Question Video: Finding the Error Bound for the Taylor Polynomial of the Square Root Function

Find the error bound 𝑅₂ when using the second Taylor polynomial for the function 𝑓(π‘₯) = √π‘₯ at π‘₯ = 4 to approximate the value √5. Approximate to five decimal places.

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Video Transcript

Find the error bound 𝑅 two when using the second Taylor polynomial for the function 𝑓 of π‘₯ is equal to the square root of π‘₯ at π‘₯ is equal to four to approximate the value of the square root of five. Approximate to five decimal places.

We recall that we can write the function 𝑓 of π‘₯ to be equal to its 𝑛 term Taylor polynomial at π‘Ž plus the remainder after 𝑛 terms if 𝑓 is 𝑛 plus one times differentiable on an interval containing π‘₯ and π‘Ž. In particular, we know if we can find an upper bound on the absolute value of our 𝑛 plus one derivative function on an interval containing both π‘₯ and π‘Ž. Then we have that the absolute value of our 𝑛th remainder term is less than or equal to the absolute value of our upper bound 𝑀 divided by 𝑛 plus one factorial multiplied by π‘₯ minus π‘Ž raised to the power of 𝑛 plus one.

The question tells us to use the error abound 𝑅 two, when using the second Taylor polynomial. So this gives us that 𝑛 is equal to two. And this is for the function 𝑓 of π‘₯ is equal to the square root of π‘₯ about π‘₯ is equal to four. So 𝑓 of π‘₯ is equal to the square root of π‘₯ and π‘Ž is equal to four. And we’re approximating the value to square root of five. So we set π‘₯ equal to five. We’ll start by calculating 𝑅 𝑛 plus one derivative functions. So 𝑓 one of π‘₯ is equal to the derivative with respect π‘₯ of the square root of π‘₯. And we notice that the square root of π‘₯ is equal to π‘₯ to the power of a half. So we can differentiate this to get one-half multiplied by π‘₯ to the power of negative one-half.

Next, to find our second derivative of 𝑓, we differentiate a half multiplied by π‘₯ to the power of negative a half. To differentiate this, we multiplied by the exponent of negative one-half and then subtracted one from the exponent. This gives us negative a quarter multiplied by π‘₯ to the power of negative three over two. Finally, we calculate our first derivative of 𝑓 with respect π‘₯. So this is the derivative of negative a quarter multiplied by π‘₯ to the power of negative three over two with respect to π‘₯. And we can calculate this to give us three-eighths multiplied by π‘₯ to the power of negative five over two.

So we found our 𝑛 plus one derivatives of our function 𝑓 and then defined on the interval containing four and five. So our function 𝑓 must be 𝑛 plus one times differentiable. The next thing to do is to try to find our upper bound 𝑀, which is an upper bound on our 𝑛 plus one derivative function on an interval containing four and five. So let’s choose the closed interval from four to five. We want to find an upper bound on the absolute value of our third derivative function on this interval, which is the absolute value of three over eight multiplied by π‘₯ to the power of negative five over two.

To find an upper bound on this, we noticed that the absolute value of π‘₯ to the power of negative five over two is equal to the absolute value of one divided by π‘₯ to the power of five over two. This fraction will be as large as possible when its denominator is as small as possible. And since we’re on an interval from four to five, this will be when π‘₯ is equal to four. So we have that this is less than or equal to three-eighths multiplied by four to the power of negative five over two, which we can evaluate to give us three divided by 256. And this is what we set the value of 𝑀 two in our error bound. We’re now ready to substitute these values into our error bound, giving us that the absolute value of 𝑅 two evaluated at five is less than or equal to the absolute value of 𝑀, which is three divided by 256.

And we divide this by 𝑛 plus one factorial, which is two plus one factorial. And then we multiply this by π‘₯ minus π‘Ž all raised to the power of 𝑛 plus one, which in our case is five minus four to the power of two plus one. And if we evaluate this, we’ll get an answer of one divided by 512 which, to five decimal places, is 0.00195. So we have shown that if we were to use a second-term Taylor polynomial for the function 𝑓 of π‘₯ is equal to the square root of π‘₯ about π‘₯ is equal to four to approximate the square root of five. Then we have found an upper bound to the absolute value of our error bound 𝑅 two which, to five decimal places, is 0.00195.

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