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estimating sign of 0.4 using a Maclaurin polynomial what is the least degree of the polynomial that assures an error smaller than 0.001 so what are we talking about here well we could take some function and estimate it with an nth degree Maclaurin polynomial in fact we could talk more generally about a Taylor polynomial but let's just say this is an nth degree Maclaurin polynomial but this isn't going to be a perfect approximation there's going to be some error some remainder and so we could call this the remainder of that nth degree Maclaurin polynomial it's going to be dependent for any given X now if we want to use the specifics of this exact problem we could phrase it this way we want to say look if we're taking the sine of 0.4 this is going to be equal to our Maclaurin our nth degree Maclaurin polynomial evaluated at 0.4 plus whatever the remainder is for that nth degree Maclaurin polynomial evaluated at 0.4 and what we really want to do is figure out for what n what is the least degree of the polynomial so what what is let me do this in a different color so we want to figure out what is the smallest n what is the smallest n such such that the remainder of our nth degree Maclaurin polynomial evaluated 0.4 is less than this number is less than 0.001 so this is just another way of rephrasing the problem and the way that we can do it is we can use something called the Lagrangian and we have other videos that prove it this is often also too called Taylor's remainder theorem and I'll first write it out and I'll try to explain it while I write it out but it'll actually become a lot more concrete when we work it out so Taylor's remainder theorem tells us or a little garage error bound tells us that if we're that if the derivative plus one derivative of our function so f so this is our n plus 1 the derivative our function if the absolute value of that is less than or equal to some M for an open interval open interval containing where our containing where our polynomial is centered in this case it's zero we're going to use the Maclaurin case so it's containing zero and the exit or zero and X the X that we care about in this particular video is zero point four but I'll say it in generally for any X so if this is true if our n plus want the derivative of our function if the absolute value of it is less than or equal to M over an open interval containing where we're centered this would be C if we're talking about the general case and X so this X right over here then then this is the Lagrange this is the part that's useful we can say that the remainder is bounded the remainder for that nth degree polynomial so this is the n plus 1 derivative that's bounded then we can say the remainder for the nth degree polynomial that approximates our function is going to be less than or equal to that M times X to the n plus 1 or at times ya times X to the n plus 1 over n plus 1 factorial so how do we apply that to this to this particular problem well think about the derivatives of sine we know that the absolute value of sine is less than or equal to 1 its derivative is cosine of X the absolute value of that is going to be bounded is going to be less than or equal to 1 so no matter how many times we take the derivative of sine of X the absolute value of that derivative is going to be less than or equal to 1 so we could write generally that for this particular f of X so let's say for this particular f of X right over here we could say that the absolute value of f the absolute value of the n plus 1 derivative evaluated any X is going to be less than or equal to and this is the case for where F is sine where f is sine of X and this is actually going to be true over any interval it doesn't even have to be over some type of a restricted interval where we can do this so we know that this is our M we know that this is our M that's sine and its derivatives are all bounded by or their absolute values are bounded by one and so then we we have our M and we can apply the Lagrange lagron Jar around so we can say that the remainder of our nth degree Maclaurin approximation at 0.4 so our X in this particular case is 0.4 we don't want it we don't have to do it generally for any X here it's going to be less than or equal to our M is 1 so I won't even write that down our x is 0.4 0.4 to the x to the n plus 1 to the n plus 1 over n plus 1 factorial and we're taking the absolute value of this whole thing this is lagron Jarrow bound and we want to figure out if we can figure out a situation where this is less than 0.001 then this for sure is going to be less than 0.001 because the remainder is less than this or less than or equal to this which is less than that so how do we do that how do we figure out the smallest n where this is going to be true well we can just try out some ends and keep increasing until this thing actually becomes smaller than that thing so let's do it all right I'm gonna set up a table here so let me make a do it relatively cleanly so let's do this is going to be our N and then this is going to be this is going to be 0.4 to the n plus 1 over n plus 1 factorial so let's try it when n is equal to 1 well then this is going to be 0.4 to the - so it's 0.4 squared over 2 factorial this is 0.16 over 2 which is equal to 0.08 that's definitely not less than one thousand here so let's try n equals two when N equals two is going to be zero point four to the third power over three factorial and that's equal to what is that zero point I'm going to need three digits behind the decimal zero point zero six four over six well this is a little bit more than one zero point zero one so this isn't where iron isn't large enough yet so let's try three so this is going to be zero point zero four to the 3 to the 3 plus one so that's going to be to the fourth power over 4 factorial and let's see that is going to be equal to this is going to be let's see it's going to have four digits behind the decimal so zero point zero two five six over 24 this is we're almost there this is a little bit this is going to be a little bit more than 0.001 so that doesn't that doesn't do the trick for us so I'm guessing already that N equals four is going to do the trick but let's verify that so this is going to be zero point zero four we're at a zero point four I should say to the fifth power over five factorial and what is this equal to let's see 4 to the fifth is 1024 I'm going to have five numbers behind the decimal and I'm going to vide it by five factorial which is 120 and let's see this right over here yes this for sure is less than 0.001 this is definitely less than a thousandth right over here so we see that when n is equal to four so we can say that the remainder for our fourth degree polynomial fourth degree Maclaurin polynomial evaluated at x equals zero point four is for sure going to be less than 0.001 so there you go that is the least degree of the polynomial that assures an error smaller than one thousandth

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