expand tanx in ascending power of x using maclaurin's series. also find the expansion of secpower2x

Question

ascending power using maclaurin's series

Answer 1

METHOD 1

The Taylor/Maclaurin series for sin(x)=x-x³/3!+x⁵/5!-... or ∑(-1)^rx^2r+1/(2r+1)! for integer r≥0.

The Maclaurin series for cos(x)=1-x²/2!+x⁴/4!-... or ∑(-1)^rx^2r/(2r)! for integer r≥0.

Since tan(x)=sin(x)/cos(x)=∑(-1)^rx^2r+1/(2r+1)!/∑(-1)^rx^2r/(2r)!.

Let's see what long division looks like:

                           x is the first term of the series for tan(x).

1-x²/2!+x⁴/4!-... | x-x³/3!+x⁵/5!-...

                           x-x³/2!+x⁵/4!-...

                              -x³(⅙-½)+x⁵(1/120-1/24)+...+(-1)^rx^2r+1(1/(2r+1)!-1/(2r)!)+... (the first difference).

Coefficient of x^2r+1 is (-1)^r(1/(2r+1)!-1/(2r)!)=(-1)^r+1/[(2r-1)!(2r+1)] for r≥1.

Method 1 is too complicated, but is still useful for a fixed number of terms. See later for a formula for the nth term of an infinite series for tan(x).

METHOD 2

f(x)=tan(x), successive derivatives will be designated in the form f⁽ⁿ⁾, where n=1 implies f'(x) and f or f⁽⁰⁾ imply f(x).

Also f₀⁽ⁿ⁾ implies f⁽ⁿ⁾(0) and f₀ implies f(0).

f⁽¹⁾=sec²(x)=1+tan²(x)=1+f²

f⁽²⁾=2ff⁽¹⁾,

f⁽³⁾=2ff⁽²⁾+2(f⁽¹⁾)²,

f⁽⁴⁾=(2ff⁽³⁾+2f⁽¹⁾f⁽²⁾)+4f⁽¹⁾f⁽²⁾=2ff⁽³⁾+6f⁽¹⁾f⁽²⁾,

f⁽⁵⁾=2(ff⁽⁴⁾+f⁽¹⁾f⁽³⁾)+6(f⁽¹⁾f⁽³⁾+(f⁽²⁾)²)=2ff⁽⁴⁾+8f⁽¹⁾f⁽³⁾+6(f⁽²⁾)²,

f⁽⁶⁾=2(ff⁽⁵⁾+f⁽¹⁾f⁽⁴⁾)+8(f⁽¹⁾f⁽⁴⁾+f⁽²⁾f⁽³⁾)+12f⁽²⁾f⁽³⁾=2ff⁽⁵⁾+10f⁽¹⁾f⁽⁴⁾+20f⁽²⁾f⁽³⁾,

f⁽⁷⁾=2(ff⁽⁶⁾+f⁽¹⁾f⁽⁵⁾)+10(f⁽¹⁾f⁽⁵⁾+f⁽²⁾f⁽⁴⁾)+20(f⁽²⁾f⁽⁴⁾+(f⁽³⁾)²)=2ff⁽⁶⁾+12f⁽¹⁾f⁽⁵⁾+30f⁽²⁾f⁽⁴⁾+20(f⁽³⁾)²,

f⁽⁸⁾=2(ff⁽⁷⁾+f⁽¹⁾f⁽⁶⁾)+12(f⁽¹⁾f⁽⁶⁾+f⁽²⁾f⁽⁵⁾)+30(f⁽²⁾f⁽⁵⁾+f⁽³⁾f⁽⁴⁾)+40f⁽³⁾f⁽⁴⁾=2ff⁽⁷⁾+14f⁽¹⁾f⁽⁶⁾+42f⁽²⁾f⁽⁵⁾+70f⁽³⁾f⁽⁴⁾,

 f⁽⁹⁾=2(ff⁽⁸⁾+f⁽¹⁾f⁽⁷⁾)+14(f⁽¹⁾f⁽⁷⁾+f⁽²⁾f⁽⁶⁾)+42(f⁽²⁾f⁽⁶⁾+f⁽³⁾f⁽⁵⁾)+70(f⁽³⁾f⁽⁵⁾+(f⁽⁴⁾)²),

f⁽⁹⁾=2ff⁽⁸⁾+16f⁽¹⁾f⁽⁷⁾+56f⁽²⁾f⁽⁶⁾+112f⁽³⁾f⁽⁵⁾+70(f⁽⁴⁾)², ...

f₀=0, f₀⁽¹⁾=1, so f₀⁽²⁾=0, f₀⁽³⁾=2, f₀⁽⁴⁾=0, so f₀⁽²ⁿ⁾=0 for all integer n≥0.

This method of calculating successive derivatives is easier than the long division method and is more closely connected to the Taylor series method. Furthermore, a pattern is emerging for the terms of the derivatives.

n (odd) f₀⁽ⁿ⁾ a_n (Taylor/Maclaurin)

    1         1                  1

    3         2               2/3!=⅓

    5        16           16/5!=2/15

    7       272        272/7!=17/315

    9     7936      7936/9!=62/2835

All even derivatives: f₀, f₀⁽²⁾, etc., are zero, so any terms containing them can be ignored.

So tan(x)=x+x³/3+2x⁵/15+17x⁷/315+62x⁹/2835+...

To confirm this we can use Method 1: (x-x³/3!+x⁵/5!-x⁷/7!+x⁹/9!)/(1-x²/2!+x⁴/4!-x⁶/6!+x⁸/8!):

                                          x+x³/3+2x⁵/15+17x⁷/315+62x⁹/2835

1-x²/2!+x⁴/4!-x⁶/6!+x⁸/8! | x-x³/3!+x⁵/5!     -x⁷/7!        +x⁹/9!

                                          x-x³/2!+x⁵/4!-     x⁷/6!         +x⁹/8!

                                             x³/3-x⁵/30    +x⁷/840       -x⁹/45360

                                             x³/3-x⁵/6      +x⁷/72         -x⁹/2160   +x¹¹/120960

                                                    2x⁵/15 -4x⁷/315      +x⁹/2268-x¹¹/...

                                                    2x⁵/15   -x⁷/15       +x⁹/180  -x¹¹/...

                                                               17x⁷/315  -29x⁹/5670

                                                               17x⁷/315  -17x⁹/630  +x¹¹/...

                                                                                 62x⁹/2835

So this division confirms the results for Method 2. But next we can look at the standard formula for the series.

tan(x)=∑[(-1)^n-12²ⁿ(2²ⁿ-1)B_2n]x^2n-1/[2n(2n-1)!], where B_2n represents the Bernoulli number.

To find the series for sec²(x) simply differentiate the series for tan(x) wrt x.

sec²(x)=1+x²+⅔x⁴+17x⁶/45+62x⁸/315+...

sec²(x)=∑[(-1)^n-12²ⁿ(2²ⁿ-1)B_2n]x^2n-2/[2n(2n-2)!].

Comments

Space for answers on this website is restricted (12000 HTML characters maximum), but there is sufficient information in the details here to continue expansion of the derivatives in similar fashion, or follow up Bernouilli's number. There are also websites where the expansion of tan(x) includes more terms than the ones given here. Suffice it to say that there is no easy way of expressing the nth coefficient of the series.