### Video Transcript

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

We recall that the Taylor series of
a function π about π is given by the sum of the πth derivative of π evaluated at
π of π factorial times π₯ minus π to the πth power for values of π between zero
and infinity. So what do we know about our
function? Itβs given as π 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 weβre going to let π be equal to four. Using the second form of the Taylor
series, we find that π of π₯ is equal to π of four plus π prime of four over one
factorial times π₯ minus four plus π double prime of four over two factorial times
π₯ minus four squared plus π triple prime of four over three factorial times π₯
minus four cubed.

We can work out π of four fairly
easily. Weβll substitute four into our
original function. But to be able to work out π prime
of four, π double prime of four, and π triple prime of four, weβre going to need
to differentiate our function π of π₯ with respect to π₯ three times. We said that π 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 π prime of π₯, the
first derivative of π is a half π₯ to the power of negative one-half. π 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, π 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.

Weβre now going to substitute π₯
equals four into our functions for π of π₯, π prime of π₯, π double prime of π₯,
and π triple prime of π₯. π of four is the square root of
four; itβs simply two. π prime of four is a half times
four to the power of negative one-half which is a quarter. π 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 π 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 π 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.