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# Worked example: Maclaurin polynomial

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.A (LO)
,
LIM‑8.A.1 (EK)
,
LIM‑8.A.2 (EK)
,
LIM‑8.B (LO)
,
LIM‑8.B.1 (EK)

## Video transcript

we're told that f of X is equal to one over the square root of x plus one and what we want to figure out is what is the second degree Maclaurin polynomial of F and like always pause this video and see if you could have a go at it so let's remind ourselves what a Maclaurin polynomial is a Maclaurin polynomial is just a Taylor polynomial centered at zero so the form of this second degree Maclaurin polynomial and we just have to find this Maclaurin expansion until our second degree term it's going to look like this so P of X I'm using P for polynomial it's going to be our F of zero plus and we could view that as F of 0 times X to the 0 power well that's just F of 0 F of 0 plus F prime of 0 X plus F prime prime of 0 divided by we could think of it as 2 factorial but it's really just 2 we could think of this dividing by 1 factorial which is just 1 this dividing by 0 factorial but that's just 1 is again so we have F prime prime of 0 the second derivative evaluated 0 divided by 2 x squared now if we wanted a higher degree we could keep on going but remember they're just asking us for the second degree so this is the form that we're going to need we're going to have these three terms so let's see if we can evaluate these let's see if we can evaluate the function and its derivatives at 0 so f of 0 f of 0 is equal to 1 over the square root of 0 plus 1 well that's 1 over the square root of 1 the principal root of 1 which is positive 1 so that's just going to be equal to 1 so that right over there is equal to 1 now let's evaluate F prime of X and then I'll evaluate F prime of 0 F prime of X is equal to well 1 over the square root of x plus 1 this is the same thing as X plus 1 let me write it this way this is the same thing as let me actually write it down this way f of X another way of writing f of X is this is the same thing as X plus 1 to the negative 1/2 and so if I'm thinking the first derivative of F well I could use the chain rule here the derivative of X plus 1 with respect to X well that's just going to be 1 and then I'll take the derivative of this whole thing with respect to X plus 1 and I'll just use the power rule there this is going to be negative 1/2 times X plus 1/2 then i decrement the exponent negative three-halves and so the first derivative evaluated at 0 is just negative 1/2 times if this is 0 0 plus 1 is just 1 1 to the negative 3 halves 1 to the negative 3 half power well that's just going to be 1 so this whole thing f prime of 0 is just negative 1/2 so that is this right over here is negative 1/2 and now let's figure out the second derivative all right I'll do this let me do this in this green color so the second derivative with respect to X well I do the same thing again the derivative of X plus 1 with respect to X is just 1 so I just have to take the derivative of the whole thing with respect to X plus 1 so take my exponent bring it out front negative three-halves times 1/2 the times negative 1/2 is going to be positive 3/4 times X plus 1 and then I decrement the exponent here by one or by two halves this is going to be negative five halves and so the second derivative evaluated at zero well if this is equal to zero you're going to have 1 to the negative 5 halves which is just 1 times 3/4 this is going to be 3/4 so this part right over here is 3/4 and so you're going to 3 fourths divided by 2 3 fourths divided by 2 is 3/8 so our Taylor or I should say our Maclaurin polynomial our second-degree Maclaurin polynomial P of X is going to be equal to and I'll do it in the same colors it's going to be equal to 1 plus or maybe I'll just write it as negative on 1/2 minus 1/2 X plus 3/8 x squared plus 3/8 x squared and we are done there you have we have our second degree Maclaurin polynomial of F which is could be used to provide an approximation for our function especially as we as especially for X is near zero
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