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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm skimming through my book and don't seem to find a concrete answer...what the difference between MacLaurin and Taylor again?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0MacLaurin is a type of TAylor?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maclourin is at 0 tho

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1A Maclauris series is a Taylor series at \(x=0\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so I have the equation f=sinx and I'm trying to approximate that line with degree 3 and number 1?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=3}^{?} \frac{sinx}{3!} (x1)^3\] this doesn't look right...where did I go wrong?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh wait :D are you in radians or degrees sofiya

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Use 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Be back in a couple minutes.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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
 3 years ago
Best ResponseYou've already chosen the best response.1To add and simplify the expanded form.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0king george check inbox

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that right what i sent you?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sin(a)+cos(a)*(xa)(1/2)*sin(a)*(xa)^2

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1The first way @MathSofiya. You don't want sigma notation in your answer.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sin(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

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1@timo86m We're told that \(a=1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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 \[0.8 \le x \le 1.2\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I 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
 3 years ago
Best 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Taylor 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
 3 years ago
Best ResponseYou've already chosen the best response.0ooohhhh can I pick any number between 0.8 and 1.2 for x to test something?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it does work for all a but it is not practical.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or 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
 3 years ago
Best ResponseYou've already chosen the best response.1That'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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\leftR_n(x) \right=\leftf(x)T_n(x)\right\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is the absolute value of the remainder

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if \[\leftf^{(n+1)}(x)\right\le M\] then \[\leftR_n(x) \right\le \frac{M}{(n+1)!}\left xa\right^{n+1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There 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
 3 years ago
Best ResponseYou've already chosen the best response.0since this is an alternating series we can use method 2 I guess

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1It 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
 3 years ago
Best ResponseYou've already chosen the best response.0It doesn't explain what M is

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1So do we want to find \(R_n(x)\)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes that would be a start sinxT_n (x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but they also said that \[f(x)\approx T_n(x)\]

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

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1So if we look at \(\sin(0.8)T_3(0.8)\), we get \(.248284...\)

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1But I don't think that's all we need.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are 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
 3 years ago
Best ResponseYou've already chosen the best response.1Since \(f(x)\approx T_n(x)\) we know that \(T_n(0.8)\approx \sin(.8)\). They aren't exactly the same, but they're about the same.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.1I'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
 3 years ago
Best ResponseYou've already chosen the best response.0I know that's not sin(x) or cosine x...but I'm trying to come as close to a line as possible

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Current 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
 3 years ago
Best ResponseYou've already chosen the best response.0Should I post it as a new question?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Might be a good idea. This thread is getting a bit cluttered.
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