Question

Help understanding Taylor polynomial

Answer 1

Sure I'm here for you, taylor series are one of the coolest things in calculus!

Answer 2

I am trying to upload the file

Answer 3

http://assets.openstudy.com/updates/attachments/520c2e01e4b0f627eb19eca6-castroseries-1376530218428-chapter9infiniteseriesbook.pdf

Answer 4

I need help on page 64, i dont understand how they bounded the remainder

Answer 5

its example 8 in section 9.7

Answer 6

Sure, so you know that sin(x) can also be written as this infinite polynomial x-x^3/3!+x^5/5!-... right?

These are exactly the same things, except it's impossible to write out all the terms of the infinite version. But we can definitely say this:

\[sin(x)=x-x^3/3!+x^5/5!-...\]

Now when you make an approximation, you're only using a few terms of this infinite polynomial version, so whatever you make might look like simply:

P=x-x^3/3!

but you're leaving out all the remainder which is everything after

R=x^5/5!-x^7/7!+...

right? So we call those two little terms up there the polynomial approximation and all the infinite remainder the remainder.

\[sinx=P+R=x-x^3/3!+x^5/5!-...\]

From here you can hopefully see better what's going on. I can explain anything that might not have answered your question if you ask.

Answer 7

I dont understand where it says 0<R{3}(0.1)=sin(z)/4!(0.1)^4<0.0001/4!

Answer 8

Oh i just fiugred it out, they are taking the max of sin(z) meaning sin(z) can at most be 1