I've attract an arbitraryfunction here. And also what we'regoing to try to carry out is almost right thisarbitrary function-- us don't know what itis-- utilizing a polynomial. We'll keep including termsto that polynomial. But to execute this,we're going to assume the we deserve to evaluatethe function at 0, the it offers us somevalue, and also that we have the right to keep taking thederivative the the duty and examining thefirst, the second, and also the third derivative, soon and so forth, at 0 together well. So we're presume thatwe know what f that 0 is. We're assuming the weknow what f prime of 0 is. We're presume that we knowthe second derivative at 0. We're suspect that we knowthe third derivative at 0. So possibly I'll writeit-- 3rd derivative. I'll just write f element primeat 0, and also so forth and so on. Therefore let's think around how wecan approximate this using polynomials that everincreasing length. Therefore we could have apolynomial of simply one term. And it would certainly justbe a continuous term. Therefore this would certainly be apolynomial of degree 0. And also if we have a constantterm, we at least might want to make the constantpolynomial-- it yes, really is just a continuous function-- equalthe role at f the 0. So at first, maybewe just want ns of 0, where p is the polynomialthat we're going come construct, we want p of 0 tobe equal to f that 0. For this reason if we want to do thatusing a polynomial of only one term, of only oneconstant term, we have the right to just collection p the xis same to f that 0. For this reason if ns were come graph it,it would look choose this. It would simply be ahorizontal line at f the 0. And also you could say, Sal, that'sa dreadful approximation. It only approximates thefunction at this point. Looks like we got lucky ata pair of various other points, however it's yes, really badeverywhere else. And now i wouldtell you, well, shot to carry out any far better usinga horizontal line. At the very least we got itright at f of 0. Therefore this is about as good as wecan perform with simply a constant. And even though-- ns justwant to repeat you-- this might not looklike a constant, however we're presume thatgiven the function, we can evaluateit at 0 and that will certainly just offer us a number. So whatever number that was, wewould placed it right over here. We'd say ns of x isequal to the number. The would just be a horizontalline appropriate there in ~ f that 0. However that obviouslyis not so great. So let's include somemore constraints. Past the reality that us wantp that 0 come be equal to f the 0, let's say that wealso desire p element at 0 to be the very same thingas f element at 0. Allow me do this in a new color. So we likewise want,in the new color, we likewise want-- that'snot a brand-new color. We also want p prime. We desire the first derivativeof our polynomial, once evaluated in ~ 0,to be the same thing together the an initial derivative that thefunction as soon as evaluated in ~ 0. And also we don't desire to losethis appropriate over here. So what if we set p of xas being equal to f the 0? for this reason we're taking our oldp of x, but now we're going to include another term sothat the derivatives match up. Plus f element of time x. So let's think aboutthis a small bit. If we usage this as our newpolynomial, what happens? What is p is 0? p of 0 is walking tobe same to-- you're walk to have f that 0 pluswhatever this f prime of 0 is times 0. If you put a 0 in because that x, thisterm is just going to it is in 0. For this reason you're going to be leftwith p of 0 is same to f of 0. That's cool. That's just as goodas our an initial version. Currently what's thederivative end here? so the derivative is pprime the x is equal to-- you take it the derivative the this. This is simply a constant,so the derivative is 0. The derivative that acoefficient times x is simply going tobe the coefficient. For this reason it's walk tobe f prime of 0. For this reason if you evaluate itat 0-- so p prime of 0. Or the derivative ofour polynomial evaluate at 0-- I know it's a littleweird because we're no using-- we're act a ns prime of xof f that 0 and all of this. However just remember, what's thevariable, what's the constant, and hopefully, it'll make sense. Therefore this is simply obviouslygoing to be f prime of 0. Its derivative isa consistent value. This is a constantvalue appropriate here. We're assuming that we have the right to takethe derivative of our duty and evaluate that point at0 to provide a constant value. Therefore if p prime the x is equalto this consistent value, obviously, p primeof x evaluated at 0 is going to be that value. However what's cool aboutthis appropriate here, this polynomial that has actually a 0degree term and a an initial degree term, is currently this polynomialis equal to our role at x is same to 0. And it likewise has thesame first derivative. It likewise has the sameslope in ~ x is same to 0. For this reason this point will look,this brand-new polynomial v two terms-- gettinga little bit better-- it will looksomething favor that. That will essentially have--it'll look favor a tangent heat at f the 0, at x is same to 0. Therefore we're act better, but stillnot a super good approximation. It sort of is going inthe same general direction as our function around 0. Yet maybe we can dobetter through making certain that they have thesame second derivative. And to shot to have actually the samesecond derivative while still having actually the same first derivativeand the exact same value at 0, let's try to dosomething interesting. Let's define p that x. Therefore let's make it clear. This to be our first try. This is ours secondtry best over here. And I'm about to embarkon our third try. For this reason in our 3rd try, my score isthat the value of my polynomial is the very same as the valueof the role at 0. They have the samederivative at 0. And also they additionally have the samesecond derivative at 0. Therefore let's specify mypolynomial come be equal to-- for this reason I'm goingto execute the very first two regards to these guysright over here. So it's going to bef that 0 to add f element of 0 times x, for this reason exactlywhat us did here. However now allow me include another term. I'll execute the otherterm in a new color. And also I'm walk toput a 1/2 the end here. And also hopefully it could makesense why I'm about to do this. Plus 1/2 time thesecond derivative of ours functionevaluated in ~ 0 x squared. And when us evaluatethe derivative that this, ns think you'll seewhy this 1/2 is there. Due to the fact that now let's evaluatethis and its derivatives in ~ 0. So if we evaluate p of 0, p of0 is going to be equal to what? Well, friend havethis constant term. If you advice it at 0,this x and this x squared are both walk to be 0. Therefore those terms aregoing to go away. So p of 0 is stillequal come f of 0. If you take the derivativeof p of x-- for this reason let me take it thederivative right here. I'll execute it in yellow. So the derivativeof my brand-new p that x is going to be equal to-- sothis ax is going to walk away. It's a constant term. It's going to beequal to f prime of 0. That's the coefficient top top this. Plus-- this is the powerrule best here-- 2 times 1/2 is just 1, plus f primeprime the 0 times x. Take it the 2, multiplyit time 1/2, and also decrement that2 appropriate there. Ns think friend now have a senseof why we put the 1/2 there. It's making it so that we don'tend up v the 2 coefficient out front. Currently what is ns prime that 0? for this reason let me create it righthere. Ns prime that 0 is what? Well, this term best hereis simply going to be 0, therefore you're left v thisconstant value best over here. Therefore it's walking tobe f prime of 0. So so far, our thirdgeneration polynomial has all the propertiesof the very first two. And also let's see how it doeson its 3rd derivative, or I need to say thesecond derivative. So ns prime prime ofx is same to-- this is a constant, soits derivative is 0. So you just take thecoefficient on the second term is equal to f element prime that 0. So what's the secondderivative of p evaluated in ~ 0? Well, it's simply going tobe this continuous value. It's walk to be fprime prime of 0. For this reason notice, by addingthis term, now, not just is ours polynomial valuethe very same thing as our duty value in ~ 0, the derivativeat 0 is the exact same thing together the derivative ofthe role at 0. And its 2nd derivativeat 0 is the very same thing as the second derivativeof the role at 0. So we're gettingpretty great at this. And also you can guess thatthere's a sample here. Every term we add, it'll allowus to collection up the case so that the n-th derivativeof ours approximation at 0 will be the exact same thingas the n-th derivative of our role at 0. Therefore in general, if wewanted to store doing this, if we had actually a lot of oftime on our hands and also we want to simply keepadding state to our polynomial, us could-- and also let medo this in a new color. Perhaps I'll perform it in acolor I currently used. We could make ourpolynomial approximation. So the an initial term, the constantterm, will just be f that 0. Climate the following term willbe f element of 0 time x. Climate the following termwill be f prime prime of 0 times1/2 time x squared. I just rewrote the in aslightly various order. Then the next term, if we wantto do their third derivative the exact same at 0, would bef prime prime prime of 0. The 3rd derivativeof the duty at 0, times 1/2 time 1/3,so 1 end 2 times 3 time x to the third. And we deserve to keep going. Maybe you you'll startto view a sample here. Plus, if we want to maketheir 4th derivatives in ~ 0 coincide, it wouldbe the 4th derivative that the function. I can put a 4 upthere, but this is yes, really emphasizing-- it's the fourthderivative at 0 times 1 over-- and I'll readjust the order. Rather of composing itin enhancing order, I'll compose it as 4 time 3times 2 times x come the fourth. And you deserve to verifyit for yourself. If we just had thisonly, and also if you to be to take it the fourthderivative that this, evaluate it in ~ 0,it'll be the exact same thing as the 4th derivative ofthe duty evaluated at 0. And also in general, youcan keep adding terms whereby the n-th termwill look prefer this. The n-th derivative of yourfunction evaluate at 0 times x come the n over n factorial. Notification this is the samething together 4 factorial. 4 factorial is same to 4times 3 times 2 times 1. Friend don't have actually towrite the 1 there, however you can put that there. This right right here is the samething together 3 factorial-- 3 time 2 times 1. I didn't placed the 1 there. This right below is the samething together 2 factorial, 2 times 1. This is the same thing. We didn't compose anything,but you could divide this by 1 factorial, whichis the same thing as 1. And you can dividethis by 0 factorial, which likewise happens to it is in 1. We won't need to studyit too lot over here. However this general collection thatI've sort of set up right below is referred to as the Maclaurin series. And you deserve to approximatea polynomial. And also we'll see it leads tosome pretty powerful results later on on. However what happens-- and also Idon't have actually the computing power in my brain to drawthe graph properly-- is that once onlythe attributes equal, you gain that horizontal line. When you do thefunction equal 0 and their firstderivatives equal at 0, then you have actually something thatlooks favor the tangent line. As soon as you add anotherdegree, it might approximate the polynomialsomething prefer this. As soon as you include another degree, itmight look something prefer that. And as you keep addingmore and much more degrees, as soon as you store addingmore and more terms, it it s okay closerand closer around, particularly as you getclose come x is equal to 0. But in theory, if friend addan infinite variety of terms, girlfriend shouldn't be able to do--I haven't proven this come you, for this reason that's why I'm speak that. I haven't showed it however to you. However if you include aninfinite variety of terms, every one of the derivativesshould be the same. And also then the functionshould pretty lot look favor each other. In the next video, I'll dothis through some actual features just so it renders alittle bit much more sense. And just so friend know,the Maclaurin collection is a special caseof the Taylor series because we're centering it in ~ 0.


You are watching: Taylor expansion of sqrt(1+x^2)


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And also when you're doinga Taylor series, you have the right to pick any kind of center point. We'll focus on theMaclaurin appropriate now.