mclaurin series and taylor inequality

**twofortwo** · May 7th 2010, 08:55 AM

use mclaurin series to compute the sin 3 degrees, correct to 5 decimal places.
use taylos inequality to show you are within 5 decimal places.

i found the bound for the sinx, its obviously one but i dont know wat that means or how to set it up.

please help!
thanx!

**twofortwo** · May 7th 2010, 09:01 AM

thanx for all help!

**twofortwo** · May 9th 2010, 05:53 PM

anyone? please help!

**AllanCuz** · May 9th 2010, 06:04 PM

Originally Posted by twofortwo

use mclaurin series to compute the sin 3 degrees, correct to 5 decimal places.
use taylos inequality to show you are within 5 decimal places.

i found the bound for the sinx, its obviously one but i dont know wat that means or how to set it up.

please help!
thanx!

The McLaurin series is

$\sum_{k=0}^{ \infty } = \frac{ f^{ (k) } (0) } {k!} x^k$

It is well known that the Mclaurin series of sinx leads to...

$sinx = \sum_{n=0}^{ \infty } \frac{ (-1)^n }{(2n+1)!} x^{2n + 1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}....$

I'm assuming your questions says evaluate until we get to the third power (not entirely clear from your post) so what you're looking for is

$x - \frac{x^3}{3!}$

But to be honest, your question is too unclear for me to get to any sort of answer here. What are you looking for?

**twofortwo** · May 9th 2010, 06:16 PM

sorry, thats exactly why im so confused cause thats how the question is worded

i think they are asking to find the sin of 3 degrees using the inequality correct to decimal places

thanx

**TheEmptySet** · May 9th 2010, 06:19 PM

Originally Posted by twofortwo

sorry, thats exactly why im so confused cause thats how the question is worded

i think they are asking to find the sin of 3 degrees using the inequality correct to decimal places

thanx

You do not need to use Taylors ineqality beucase it is an alternating series. The error in an alternating series is always less than the value of the next term in the series.

**HallsofIvy** · May 10th 2010, 03:05 AM

Taylor's inequality says that the error in approximating the infinite series (Taylor's series) $\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(x- x_0)^n$ by truncating at the Nth power is less than $\frac{M}{(N+1)!}|x- x_0|^{N+1}$ where M is an upper bound on the absolute value of the Mth dervivative of f between x and $x_0$ .

All derivatives of sin(x) are $\pm sin(x)$ or $\pm cos(x)$ both of which have maximum value over all x of 1, as you say. Of course, the derivative of sin(x) is cos(x) and the derivative or cos(x) is -sin(x) only when x is measured in radians so $x_0= 0$ and $x= \frac{3}{180}\pi= 0.03491$ , approximately.

So the errror is less than $\frac{1}{(N+1)!}(.03491)^{N+1}$ and you want to make that .00001.

I would recommend just evaluating that at N= 3, 4, 5, ... until it gets small enough.

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