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What are the first four terms of the Taylor series of the function 𝑓(π‘₯) = √π‘₯ about π‘₯ = 4?

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Video Transcript

What are the first four terms of the Taylor series of the function 𝑓 of π‘₯ equals the square root of π‘₯ about π‘₯ equals four?

We recall that the Taylor series of a function 𝑓 about π‘Ž is given by the sum of the 𝑛th derivative of 𝑓 evaluated at π‘Ž of 𝑛 factorial times π‘₯ minus π‘Ž to the 𝑛th power for values of 𝑛 between zero and infinity. So what do we know about our function? It’s given as 𝑓 of π‘₯ equals the square root of π‘₯, which we might alternatively say is π‘₯ to the power of one-half. We want to find the Taylor series about π‘₯ equals four, so we’re going to let π‘Ž be equal to four. Using the second form of the Taylor series, we find that 𝑓 of π‘₯ is equal to 𝑓 of four plus 𝑓 prime of four over one factorial times π‘₯ minus four plus 𝑓 double prime of four over two factorial times π‘₯ minus four squared plus 𝑓 triple prime of four over three factorial times π‘₯ minus four cubed.

We can work out 𝑓 of four fairly easily. We’ll substitute four into our original function. But to be able to work out 𝑓 prime of four, 𝑓 double prime of four, and 𝑓 triple prime of four, we’re going to need to differentiate our function 𝑓 of π‘₯ with respect to π‘₯ three times. We said that 𝑓 of π‘₯ was equal to π‘₯ to the power of one-half and we recall that to differentiate a polynomial term, we multiply the entire term by the exponent and then reduce that exponent by one. This means 𝑓 prime of π‘₯, the first derivative of 𝑓 is a half π‘₯ to the power of negative one-half. 𝑓 double prime of π‘₯ is negative a half times a half π‘₯ to the power of negative three over two, which simplifies to negative one-quarter π‘₯ to the power of negative three over two. Finally, 𝑓 triple prime of π‘₯ is negative three over two times negative a quarter π‘₯ to the power of negative five over two, which is three-eighths π‘₯ to the power of negative five over two.

We’re now going to substitute π‘₯ equals four into our functions for 𝑓 of π‘₯, 𝑓 prime of π‘₯, 𝑓 double prime of π‘₯, and 𝑓 triple prime of π‘₯. 𝑓 of four is the square root of four; it’s simply two. 𝑓 prime of four is a half times four to the power of negative one-half which is a quarter. 𝑓 double prime of four is negative one-quarter times four to the power of negative three over two, which is negative one over 32. And 𝑓 triple prime of four is three-eighths times four to the power of negative five over two, which is three over 256. Our final job is simply to substitute these into our expansion. When we do, we find the first four terms of the Taylor series of the function 𝑓 of π‘₯ equals the square root of π‘₯ about π‘₯ equals four to be two plus a quarter times π‘₯ minus four minus one over 64 times π‘₯ minus four squared plus one over 512 times π‘₯ minus four cubed.

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