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Question Video: Using the Lagrange Error Bound to Approximate the Value of a Logarithmic Function at a Point Mathematics • Higher Education

Find the error bound when using the third Taylor polynomial for the function 𝑓(π‘₯) = ln 3π‘₯ at π‘₯ = 1/3 to approximate the value of 𝑓(1/2). Give your answer in scientific form to three significant figures.

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Video Transcript

Find the error bound when using the third Taylor polynomial for the function 𝑓 of π‘₯ is equal to the natural logarithm of three π‘₯ at π‘₯ is equal to one-third to approximate the value of 𝑓 evaluated at one-half. Give your answer in scientific form to three significant figures.

The question gives us a function 𝑓 of π‘₯, and it wants us to determine the error bound if we were to approximate this function by using the third Taylor polynomial. We want to center our Taylor polynomial at one-third. And we’re going to use this to approximate the value of 𝑓 evaluated at one-half. Finally, we need to give our answer in scientific form to three significant figures.

To start, recall if we can approximate 𝑓 of π‘₯ by using an 𝑛-term Taylor polynomial, we can also add on a remainder term to help us see the difference between our function 𝑓 of π‘₯ and our approximation 𝑇 𝑛 of π‘₯. And we can actually find a bound on this remainder term. We need to recall the following. For the 𝑛th Taylor polynomial centered at a value of π‘Ž where we want to approximate at a value of π‘₯, we have the absolute value of 𝑅 𝑛 of π‘₯ is less than or equal to the absolute value of 𝑀 times π‘₯ minus π‘Ž all raised to the power of 𝑛 plus one divided by 𝑛 plus one factorial. And this value of 𝑀 will be an upper bound on the absolute value of the 𝑛 plus oneth derivative of 𝑓 of π‘₯ on an interval containing both π‘Ž and π‘₯.

And it’s worth pointing out there are a few pieces of information we’re missing about this, for example, how many times can we differentiate 𝑓 of π‘₯ and are these derivatives continuous on the interval containing π‘₯ and π‘Ž. These are definitely worth thinking about. However, usually problems of this will arise when we try and find our value of 𝑀 or we try and construct our Taylor polynomial. So we’ll ignore this and move on to find our bound. This is a very complicated-looking expression. However, it can be made easier by substituting in our values of π‘Ž and π‘₯.

First, we’re told to center our Taylor polynomial at π‘₯ is equal to one-third. So we’ll set our value of π‘Ž equal to one-third and update all of our bounds. Next, we see we’re using this to estimate the value of 𝑓 at one-half. So we can set our value of π‘₯ equal to one-half. And once again we can update our bound to include the value of π‘₯ at one-half. We can also update our equation for 𝑓 of π‘₯ involving the Taylor polynomial. However, this is not necessary. Finally, the question tells us to use the third Taylor polynomial. So we’ll set our value of 𝑛 equal to three. And once again, we can update all of our bounds with the value of 𝑛 set to be three.

The question is asking us to find a bound on our error, which is the following inequality. In other words, all we need to do is calculate the absolute value of 𝑀 times one-half minus one-third all raised to the power of three plus one divided by three plus one factorial. And the only part of this we don’t know is the value of 𝑀. So all we need to do is find our value of 𝑀. And we know that 𝑀 is an upper bound on the 𝑛 plus oneth derivative of π‘₯ on an interval.

In this case, since 𝑛 is three, this will be the fourth derivative of 𝑓 of π‘₯ with respect to π‘₯. We know this needs to be an interval containing both π‘Ž and π‘₯. So we need an interval containing one-third and one-half. When we’re only using this on individual values β€” for example, in this case, we’re only using this to approximate 𝑓 evaluated at one-half β€” we can just choose the closed interval between π‘Ž and π‘₯. In other words, we can just choose the closed interval from one-third to one-half.

We’re now almost ready to find our value of 𝑀. We just need to find an expression for the fourth derivative of 𝑓 of π‘₯ with respect to π‘₯. To do this, we just need to differentiate 𝑓 of π‘₯ four times. First, we need to find 𝑓 prime of π‘₯. That’s the derivative of the natural logarithm of three π‘₯ with respect to π‘₯.

There’s a few different ways of doing this. One way is to use the product rule for logarithms to rewrite 𝑓 of π‘₯ as the natural algorithm of three plus the natural logarithm of π‘₯. Then, the natural logarithm of three is a constant. So its derivative is just going to be equal to zero. And we know the derivative of the natural algorithm of π‘₯ with respect to π‘₯ is the reciprocal function, one over π‘₯. So 𝑓 prime of π‘₯ is one over π‘₯. Because we’re going to differentiate this again, we’ll write this as π‘₯ to the power of negative one.

We can now find all of our remaining derivatives by using the power rule for differentiation. We want to multiply by our exponent of π‘₯ and then reduce this exponent by one. We get 𝑓 double prime of π‘₯ will be equal to negative one times π‘₯ to the power of negative two. Doing the same again, we get 𝑓 triple prime of π‘₯ will be equal to two times π‘₯ to the power of negative three. And doing this one more time, we get the fourth derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to negative six times π‘₯ to the power of negative four. And we’ll use our laws of exponents to rewrite this as negative six divided by π‘₯ to the fourth power. The reason we do this is because we’re going to want to find an upper bound for this expression. And it’s easier to do this in this form.

First, it’s important to realize we don’t want an upper bound on the fourth derivative of 𝑓 of π‘₯. We want an upper bound on the sides of 𝑓 of π‘₯. So we need to take the absolute value of this expression, so we’ll just take the absolute value of this. This gives us the absolute value of the fourth derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to the absolute value of negative six divided by π‘₯ to the fourth power.

And we can simplify this. First, in our denominator, π‘₯ to the fourth power is equal to π‘₯ squared all squared. And we know this is greater than or equal to zero for all values of π‘₯. So taking the absolute value of our denominator is not going to change its value. This means all we need to do is take the absolute value of our numerator. And of course, we know the absolute value of negative six is just equal to six. So we can just simplify this expression to get six divided by π‘₯ to the fourth power.

Now we’re ready to find our value of 𝑀. We want to find an upper bound for this expression on the closed interval from one-third to one-half. We can see on this interval, we have a positive number divided by another positive number. We want to make this as big as possible. So to make this number as big as possible, we want to divide by the smallest number we can. The smallest number on our interval is when π‘₯ is equal to one-third. Therefore, on the closed interval from one-third to one-half, the absolute value of the fourth derivative of 𝑓 of π‘₯ with respect to π‘₯ will be less than or equal to six divided by one-third raised to the fourth power.

All we’re really saying here is we’re dividing by the smallest positive number we can on our interval, and this will be our value of 𝑀. And we can just calculate this expression. Six divided by one-third raised to the fourth power is equal to 486. Now, all we need to do is substitute this into our error bound. Substituting 𝑀 is equal to 486, π‘Ž is equal to one-third, π‘₯ is equal to one-half, and 𝑛 is equal to three into our error bound formula, we get the absolute value of 𝑅 three will be less than or equal to the absolute value of 486 times one-half minus one-third all raised to the power of three plus one divided by three plus one factorial. And it’s worth pointing out since we’re only interested in the error at our point, we will simplify our remainder polynomial to just be 𝑅 three.

Now, all that’s left to do is evaluate this expression. First, one-half minus one-third is one-sixth, three plus one is equal to four, and we know that four factorial is equal to 24. Then, if we just calculate this expression, we get one divided by 64. But remember, the question doesn’t want us to give this as an exact number. It wants us to give this in scientific form to three significant figures. To do this, we’ll start by writing this out as a decimal expansion. It’s equal to 0.015625. Next, we want this to three significant figures. We need to remember what this means. First, the initial zeros will never count. So in this case, our three significant figures will be the one, five, and six.

Next, we need to check the first emitted digit to see if we need to round up. In this case, it’s two, so we don’t need to round up. It’s less than five. So to three significant figures, the absolute value of 𝑅 three is approximately 0.0156. Finally, we need to write this in scientific notation. We start with a number between negative 10 and 10 and multiply this by 10 to the power of some integer. In this case, by looking at our three significant figures, we can see we need to start with 1.56. Then, we can see that 0.0156 is just equal to 1.56 times 10 to the power of negative two, which is our final answer.

Therefore, we were able to find the error bound when using the third Taylor polynomial for the function 𝑓 of π‘₯ is equal to the natural logarithm of three π‘₯ at π‘₯ is equal to one-third to approximate the value of 𝑓 of one-half. In scientific form to three significant figures, we got 1.56 times 10 to the power of negative two.

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