Question Video: Finding a Bound on the Error When Approximating an Exponential Function at a Point with Its Maclaurin Series | Nagwa Question Video: Finding a Bound on the Error When Approximating an Exponential Function at a Point with Its Maclaurin Series | Nagwa

# Question Video: Finding a Bound on the Error When Approximating an Exponential Function at a Point with Its Maclaurin Series Mathematics • Higher Education

Find the error bound when using the third Maclaurin polynomial for the function π(π₯) = π^(2π₯) to approximate the value of π(0.2). Give your answer to five decimal places.

06:04

### Video Transcript

Find the error bound when using the third Maclaurin polynomial for the function π of π₯ is equal to π to the power of two π₯ to approximate the value of π evaluated at 0.2. Give your answer to five decimal places.

In this question, we need to find the error bound, and we want to approximate the function π to the power of two π₯ by using the third Maclaurin polynomial at π₯ is equal to 0.2. We need to give our answer to five decimal places. To start, weβre going to need to recall how we find the error bound of a Taylor polynomial approximation. We recall that a function π of π₯ is equal to its πth term Taylor approximation plus the remainder term π π of π₯.

We recall we can find a bound on this remainder term π π of π₯ for a Taylor polynomial centered at π. The absolute value of π π of π₯ will be less than or equal to the absolute value of π times π₯ minus π all raised to the power of π plus one divided by π plus one factorial. And this is where π is an upper bound on the absolute value of the π plus oneth derivative of π on an interval containing both π and π₯. In our case though, weβre not using a Taylor polynomial. Instead, weβre using a Maclaurin polynomial. And we need to remember that a Maclaurin polynomial is exactly the same as a Taylor polynomial except Maclaurin polynomial is centered at π is equal to zero. So we set our center to be zero. And in fact, weβll update our error bound so that π is zero.

Next, in the question, weβre using the third Maclaurin polynomial for our approximation. So we need to set the value of π equal to three. Then, just as we did before, we can update our bound of the value of π set to be three. Finally, in the question, weβre using our third Maclaurin polynomial to approximate π at 0.2. So we need to set the value of π₯ to be 0.2. And once again, weβll update our bound so the value of π₯ is set to be 0.2. And now we can see weβve simplified the expression for our error bound. The only part of this expression we donβt know is the value of π. So we need to find the value of π.

First, weβre going to need to choose an interval which contains both zero and 0.2. This is our value of π and our value of π₯. Weβll choose the closed interval from zero to 0.2 since this is the smallest interval which contains both of these values. And by choosing the smallest interval we can, weβll get a better error bound. We now want to find our value of π. We can see from the definition of π itβs an upper bound on the absolute value of the fourth derivative of π of π₯ on the closed interval from zero to 0.2. So to find this, weβre going to need to find an expression for the fourth derivative of π of π₯ with respect to π₯.

Letβs start by finding the first derivative of π of π₯ with respect to π₯. Thatβs the derivative of π to the power of two π₯ with respect to π₯. We know how to do this by using our rules for differentiating exponential functions. We just need to multiply by the coefficient of π₯ in our exponent. We get π prime of π₯ is two times π to the power of two π₯. We can do exactly the same to find our second derivative. We need to multiply by the coefficient of π₯, which is two. This gives us π double prime of π₯ is equal to four times π to the power of two π₯. Doing this again, we get the third derivative of π of π₯ with respect to π₯ is equal to eight times π to the power of two π₯. And weβll do this one more time to find the fourth derivative of π of π₯ with respect to π₯. Itβs equal to 16 times π to the power of two π₯.

We see in our expression for π, weβre going to need to take the absolute value of the fourth derivative of π of π₯. So by taking the absolute value of both sides of this equation, we get the absolute value of the fourth derivative of π of π₯ with respect to π₯ is equal to the absolute value of 16 times π to the power of two π₯. But we can simplify this expression. First, we know π to the power of two π₯ is greater than zero for all values of π₯. And of course, we know that 16 is a positive constant. So this is the product of two positive numbers. So, the absolute value of this wonβt change its value. So we can just remove the absolute value symbol from this expression.

Weβre now ready to find our value of π. Remember, we need to maximize the absolute value of the fourth derivative of π of π₯ on the closed interval from zero to 0.2, so we need to maximize 16 times π to the power of two π₯ on the closed interval from zero to 0.2. And to make 16 times π to the power of two π₯ as big as possible, weβre going to need to make π to the power of two π₯ as big as possible. And we know something interesting about π to the power of two π₯. Itβs an increasing function. The larger our input is, the bigger our output is. In other words, we want to input as bigger value of π₯ as possible. And of course, the biggest input we can use on the closed interval from zero to 0.2 is 0.2.

So on this interval, 16 times π to the power of two π₯ is less than or equal to 16 times π to the power of two times 0.2. And we can simplify this slightly. Two times 0.2 is equal to 0.4. So weβve shown, on the closed interval from zero to 0.2, the absolute value of the fourth derivative of π of π₯ with respect to π₯ is less than or equal to 16 times π to the power of 0.4. This means this is going to be our value of the constant π. Weβre now ready to go back and calculate our error bound since we found the value of the constant π.

Substituting in π is equal to 16 times π to the power of 0.4 into our error bound, we get the absolute value of our remainder term π three is less than or equal to the absolute value of 16 times π to the power of 0.4 multiplied by 0.2 minus zero all raised to the power of three plus one all divided by three plus one factorial.

And now we can just calculate this expression. Doing this, we get two times π to the power of 0.4 all divided by 1875. But weβre not done yet. Remember, the question wants us to write our answer to five decimal places. And if we do this, we get the absolute value of our remainder term π three is less than or equal to 0.00159, which is our final answer. And itβs worth pointing out that 0.00159 is a very small number. And remember, this is a bound on the error of our approximation. When this is small, this means our approximation is very good. So by taking the third Maclaurin polynomial for this approximation, we were able to get our answer to within 0.00159.

We can in fact use a similar line of reasoning to work out how many terms of our Taylor polynomial we would need to get within a certain level of accuracy. However, we donβt need this to answer this question. Therefore, we were able to show the error bound when using the third Maclaurin polynomial for the function π of π₯ is equal to π to the power of two π₯ to approximate the value of π evaluated at 0.2 to five decimal places is 0.00159.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions