylor Series Expansion?

[silverfury]

Homework Statement

I have tried differentiating many times in order to find a simpler diffrential eq to solve it further but didn’t have any luck there. Pls help

Relevant Equations

To prove:
ln(sec x + tan x) = x + x^3/6 + x^5/24 ......

I tried diffrentiating upto certain higher orders but didn’t find any way.. is there a trick or a transformation involved to make this task less hectic? Pls help

 

[fresh_42]

I have only the idea of using three series and calculate the first terms.
We have $\log(\sec x +\tan x)= \log (1+\sin x)-\log(1+(\cos x -1))$ and the series
\begin{align*}
\sin x &= x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} \pm \ldots\\
\cos x -1 &= - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} -\dfrac{x^6}{6!} \pm \ldots \\
\log (1+x)&= x -\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4} \pm \ldots
\end{align*}
This works only for a few series elements but seems to be less work than differentiating. However, if you need the correct formula with all series members, things are really ugly.

 

[pasmith]

Problem Statement: I have tried differentiating many times in order to find a simpler diffrential eq to solve it further but didn’t have any luck there. Pls help
Relevant Equations: To prove:
ln(sec x + tan x) = x + x^3/6 + x^5/24 ...

I tried diffrentiating upto certain higher orders but didn’t find any way.. is there a trick or a transformation involved to make this task less hectic? Pls help

You can say that $$1 + \sin x = 1 + x - \tfrac16x^3 + \tfrac1{120}x^5 + O(x^7)\\<br /> \sec x = 1 + \tfrac12 x^2 + \tfrac5{24}x^4 + O(x^6) \\<br /> \log(1 + x) = x - \tfrac12x^2 + \tfrac13 x^3 - \tfrac14 x^4 + \tfrac15x^5 + O(x^6).$$ Now multiply $\sec x$ by $1 + \sin x$ retaining all terms up to and including $O(x^5)$. The constant term is 1, so you can drop this and substitute the remaining terms directly into the log series and expand, retaining only terms up to and including $O(x^5)$.

You can see that getting higher order terms out of this involves a lot of algebra, so differentiation might actually be easier.

 

[jim mcnamara]

Suppose $y=ln(f(x))$

Chain rule, $y'=1f(x)⋅f'(x)$
y'=1sec(x)+tan(x)⋅(sec(x)+tan(x))'
y'=1sec(x)+tan(x)⋅(sec(x)tan(x)+sec^2(x))
y'=1sec(x)+tan(x)⋅sec(x)(sec(x)+tan(x))
y'=sec(x)

Is this any help?

Wrong. Thanks @SammyS

 

[fresh_42]

I think you have forgotten the division signs: $\dfrac{1}{f(x)}$ and $\dfrac{1}{\sec x +\tan x}.$

 

[SammyS]

Suppose $y=ln(f(x))$

Chain rule, $y'=1f(x)⋅f'(x)$
$$y'=1sec(x)+tan(x)⋅(sec(x)+tan(x))'$$
$$y'=1sec(x)+tan(x)⋅(sec(x)tan(x)+sec^2(x))$$
$$y'=1sec(x)+tan(x)⋅sec(x)(sec(x)+tan(x))$$
$$y'=sec(x)$$

Is this any help?

To be more explicit regarding what @fresh_42 states in the previous post, the above should be:
Chain rule, $y'=\dfrac{1}{f(x)}\cdot f'(x)$

$y'=\dfrac{1} {\sec(x)+\tan(x)}\cdot \dfrac{d}{dx}(\sec(x)+\tan(x))$​

...

$y'=\sec(x)$​

 

[jim mcnamara]

Thanks for the correction.