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Which part is giving you trouble?

It looks like you made some mistakes in constructing the Taylor Series.

where?

i see it

Kind of a lot of mistakes. Let me see you try again.

I fixed it

That's just by the definition of the Taylor series.

Much better =)

Okay, that's the first part. Now for the second part, we need to get M.

\[f^4(x)=sinx\le M\]

\[\left|R_3(x)\right|\le \frac{sinx}{4!} {\left|x-1\right|}^4\]
shoot me

Hold on. We need to get an actual value of M. Remember how we get that?

it's the absolute value of f^4 (x)

don't give up on me quite yet please.....
sin(1)
or
I have to do something with \[x \ge 0.8\]

Let's step back, slow down, and understand what it is that we're doing. Just some theory.

\[f^{(n+1)}\le f(lower limit)<\]

\[f^{(n+1)}\le f(lower limit)

ceiling*

yep. So we pick the lowest limit that we're allowed 0.8 in this case

I don't think you're right.

Let me talk about the maximum of a function for a bit.

ok

Okay, I'm thinking just a few examples will help you to see what it is we're doing.

|dw:1342915184406:dw|

Text on right says max=10

|dw:1342915323520:dw|

|dw:1342915451038:dw|

Tell me that you understand. Because I can't make the words to explain this.

That M allows us to put a cap on the error

That value M is greater than the next derivative up.

hold on smooth

Well, let me plot the 4th derivative for us.

no hold on

http://www.wolframalpha.com/input/?i=plot+sin%28x%29+on+x%3D.5+to+x%3D1.5

x=1.2 will give me the ultimate cap.
\[f^4(x)=sinx\le sin(1.2)<1.932\]

ultimate cap for our interval

sin(1.2) is not 1.932.

0.932 :P

That's impossible, since sin oscillates between 1 and -1.

Good good =)

typo dude...chill out

and I didn't look at your wolfram ;P

Victory?

Almost there.

remember the interval \[7 \le x \le 9\]

I do remember. Please realize that the maximum is not always going to be at one of the endpoints.

I understand that now

Yes sir.

Okay good =)

\[\left|R_3(x)\right|\le \frac{.932}{4!} {\left|1.2-1\right|}^4\]

yes?

\[\left|R_3(x)\right|\le 0.000054\]

Looks good, pal.

Thank ya! You're amazingly patient. Thank you soo much John. :)

My pleasure! =D