Voofie | Finding nth derivative of the function sec x + tan x and partial difference equation

Finding nth derivative of the function sec x + tan x and partial difference equation

27 Jul 10

Content

1 Introduction
2 The pattern
3 Relationship between the coefficients

Introduction

I am going to find the n-th derivative of the function:

$\f$f(x) = \sec x + \tan x\f$$

The pattern

First, we find the first few derivatives of f:

$\f$f'(x)=\sec^2 x +\sec x \tan x\f$$

$\f$f^{(2)}(x)=\sec^3 x +2 \sec^2 x \tan x+\sec x \tan^2 x\f$$

$\f$f^{(3)}(x)=2 \sec^4 x+5 \sec^3 x \tan x + 4 \sec^2 x \tan^2 x +\sec x \tan^3 x\f$$

It is clear that we can write n-th derivatives of f as:

$\f$f^{(n)}(x) = \sum _{k=0}^{n+1} A_k^n\sec^{n+1-k} x \tan^k x\f$$

Relationship between the coefficients

Now, we need to solve for $\f$A_k^n\f$$ by first finding a partial difference equation.

$\f$\begin{align*} f^{(n+1)}(x) &= \frac{d}{d x}\left(\sum _{k=0}^{n+1} A_k^n\sec^{n+1-k} x \tan^k x \right)\\ &= \sum _{k=0}^{n+1} A_k^nk \sec^{n+3-k} x \tan^{k-1} x +\sum _{k=0}^{n+1} A_k^n(n+1-k) \sec^{n+1-k} x \tan^{k+1} x\\ &= \sum _{k=0}^{n} A_{k+1}^n(k+1) \sec^{n+2-k} x \tan^k x +\sum _{k=1}^{n+2} A_{k-1}^n(n+2-k) \sec^{n+2-k} x \tan^k x \end{align*}\f$$

Therefore, we get:

$\f$A_k^{n+1}=(k+1)A_{k+1}^n+(n+2-k)A_{k-1}^n\f$$

With initial condition:

$\f$\begin{cases} A_0^0 &= 1 \\ A_1^0 &= 1 \\ A_k^0 &=0 \text{ otherwise} \end{cases}\f$$

An attempt to solve the partial difference equation is given in:

Reducing a partial difference equation into a partial differential equation and solving for the generating function using method of characteristics

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Finding nth derivative of the function sec x + tan x and partial difference equation

Introduction

The pattern

Relationship between the coefficients

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