Can someone help me with this 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

Mentor
2022 Award
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.

Homework Helper
2022 Award
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 $$1 + \sin x = 1 + x - \tfrac16x^3 + \tfrac1{120}x^5 + O(x^7)\\ \sec x = 1 + \tfrac12 x^2 + \tfrac5{24}x^4 + O(x^6) \\ \log(1 + x) = x - \tfrac12x^2 + \tfrac13 x^3 - \tfrac14 x^4 + \tfrac15x^5 + O(x^6).$$ 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.

Mentor
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

Last edited:
Mentor
2022 Award
I think you have forgotten the division signs: ##\dfrac{1}{f(x)}## and ##\dfrac{1}{\sec x +\tan x}.##

Staff Emeritus
Homework Helper
Gold Member
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
Mentor
Thanks for the correction.