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

Mathematics
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I'm skimming through my book and don't seem to find a concrete answer...what the difference between MacLaurin and Taylor again?
MacLaurin is a type of TAylor?
maclourin is at 0 tho

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Other answers:

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?
To approximate one with a Taylor series, use the formula \[\sum_{i=0}^n \frac{f^{(n)}(a)}{n!}(x-a)^n\]
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?
\[\sin \left( 1 \right) +\cos \left( 1 \right) \left( x-1 \right) -1/2 \,\sin \left( 1 \right) \left( x-1 \right) ^{2} \]
So you have \(a=1\), and \[f^{(0)}=\sin(x)\]\[f^{(1)}=\cos(x)\]\[f^{(2)}=-\sin(x)\]\[f^{(3)}=-\cos(x)\]
oh wait :D are you in radians or degrees sofiya
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.
Be back in a couple minutes.
k
\[\sin \left( a \right) +\cos \left( a \right) \left( x-a \right) -1/2 \,\sin \left( a \right) \left( x-a \right) ^{2} \]
Remember we're going to \(n=3\) and not \(n=2\).
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?
To add and simplify the expanded form.
king george check inbox
is that right what i sent you?
One minute.
\[\frac{\sin(1)}{0!}(x-1)^0+\frac{\cos(1)}{1!}(x-1)^1-\frac{\sin(1)}{2!}(x-1)^2-\frac{\cos(1)}{3!}(x-1)^3\] = \[sin(1)+cos(1)(x-1)-\frac{sin(1)(x-1)^2}{2}-\frac{cos(1)(x-1)^3}{6}\] Like this or in the\[\sum\] form?
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.
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
@timo86m We're told that \(a=1\)
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 \[0.8 \le x \le 1.2\]
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.
@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.
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 \(\sin(1)\).
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
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"
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.
\[\left|R_n(x) \right|=\left|f(x)-T_n(x)\right|\]
is the absolute value of the remainder
a 1 could be what they wanted sin(x) evaluated at. it makes no sense to have cos1 sin1 because it has irrational number.
That's why you don't evaluate it. We merely leave it as \(\sin(1)\) or \(\cos(1)\).
it said approximate
A Taylor series of finite length is by definition an approximation, whether we evaluate \(\sin(1)\) or not.
if \[\left|f^{(n+1)}(x)\right|\le M\] then \[\left|R_n(x) \right|\le \frac{M}{(n+1)!}\left| x-a\right|^{n+1}\]
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
since this is an alternating series we can use method 2 I guess
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)
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 \[f(x)\approx T_n(x)\]
I think we should be using \(x=0.8\) to get these approximations since \(|\sin(1)-\sin(.8)|>|\sin(1)-\sin(1.2)|\).
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.
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)?
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.
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
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.
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.
Current question I'm unable to help on is "Use Taylors Inequality to estimate the accuracy of the approximation \(T_n\) when \(0.8\leq x\leq 1.2\)?"
Should I post it as a new question?
Might be a good idea. This thread is getting a bit cluttered.
sure

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