Use the Maclaurin series for e^x to calculate 1/ √(e) correct to five decimal places.

Name: Problem 40, Chapter 8: Essential Calculus Early Transcendentals
Uploaded: 2020-05-22T19:29:56-08:00
Duration: 5 min 44 s
Channel: Yaw Asomani
Description: Use the Maclaurin series for e^x to calculate 1/ √(e) correct to five decimal places.

Use the Maclaurin series for e^x to calculate 1/ √(e)...

Key Concepts

Exponentiation Rules

Exponentiation rules allow expressions involving radicals and reciprocals to be rewritten in exponential form, thereby making them amenable to series expansion techniques. For example, rewriting a radical expression like 1/?[10]{e} as e^(–1/10) facilitates the direct application of the Maclaurin series for e^x.

Numerical Approximation using Series

Numerical approximation using a series involves truncating an infinite series to a finite number of terms to estimate a function's value. Ensuring that the error from truncation is within acceptable limits is key, especially when a specific degree of accuracy, like five decimal places, is required.

Exponential Function Series

The exponential function e^x has an elegant and universally convergent Maclaurin series: e^x = 1 + x + x²/2! + x³/3! + …. Because its series expansion is simple and converges for all real x, it serves as a fundamental example in both theoretical and applied mathematics for approximating function values.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series centered at zero. It expresses a function as an infinite sum of terms calculated from the derivatives of the function at zero, making it a powerful tool for approximating functions by a finite polynomial. This concept is widely used in numerical methods to estimate the values of functions like exponentials.

Transcript

00:01 So if you've not watched the last tutorial before this tutorial, please watch that one because we're going to do exactly the same thing, right? let's say, for instance, we're going to use the mclaurian series to find e to the power.

00:16 Point one, remember, e to the power point one is one over e to the power is 0 .1.

00:22 What is 0 .1? 0 .1 is the same as 1 over 10, right? and this one over 10 means that it is the 10th root of e, right? so this is just the same as this one.

00:37 I was just breaking it down for you to see.

00:40 So once you have this one, what is the mclaurin series formula for e to the x? well, for e to the x, you have x to the n, right? you have x to the n over two, over n factor.

01:01 Right you have x to the n over n factorial so now whenever we see x we're going to replace this with that with that x right so what are we going to do this is going to be e to the negative 0 .1 this is going to be summation n to the 0 to infinity negative 1 to the n over n factorial right this is also an alternating series right because it's an alternate it's going to be negative and positive and so it's alternating, right? so we're still going to use the alternating series test.

01:37 Please, once again, if we didn't watch the tutorial before this one, please watch it because we're still going to be using that same procedure to find this one, right? using that same procedure.

01:51 So this is what we have.

01:53 So we have to find the terms such that we get the final term that we're going to add should be less than this, right? so we have to find the series of this one.

02:08 As usual, when n is zero, you have one.

02:11 When n is 1, you have negative, right? so what is here is supposed to be negative 0 .1 to the power in, right? remember we were replacing every instance of x by negative .1, right? so that is what we just did.

02:39 Good.

02:40 So when we put zero, as usual, zero is one, right? when you put n is zero, put zero in each and every here, term here.

02:49 It is going to be one.

02:50 The second one is going to be 0 .1 to the power of one, right? 0 .1 to the power of one because n is one, right? but i'm not going to write the one.

03:01 So 0 .1 over 1 factor is still 1.

03:05 So it's just 0 .1, right? so this 0 .1 is not less than this, right? it's not...

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