Question

How do you find the Maclaurin Series for `sin(3x^2)`?

Answer 1

The Maclaurin series is just another name for a Taylor series at `a = 0`.

I got `3x^2 - (9x^6)/2 + (81x^10)/40 + . . .`... eventually.

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A Taylor series expansion around `a` is written as:

`\mathbf(sum_(n=0)^(N) (f^((n))(a))/(n!)(x-a)^n)`

where:

• `f^((n))(a)` is the `\mathbf(n)`th derivative of `f(x)` evaluated at `x = a`, where `a` is normally some chosen constant called the expansion point of the Taylor series.
• `n` is the dummy index going from `0` to some arbitrary finite final index `N`.
• `n!` is the factorial of each index `n`.

Like I mentioned earlier, for a Maclaurin series, `a = 0`. So, first let's take four derivatives of `sin(3x^2)`, since this will take a lot of product rule and chain rule maneuvering.

Recall that `sin` is cyclic after four derivatives (`sin -> cos -> -sin -> -cos -> sin`). We can plug in `a = 0` later.

IMPORTANT: Note that `x - a` becomes `x`, not `0`. Only `f(a)` becomes `f(0)`, even though for `f(a)`, we stated that `x = a` and `a = 0`.

`color(green)(f^((0))(x)) = f(x) = color(green)(sin(3x^2))`

`color(green)(f'(x) = 6xcos(3x^2))`

`color(green)(f''(x))`

`= (6x)(-6xsin(3x^2) + 6cos(3x^2)`

`= color(green)(-36x^2sin(3x^2) + 6cos(3x^2))`

`color(green)(f'''(x))`

`= [(-36x^2)(6xcos(3x^2)) + sin(3x^2)(-72x)] - 36xsin(3x^2)`

`= color(green)(-216x^3cos(3x^2) - 108xsin(3x^2))`

`color(green)(f''''(x))`

`= [(-216x^3)(-6xsin(3x^2)) + cos(3x^2)(-648x^2)] - [(108x)(6xcos(3x^2) + 108sin(3x^2)]`

`= [1296x^4sin(3x^2) - 648x^2cos(3x^2)] - [648x^2cos(3x^2) + 108sin(3x^2)]`

`= color(green)(1296x^4sin(3x^2) - 1296x^2cos(3x^2) - 108sin(3x^2))`

You can see why I'm stopping here. ;)

(Feel free to use Wolfram Alpha until the 10th order term, because the derivatives get repetitive and cumbersome at this point, and by this point you should know how to do the `n`th derivative of `f(x)`.)

So, the Taylor series would have been:

`= sum_(n=0)^(10) (f^((n))(a))/(n!)(x-a)^n`

`= (f(a))/(0!)(x-a)^0 + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + (f''''(a))/(4!)(x-a)^4 + . . .`

`= sin(3a^2) + 6acos(3a^2)(x-a) + (-36a^2sin(3a^2) + 6cos(3a^2))/(2)(x-a)^2 + (-216a^3cos(3a^2) - 108asin(3a^2))/(6)(x-a)^3 + (1296a^4sin(3a^2) - 1296a^2cos(3a^2) - 108sin(3a^2))/(24)(x-a)^4 + . . .`

Fortunately, for a Maclaurin series, since `a = 0`, and since `0*"stuff" = 0`, `cos(0) = 1`, and `sin(0) = 0`, we can cross out many of these terms.

`=> cancel(sin(3a^2))^(0) + cancel(6acos(3a^2)(x-a))^(0) + (cancel(-36a^2sin(3a^2))^0 + 6cancel(cos(3a^2))^(1))/(2)(x-a)^2 + (cancel(-216a^3cos(3a^2))^(0) - cancel(108asin(3a^2))^(0))/(6)(x-a)^3 + (cancel(1296a^4sin(3a^2))^(0) - cancel(1296a^2cos(3a^2))^(0) - cancel(108sin(3a^2))^(0))/(24)(x-a)^4 + . . .`

Interestingly enough, it isn't until you get to the 6th and 10th order terms that you see the next nonzero terms after the 2nd order term...

You could have predicted that it would be these terms that are nonzero since in the Taylor series for `sin(x)`, the 1st, 3rd, and 5th order terms are nonzero.

Hence, the Taylor series for `sin(x^2)` should have nonzero `("1st")^2 = "2nd"`, `("3rd")^2 = "6th"`, and `("5th")^2 = "10th"` order terms.

`= color(blue)(3x^2 - (9x^6)/2 + (81x^10)/40 + . . . )`