- Maclaurin series of cos(x)
- Maclaurin series of sin(x)
- Maclaurin series of eˣ
- Worked example: power series from cos(x)
- Worked example: cosine function from power series
- Worked example: recognizing function from Taylor series
- Maclaurin series of sin(x), cos(x), and eˣ
- Visualizing Taylor series approximations
- Euler's formula & Euler's identity
Worked example: recognizing function from Taylor series
Sample AP Calculus question asking to recognize a function from its Taylor series.
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- The series can also be written as Sigma n = 0 to infinity (-1) to the power n, multiplied with e power x ? isnt it?(1 vote)
- Perhaps for the first constant, but after that the series just diverges into bits of -1 and 1, so your hypothesis is probably not applicable here.(1 vote)
- Isn't it easier to recognize that ((-1)^n)*(x^n) = ((-1)*x)^n, then if we consider f(x)=e^x one can realize that the series is the expansion of f(-x)=e^(-x)?(1 vote)
- [Instructor] 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 zero, it'd be negative one to the zero power which is one times x to the zero which is one over zero factorial which is one so it'll be one plus and then when n is one, well then this is going to be negative so it's going to be minus and then x to the first over one factorial over, well, I could just write that as x over here and then when n is two, the negative one squared, that's gonna be positive x squared over two and there's going to be minus x to the third over three factorial and then it's going to be plus and I can keep going. You've seen this before. x to the fourth over four factorial and it's gonna keep going. Minus, plus. It's going to keep alternating on and on and on. Now, our general form for a Taylor series about zero which we could also call a Maclaurin series would be, our general form would be f of zero plus f prime of zero times x plus f prime prime of zero times x squared over two plus the the third derivative at zero times x to the third over three factorial plus the fourth derivative, you get the idea, evaluated at zero times x to the fourth over four factorial and we would just go on and on and on. Now, to figure out which function, in order what I wrote in blue to be the Maclaurin or to be the Taylor series about zero or in order to be the Maclaurin series, that means that, that means that f of zero needs to be equal to one. It means that f prime of zero, actually let me write this down. It means that f of zero needs to be equal to one. f of zero is equal to one. It means that f prime of zero needs to be the coefficient on the x here which is negative one and we could keep going. It means that, it means that the second derivative at zero, well, that's going to be the coefficient on this x squared over two so that's gotta be equal to one and you see the general idea that the third derivative at zero is equal to negative one. It's the coefficient on the x to the third over three factorial which is negative one 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. Let's evaluate all of these functions at zero and see which of these are one. So sine of zero. Well, that's zero. Just by looking at this first constraint, sine of zero isn't one. We can rule that out. Cosine of zero is one so that's still in the running. e to the zero is one. e to the zero is one and then the natural log of one plus zero, that's the natural log of one which is a zero so that's out of the running. So just from that first constraint knowing that f of zero is equal to one, we're able to rule out two of the choices then knowing that the first derivative evaluated at zero is going to be negative one but what's the first derivative of cosine of x? Well, it's the negative sine of x. If we evaluate that at zero, we're not gonna get negative one. We're gonna get zero 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 zero, we're gonna get one, not negative one 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 zero is going to be, is going to be negative e to the zero or negative one so it meets that one and if you were curious, you could keep going and 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 zero and the first derivative at zero.