# Is There a Trick to Simplify Taylor Series Expansion?

• silverfury
In summary: I am not able to edit my post, so I am posting the correct version here: Suppose ##y=ln(f(x))##Chain rule, ##y'=\dfrac{1}{f(x)}\cdot f'(x)## y'=\dfrac{1}{\sec(x)+\tan(x)}\cdot (\sec(x)+\tan(x))' y'=\dfrac{1}{\sec(x)+\tan(x)}\cdot (\sec(x)\tan(x)+\sec^2(x)) y'=\dfrac{1}{\sec(x)+\tan(x)}\cdot \sec(x)(\sec(x)+\tan(x)) y'=\sec(x)Is
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

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.

silverfury said:
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.

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

jim mcnamara said:
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.

## 1. What is a Taylor series expansion?

A Taylor series expansion is a mathematical representation of a function as an infinite sum of terms, where each term is a polynomial approximation of the function at a specific point.

## 2. Why do we use Taylor series expansions?

Taylor series expansions are useful for approximating complicated functions with simpler polynomial expressions, making it easier to analyze and manipulate the function. They are also used in numerical analysis and in solving differential equations.

## 3. How do you find the coefficients in a Taylor series expansion?

The coefficients in a Taylor series expansion can be found by taking derivatives of the function at the expansion point and plugging them into the general formula for a Taylor series.

## 4. What is the difference between a Taylor series and a Maclaurin series?

A Taylor series expansion is a generalization of a Maclaurin series, which is a special case where the expansion point is 0. In other words, a Maclaurin series is a type of Taylor series expansion with a specific expansion point.

## 5. Can Taylor series expansions be used for any function?

Technically, yes. However, for a Taylor series to accurately approximate a function, the function must be infinitely differentiable at the expansion point. Otherwise, the Taylor series may not converge to the original function.

• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
13
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
3K
• Calculus and Beyond Homework Help
Replies
14
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
4K
• Calculus and Beyond Homework Help
Replies
3
Views
3K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus
Replies
2
Views
1K