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# Worked example: recognizing function from Taylor series

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.E (LO)
,
LIM‑8.E.1 (EK)
,
LIM‑8.F (LO)
,
LIM‑8.F.2 (EK)

## Video transcript

so we're given this expression is the Taylor series about zero for which of the following functions and they give us some choices here so let's just think a little bit about this series that they gave us so if we were to expand it out let's see when n is equal to 0 would be negative 1 to the 0 power which is 1 times X to 0 which is 1 over 0 factorial which is 1 so it'll be 1 plus and then when n is 1 well then this is going to be negative so it's going to be minus and then X to the first over 1 factorial over well I could just write that as X right over here and then when n is to the negative negative 1 squared that's going to be positive x squared over 2 and that's going to be minus X to the third over 3 factorial and then it's going to be plus and I can keep going you've seen this before X to the fourth over 4 factorial it's gonna keep going - plus it's going to keep alternating on and on and on now our general form for our Taylor series about 0 which we could also call them Maclaurin series would be our general form would be f of 0 plus F prime of 0 times X plus F prime prime of 0 times x squared over 2 plus the third derivative at 0 times X to the third over 3 factorial plus the fourth derivative if you get the idea evaluated at 0 times X to the fourth over 4 factorial and we would just go on and on and on now to figure out which function in order for this in order what I wrote in blue to be the mcclure or to be the Taylor series about zero room in order to be the Maclaurin series and that means that that means that F of 0 needs to be equal to 1 it means that F prime of 0 actually let me write this down it means that F of 0 needs to be equal to 1 f of 0 is equal to 1 it means that F prime of 0 needs to be the coefficient on the X here which is negative 1 now we could keep going it means that it means that the second derivative at 0 well that's going to be the coefficient on this x squared over 2 so that's got to be equal to 1 and you see the general idea the third derivative at 0 is equal to negative 1 it's the coefficient on the X third x to the third over 3 factorial which is negative 1 right over here and so just using this information can we figure out which of these it is so you could do a little bit of deductive reasoning here for let's all evaluate all of these functions at 0 and see which of these are 1 so sine of 0 well that's 0 just by looking at this first constraint sine of 0 isn't 1 we can rule that out cosine of 0 is 1 so that's still in the running e to the 0 is 1 e to the 0 is 1 and then the natural log of 1 plus 0 that's the natural log of 1 which is a 0 so that's out of the running so just from that first constraint knowing that F of 0 is equal to 1 were able to rule out two of the choices then knowing that the first derivative evaluated at 0 is going to be negative 1 well what's the first derivative of cosine of X what's negative sine of X if we evaluate that at 0 we're not going to get negative 1 we're going to get 0 so we can rule this out now the first derivative of e to the X is e to the X if we evaluate that at 0 we're gonna get 1 not negative 1 so we can rule that out and so not even looking at anything else we have a pretty good sense that D is probably our answer but we could check the first derivative here F prime of X here is going to be negative e to the negative x so f prime of 0 is going to be is going to be negative e to the 0 or negative 1 so it meets that one and if we if you were curious you could keep going and see that see that it meets all the other constraints but choice D is the only one that meets even the first two constraints for the function at 0 and the first derivative at zero