Taylor polynomials... no way seems practical..

**OneMileCrash** · November 22nd 2011, 04:55 PM

e^(t)cos(t), find first 4 terms centered at 0.

Alright, I know of two methods for doing this, and both seem remarkably painful.

I thought of listing 4 terms of each series (e^t and cost) and just multiplying them as polynomials and combining like terms. This wildly confuses me..

Then, someone recommended that I use derivatives to find the coefficients. Differentiating that function 4 times would be outright terrible.

Which way is the recommended?

**Prove It** · November 22nd 2011, 05:10 PM

Originally Posted by OneMileCrash

e^(t)cos(t), find first 4 terms centered at 0.

Alright, I know of two methods for doing this, and both seem remarkably painful.

I thought of listing 4 terms of each series (e^t and cost) and just multiplying them as polynomials and combining like terms. This wildly confuses me..

Then, someone recommended that I use derivatives to find the coefficients. Differentiating that function 4 times would be outright terrible.

Which way is the recommended?

The first way is the easiest...

**OneMileCrash** · November 22nd 2011, 05:17 PM

I went with the second way, finding the fourth derivative wasn't nearly as bad as I thought.

First way seems wildly harder still. Is it a matter of opinion or would you say that I'm just doing the first way incorrectly if I think that?

**skeeter** · November 22nd 2011, 05:41 PM

matter of opinion ... I just multiplied the polynomials up to $t^4$

$e^t\cos{t} \approx 1 + t - \frac{t^3}{3} - \frac{t^4}{6}$

**Deveno** · November 22nd 2011, 06:11 PM

well you don't want "the first four terms" of each series for the first method, you want the product of any terms that could contribute to the first 4 terms of the product.

so anything higher than a term of $t^3$ can be disregarded (but see below). if we write the coefficents of the terms for $e^t$ as $a_i$ and the terms for $\cos(t)$ as $b_i$ , and the terms of $e^t\cos(t)$ as $c_i$ , then:

$c_0 = a_0b_0$
$c_1 = a_0b_1 + a_1b_0$
$c_2 = a_0b_2 + a_1b_1 + a_2b_0$
$c_3 = a_0b_3 + a_1b_2 + a_2b_1 + a_3b_0$

note that all the odd terms $b_i$ are 0, since cosine is an even function. we have:

$a_0 = 1, a_1 = 1, a_2 = \frac{1}{2}, a_3 = \frac{1}{6}$
$b_0 = 1, b_2 = -\frac{1}{2}$ so:

$c_0 = (1)(1) = 1$
$c_1 = (1)(1) = 1$
$c_2 = (1)(-\frac{1}{2}) + (\frac{1}{2})(1) = 0$

$c_3 = (1)(-\frac{1}{2}) + (\frac{1}{6})(1) = -\frac{1}{3}$

if we must calculate the first 4 non-zero terms, we need to calculate:

$c_4 = a_0b_4 + a_1b_3 + a_2b_2 + a_3b_1 + a_4b_0$

$= (1)(\frac{1}{24}) + (\frac{1}{2})(-\frac{1}{2}) + (\frac{1}{24})(1) = -\frac{1}{6}$

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