Trig to Polynomial

Calculus Level 4

Which of these are the first three terms of the Maclaurin series of the function below?

cos ⁡ ( x ) x 2 + 3 x − 2 \large \dfrac{\cos(x)}{x^{2} + 3x - 2}

Hint: Use long division to save time

− 1 2 − 1 2 ⋅ 1 2 ! x 2 − 1 2 ⋅ 1 4 ! x 4 -\frac12 - \frac{1}{2} \cdot \frac1{2!}x^{2} - \frac{1}{2}\cdot\frac1{4!} x^{4} − 1 2 − 3 3 x − 9 4 x 2 -\frac12 - \frac33x - \frac94x^2 − 1 2 − 1 2 ! x 2 − 1 4 ! x 4 -\frac12 - \frac1{2!} x^{2} - \frac1{4!} x^{4} − 1 2 − 3 4 x − 9 8 x 2 -\frac12 - \frac34x - \frac98x^2 − 1 2 + 1 2 ! x 2 − 1 4 ! x 4 -\frac12 + \frac1{2!}x^{2} - \frac1{4!}x^{4} − 1 2 − 3 4 x − 9 4 x 2 -\frac12 - \frac34x - \frac94x^2

Brandon Stocks by Brandon Stocks , 45, USA

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1 solution

Brandon Stocks
Apr 29, 2016

Its long division with series. I don't know how to type out long division here but the basic idea is that you knock off one term at a time from the dividend, working your way through terms with increasingly large powers of x. The dividend is the power series for cos(x)

c o s ( x ) = 1 − x 2 / 2 ! + x 4 / 4 ! − x 6 / 6 ! + . . . . cos(x) = 1 - x^{2}/2! + x^{4}/4! - x^{6}/6! + ....

The divisor is − 2 + 3 x + x 2 -2 + 3x + x^{2} . The first term is -1/2 because (-1/2)(-2) = 1. After you subtract the remainder is 3 x / 2 + x 4 / 4 ! − x 6 / 6 ! 3x/2 + x^{4}/4! - x^{6}/6! . The next term is -3x/4 because (-3x/4)(-2) = 3x/2. After you subtract the remainder is 9 x 2 / 4 + 9x^{2}/4 + ... The next term is − 9 x 2 / 8 -9x^{2}/8 because ( − 9 x 2 / 8 ) ( − 2 ) = 9 x 2 / 4 (-9x^{2}/8)(-2) = 9x^{2}/4 .

The direct method of taking derivatives isn't too difficult for the first two or three terms, but if you want more terms then the division method is definitely easier.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

I disagree that long division is the straight-forward approach because we still have to figure out how to pair up the terms and find the remainder. In particular, to find the coefficient of x n x^n , we have to find the coefficient of every term smaller than it.

Instead, we should just treat these as a Maclaurin expansion, which are just power series which we can manipulate easily. As mentioned, we have

c o s ( x ) = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! + … cos(x) = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \ldots

We then need to find the power series expansion of \frac{1}{ -2 + 3x + x^2} = \frac{1}{-2} \times \frac{ 1}{ 1 - ( \frac{3}{2}x + \frac{1}{2} x^2 ) . This is just

1 − 2 + 3 x + x 2 = 1 − 2 × ( 1 + ( 3 2 x + 1 2 x 2 ) + ( 3 2 x + 1 2 x 2 ) 2 + … ) \frac{1}{ -2 + 3x + x^2} = \frac{1}{-2} \times \left( 1 + ( \frac{3}{2}x + \frac{1}{2} x^2 ) + ( \frac{3}{2}x + \frac{1}{2} x^2 ) ^2 + \ldots \right)

Now, we just multiply these power-series and we are done.

Thank you for your feed back. I think the term in the dividend with the smallest power of x is divided by the term in the divisor having the smallest power of x. After that its the term in the remainder having the smallest power of x that's divided by the term in the divisor having the smallest power of x. If its done right then increasingly large powers of x will be eliminated from the remainder. What method did you use to expand ( − 2 + 3 x + x 2 ) − 1 (-2 + 3x + x^{2})^{-1} ?

Brandon Stocks - 5 years, 1 month ago

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