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

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MathSofiyaBest ResponseYou've already chosen the best response.1
I'm skimming through my book and don't seem to find a concrete answer...what the difference between MacLaurin and Taylor again?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
MacLaurin is a type of TAylor?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
A Maclauris series is a Taylor series at \(x=0\)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
so I have the equation f=sinx and I'm trying to approximate that line with degree 3 and number 1?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
To approximate one with a Taylor series, use the formula \[\sum_{i=0}^n \frac{f^{(n)}(a)}{n!}(xa)^n\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Excuse me, \[\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(xa)^i\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=3}^{?} \frac{sinx}{3!} (x1)^3\] this doesn't look right...where did I go wrong?
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
\[\sin \left( 1 \right) +\cos \left( 1 \right) \left( x1 \right) 1/2 \,\sin \left( 1 \right) \left( x1 \right) ^{2} \]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
So you have \(a=1\), and \[f^{(0)}=\sin(x)\]\[f^{(1)}=\cos(x)\]\[f^{(2)}=\sin(x)\]\[f^{(3)}=\cos(x)\]
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
oh wait :D are you in radians or degrees sofiya
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Use the formula, and we get \[\frac{\sin(1)}{0!}(x1)^0+\frac{\cos(1)}{1!}(x1)^1\frac{\sin(1)}{2!}(x1)^2\frac{\cos(1)}{3!}(x1)^3\]This can, of course, be simplified a bit.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Be back in a couple minutes.
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
\[\sin \left( a \right) +\cos \left( a \right) \left( xa \right) 1/2 \,\sin \left( a \right) \left( xa \right) ^{2} \]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Remember we're going to \(n=3\) and not \(n=2\).
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
To add and simplify the expanded form.
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
king george check inbox
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
is that right what i sent you?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\frac{\sin(1)}{0!}(x1)^0+\frac{\cos(1)}{1!}(x1)^1\frac{\sin(1)}{2!}(x1)^2\frac{\cos(1)}{3!}(x1)^3\] = \[sin(1)+cos(1)(x1)\frac{sin(1)(x1)^2}{2}\frac{cos(1)(x1)^3}{6}\] Like this or in the\[\sum\] form?
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
sin(a)+cos(a)*(xa)(1/2)*sin(a)*(xa)^2
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
The first way @MathSofiya. You don't want sigma notation in your answer.
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
sin(a)+cos(a)*(xa)(1/2)*sin(a)*(xa)^2(1/6)*cos(a)*(xa)^3 with 4th approximation choose a=0 and x(1/6)*x^3
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
@timo86m We're told that \(a=1\)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
so this way would be correct I guess \[sin(1)+cos(1)(x1)\frac{sin(1)(x1)^2}{2}\frac{cos(1)(x1)^3}{6}\] and i have to find a way to add this fact to the equation \[0.8 \le x \le 1.2\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
@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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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 \(\sin(1)\).
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
ooohhhh can I pick any number between 0.8 and 1.2 for x to test something?
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
it does work for all a but it is not practical.
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
i think it is just f(x)=x(1/6)*x^3 and the answer is f(1)=5/6
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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"
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\leftR_n(x) \right=\leftf(x)T_n(x)\right\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
is the absolute value of the remainder
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
a 1 could be what they wanted sin(x) evaluated at. it makes no sense to have cos1 sin1 because it has irrational number.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
That's why you don't evaluate it. We merely leave it as \(\sin(1)\) or \(\cos(1)\).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
A Taylor series of finite length is by definition an approximation, whether we evaluate \(\sin(1)\) or not.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
if \[\leftf^{(n+1)}(x)\right\le M\] then \[\leftR_n(x) \right\le \frac{M}{(n+1)!}\left xa\right^{n+1}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.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
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
since this is an alternating series we can use method 2 I guess
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
It doesn't explain what M is
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
So do we want to find \(R_n(x)\)?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yes that would be a start sinxT_n (x)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
but they also said that \[f(x)\approx T_n(x)\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
I think we should be using \(x=0.8\) to get these approximations since \(\sin(1)\sin(.8)>\sin(1)\sin(1.2)\).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
So if we look at \(\sin(0.8)T_3(0.8)\), we get \(.248284...\)
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
But I don't think that's all we need.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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)?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I know that's not sin(x) or cosine x...but I'm trying to come as close to a line as possible
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Current question I'm unable to help on is "Use Taylors Inequality to estimate the accuracy of the approximation \(T_n\) when \(0.8\leq x\leq 1.2\)?"
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Should I post it as a new question?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Might be a good idea. This thread is getting a bit cluttered.
 one year ago
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