Approximate f by a Taylor polynomial with degree n at the number a.
\[f(x)=sinx\]
a=1 n=3
\[0.8 \le x \le 1.2\]

- anonymous

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- anonymous

I'm skimming through my book and don't seem to find a concrete answer...what the difference between MacLaurin and Taylor again?

- anonymous

MacLaurin is a type of TAylor?

- anonymous

maclourin is at 0 tho

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## More answers

- KingGeorge

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

- anonymous

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

- KingGeorge

To approximate one with a Taylor series, use the formula \[\sum_{i=0}^n \frac{f^{(n)}(a)}{n!}(x-a)^n\]

- KingGeorge

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

- anonymous

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

- anonymous

\[\sin \left( 1 \right) +\cos \left( 1 \right) \left( x-1 \right) -1/2 \,\sin \left( 1 \right) \left( x-1 \right) ^{2} \]

- KingGeorge

So you have \(a=1\), and \[f^{(0)}=\sin(x)\]\[f^{(1)}=\cos(x)\]\[f^{(2)}=-\sin(x)\]\[f^{(3)}=-\cos(x)\]

- anonymous

oh wait :D are you in radians or degrees sofiya

- KingGeorge

Use the formula, and we get \[\frac{\sin(1)}{0!}(x-1)^0+\frac{\cos(1)}{1!}(x-1)^1-\frac{\sin(1)}{2!}(x-1)^2-\frac{\cos(1)}{3!}(x-1)^3\]This can, of course, be simplified a bit.

- KingGeorge

Be back in a couple minutes.

- anonymous

k

- anonymous

\[\sin \left( a \right) +\cos \left( a \right) \left( x-a \right) -1/2 \,\sin \left( a \right) \left( x-a \right) ^{2} \]

- KingGeorge

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

- anonymous

I'm kinda lost...what's my goal here? to write it in the condensed form \[\sum\] or to add or simplify the written out form?

- KingGeorge

To add and simplify the expanded form.

- anonymous

king george check inbox

- anonymous

is that right what i sent you?

- KingGeorge

One minute.

- anonymous

\[\frac{\sin(1)}{0!}(x-1)^0+\frac{\cos(1)}{1!}(x-1)^1-\frac{\sin(1)}{2!}(x-1)^2-\frac{\cos(1)}{3!}(x-1)^3\]
=
\[sin(1)+cos(1)(x-1)-\frac{sin(1)(x-1)^2}{2}-\frac{cos(1)(x-1)^3}{6}\]
Like this or in the\[\sum\] form?

- anonymous

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

- KingGeorge

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

- anonymous

sin(a)+cos(a)*(x-a)-(1/2)*sin(a)*(x-a)^2-(1/6)*cos(a)*(x-a)^3
with 4th approximation choose a=0 and
x-(1/6)*x^3

- KingGeorge

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

- anonymous

so this way would be correct I guess
\[sin(1)+cos(1)(x-1)-\frac{sin(1)(x-1)^2}{2}-\frac{cos(1)(x-1)^3}{6}\]
and i have to find a way to add this fact to the equation
\[0.8 \le x \le 1.2\]

- KingGeorge

I have no idea what \(0.8\leq x\leq 1.2\) is supposed to do to our solution. As far as I'm concerned, that's irrelevant since \(x\) is a variable. We've just restricted the domain.

- anonymous

@KingGeorge
Taylor is all about knowing a value at a such as 0 90 and such for sin(x) so that you get an appoximation. nobody knows what sin(1) is but we know cos(0) sin(o) and such.

- KingGeorge

Taylor series works with all \(a\). Whether we can compute \(\sin(1)\) by hand or not is irrelevant. If we can't compute it, leave it as \(\sin(1)\).

- anonymous

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

- KingGeorge

Sure.

- anonymous

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

- anonymous

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

- anonymous

or that might be to estimate the accuracy...the next question reads:
"Use Taylors Inequality to estimate the accuracy of the approximation \[f(x)\approx T_n(x)\]when x lies in the given interval"

- KingGeorge

That's probably what it's referring to in that case. I can't say I'm familiar with Taylor's Inequality though.
@timo86m It specifically asks for \(a=1\), and not 0.

- anonymous

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

- anonymous

is the absolute value of the remainder

- anonymous

a 1 could be what they wanted sin(x) evaluated at.
it makes no sense to have cos1 sin1 because it has irrational number.

- KingGeorge

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

- anonymous

it said approximate

- KingGeorge

A Taylor series of finite length is by definition an approximation, whether we evaluate \(\sin(1)\) or not.

- anonymous

if \[\left|f^{(n+1)}(x)\right|\le M\] then
\[\left|R_n(x) \right|\le \frac{M}{(n+1)!}\left| x-a\right|^{n+1}\]

- anonymous

There are three possible methods for estimating the size of error:
1. The method above
2. Alternating Series Estimation Theorem...if it's an alternating series
3. Use graphing device

- anonymous

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

- KingGeorge

It says use Taylor's inequality though, so I'm inclined to think method 1 is probably what we want. (as long as that is Taylor's inequality)

- anonymous

sure

- anonymous

It doesn't explain what M is

- KingGeorge

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

- anonymous

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

- anonymous

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

- KingGeorge

I think we should be using \(x=0.8\) to get these approximations since \(|\sin(1)-\sin(.8)|>|\sin(1)-\sin(1.2)|\).

- KingGeorge

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

- KingGeorge

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

- anonymous

Are we assuming that f(x) is approximately T_n(x) therefore its sin(x)-sin(x) and in
f(1)=sin(1) and T_n(.8)=sin(.8)?

- KingGeorge

Since \(f(x)\approx T_n(x)\) we know that \(T_n(0.8)\approx \sin(.8)\). They aren't exactly the same, but they're about the same.

- anonymous

The above statement makes sense...but when do I know to plug 0.8 in or 1.2 in...I guess I'm missing the bigger picture

- KingGeorge

I'm really not sure where to go from here. Let me see if I can get someone else to help out with this last part.

- anonymous

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

- anonymous

ok sounds good.

- KingGeorge

Current question I'm unable to help on is "Use Taylors Inequality to estimate the accuracy of the approximation \(T_n\) when \(0.8\leq x\leq 1.2\)?"

- anonymous

Should I post it as a new question?

- KingGeorge

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

- anonymous

sure

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