Check out some similar questions!

bayley86 · Jan 7, 2010, 02:16 PM

using the power series, estimate the values of the following engineering variables:

e^3
sin4
cosh3
cos2

any help with this would be much appreciated

thanks

Andy

galactus · Jan 7, 2010, 02:36 PM

The Taylor series fore e, cos, sin, etc can be found in any calc book. Do you have one?

See here:

http://en.wikipedia.org/wiki/Taylor_series

all the series are listed. Just plug in the given values and estimate.

For instance, $e^{3}$

$e^{x}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{3!}+\frac{x ^{4}}{4!}+...$

Plug in x=3 and find e^3 out to however many decimal places accuracy is specified.

Using the terms out to x^4, we get 16.375

The actual value is 20.085536...

The farther you go out, the more accurate it becomes.

bayley86 · Jan 9, 2010, 05:40 AM

I don't have a calc book do you no what the best 1 to get is for someone who isn't very good at this level of maths.

How many decimal places do you think I should go to because it just says estimate and could you start the sin4 one for me please.

Thanks a lot

Andy

galactus · Jan 9, 2010, 09:39 AM

i don't have a calc book do you no what the best 1 to get is for someone who isn't very good at this level of maths.

Calc books of all sorts can be found on Amazon, along with solutions manuals. Larson and Hostetler publish a very nice calc text.

could you start the sin4 one for me please.

Do you have a nice calculator? Here is a screen capture from my TI-92. Just plug in x=4. As I said, the series for sin, cos, tan, e, etc can be found online and in any calc book. Surely, you can find them. I gave a link to Wikipedia.

Andy[/QUOTE]

ebaines · Jan 12, 2010, 12:22 PM

The Taylor series is found from the following:

f(x) = f(0) + x f'(0) + x^2 f''(0)/2! + x^3 f'''(0)/3! +...

For f(x) = sin(x), then you get:

sin(x) = x - x^3/3! + x^5/5! - x^7/7! +...

If you put in x = 4, you'll find that this converges to an answer with about 4 decimal places of accuracy in about 6 or 7 steps - that should certainly be enough.