anonymous
  • anonymous
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
chestercat
  • chestercat
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anonymous
  • 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
  • anonymous
MacLaurin is a type of TAylor?
anonymous
  • anonymous
maclourin is at 0 tho

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KingGeorge
  • KingGeorge
A Maclauris series is a Taylor series at \(x=0\)
anonymous
  • anonymous
so I have the equation f=sinx and I'm trying to approximate that line with degree 3 and number 1?
KingGeorge
  • KingGeorge
To approximate one with a Taylor series, use the formula \[\sum_{i=0}^n \frac{f^{(n)}(a)}{n!}(x-a)^n\]
KingGeorge
  • KingGeorge
Excuse me, \[\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i\]
anonymous
  • anonymous
\[\sum_{n=3}^{?} \frac{sinx}{3!} (x-1)^3\] this doesn't look right...where did I go wrong?
anonymous
  • 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
  • KingGeorge
So you have \(a=1\), and \[f^{(0)}=\sin(x)\]\[f^{(1)}=\cos(x)\]\[f^{(2)}=-\sin(x)\]\[f^{(3)}=-\cos(x)\]
anonymous
  • anonymous
oh wait :D are you in radians or degrees sofiya
KingGeorge
  • 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
  • KingGeorge
Be back in a couple minutes.
anonymous
  • anonymous
k
anonymous
  • 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
  • KingGeorge
Remember we're going to \(n=3\) and not \(n=2\).
anonymous
  • 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
  • KingGeorge
To add and simplify the expanded form.
anonymous
  • anonymous
king george check inbox
anonymous
  • anonymous
is that right what i sent you?
KingGeorge
  • KingGeorge
One minute.
anonymous
  • 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
  • anonymous
sin(a)+cos(a)*(x-a)-(1/2)*sin(a)*(x-a)^2
KingGeorge
  • KingGeorge
The first way @MathSofiya. You don't want sigma notation in your answer.
anonymous
  • 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
  • KingGeorge
@timo86m We're told that \(a=1\)
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
ooohhhh can I pick any number between 0.8 and 1.2 for x to test something?
KingGeorge
  • KingGeorge
Sure.
anonymous
  • anonymous
it does work for all a but it is not practical.
anonymous
  • anonymous
i think it is just f(x)=x-(1/6)*x^3 and the answer is f(1)=5/6
anonymous
  • 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
  • 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
  • anonymous
\[\left|R_n(x) \right|=\left|f(x)-T_n(x)\right|\]
anonymous
  • anonymous
is the absolute value of the remainder
anonymous
  • 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
  • KingGeorge
That's why you don't evaluate it. We merely leave it as \(\sin(1)\) or \(\cos(1)\).
anonymous
  • anonymous
it said approximate
KingGeorge
  • KingGeorge
A Taylor series of finite length is by definition an approximation, whether we evaluate \(\sin(1)\) or not.
anonymous
  • 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
  • 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
  • anonymous
since this is an alternating series we can use method 2 I guess
KingGeorge
  • 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
  • anonymous
sure
anonymous
  • anonymous
It doesn't explain what M is
KingGeorge
  • KingGeorge
So do we want to find \(R_n(x)\)?
anonymous
  • anonymous
yes that would be a start |sinx-T_n (x)|
anonymous
  • anonymous
but they also said that \[f(x)\approx T_n(x)\]
KingGeorge
  • 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
  • KingGeorge
So if we look at \(|\sin(0.8)-T_3(0.8)|\), we get \(.248284...\)
KingGeorge
  • KingGeorge
But I don't think that's all we need.
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
ok sounds good.
KingGeorge
  • 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
  • anonymous
Should I post it as a new question?
KingGeorge
  • KingGeorge
Might be a good idea. This thread is getting a bit cluttered.
anonymous
  • anonymous
sure

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