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MacLaurin is a type of TAylor?

maclourin is at 0 tho

A Maclauris series is a Taylor series at \(x=0\)

so I have the equation f=sinx and I'm trying to approximate that line with degree 3 and number 1?

Excuse me, \[\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i\]

\[\sum_{n=3}^{?} \frac{sinx}{3!} (x-1)^3\]
this doesn't look right...where did I go wrong?

oh wait :D are you in radians or degrees sofiya

Be back in a couple minutes.

Remember we're going to \(n=3\) and not \(n=2\).

To add and simplify the expanded form.

king george check inbox

is that right what i sent you?

One minute.

sin(a)+cos(a)*(x-a)-(1/2)*sin(a)*(x-a)^2

The first way @MathSofiya. You don't want sigma notation in your answer.

@timo86m We're told that \(a=1\)

ooohhhh can I pick any number between 0.8 and 1.2 for x to test something?

Sure.

it does work for all a but it is not practical.

i think it is just
f(x)=x-(1/6)*x^3
and the answer is f(1)=5/6

\[\left|R_n(x) \right|=\left|f(x)-T_n(x)\right|\]

is the absolute value of the remainder

That's why you don't evaluate it. We merely leave it as \(\sin(1)\) or \(\cos(1)\).

it said approximate

since this is an alternating series we can use method 2 I guess

sure

It doesn't explain what M is

So do we want to find \(R_n(x)\)?

yes that would be a start
|sinx-T_n (x)|

but they also said that \[f(x)\approx T_n(x)\]

So if we look at \(|\sin(0.8)-T_3(0.8)|\), we get \(.248284...\)

But I don't think that's all we need.

I know that's not sin(x) or cosine x...but I'm trying to come as close to a line as possible

ok sounds good.

Should I post it as a new question?

Might be a good idea. This thread is getting a bit cluttered.

sure