Worked example: Maclaurin polynomial | Series | AP Calculus BC | Khan Academy

We're told that ( f(x) ) is equal to one over the square root of ( x + 1 ), 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(x) ), I'm using ( P ) for a polynomial, it's going to be our ( f(0) ) plus we could view that as ( f(0) ) times ( x^{0} ) power; well, that's just ( f(0) ). ( f(0) ) plus ( f'(0) x ) plus ( \frac{f''(0)}{2} x^{2} ).

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(0) ) is equal to ( \frac{1}{\sqrt{0 + 1}} ). Well, that's ( \frac{1}{\sqrt{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' ) of ( x ) and then I'll evaluate ( f' ) of 0. ( f'(x) ) is equal to... well, ( \frac{1}{\sqrt{x + 1}} ). This is the same thing as ( (x + 1)^{-\frac{1}{2}} ).

So if I'm thinking of the first derivative of ( f ), well I could use the chain rule here. The derivative of ( x + 1 ) with respect to ( x ) well that's just going to be 1. Then I'll take the derivative of this whole thing with respect to ( x + 1 ) and I'll just use the power rule there.

So it's going to be ( -\frac{1}{2} ) times ( (x + 1)^{-\frac{3}{2}} ) and so the first derivative evaluated at zero is just ( -\frac{1}{2} ) times one, one to the negative three halves; one to the negative three half power well that's just going to be 1.

So this whole thing ( f'(0) ) is just ( -\frac{1}{2} ). So that is this right over here is ( -\frac{1}{2} ), and now let's figure out the second derivative.

Alright, 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 + 1 ) with respect to ( x ) is just one, so I just have to take the derivative of the whole thing with respect to ( x + 1 ).

So I take my exponent, bring it out front, ( -\frac{3}{2} ) times ( -\frac{1}{2} ) is going to be ( \frac{3}{4} ) times ( (x + 1) ), and then I decrement the exponent here by 1 or by two halves, so it's going to be ( -\frac{5}{2} ).

So the second derivative evaluated at zero, well if this is equal to zero, you're going to have one to the negative five halves, which is just one times ( \frac{3}{4} ) is going to be ( \frac{3}{4} ). So this part right over here is ( \frac{3}{4} ), and so you're going to have ( \frac{3}{4} \div 2 ). ( \frac{3}{4} \div 2 ) is ( \frac{3}{8} ).

So our Taylor or actually our Maclaurin polynomial, our second degree Maclaurin polynomial ( P(x) ) is going to be equal to, and I'll do it in the same colors, it's going to be equal to ( 1 - \frac{1}{2} x + \frac{3}{8} x^{2} ).

And we are done! There you have, we have our second degree Maclaurin polynomial of ( f ), which could be used to provide an approximation for our function, especially as we—as especially for ( x )'s near zero.