Use a Taylor series to verify the given formula. ∑i=0^∞ ((-1)^k π^2 k+1)/((2 k+1) !)=0

Name: Problem 30, Chapter 8: Calculus: Early Transcendental Functions
Uploaded: 2020-07-18T21:19:10-08:00
Duration: 9 min 19 s
Channel: Carrie Hicks
Description: Use a Taylor series to verify the given formula. ∑i=0^∞ ((-1)^k π^2 k+1)/((2 k+1) !)=0

Use a Taylor series to verify the given formula. ∑i=0^∞...

Key Concepts

Maclaurin Series

The Maclaurin series is a specific type of Taylor series where the expansion is done around zero. It expresses functions as a power series in x, which is particularly useful in approximating and understanding the behavior of standard functions like sine, cosine, and exponential functions near the origin.

Sine Function Series

The sine function can be expanded into its Maclaurin series form as sin(x) = ? (-1)^k x^(2k+1)/(2k+1)!, where k starts at 0. This alternating series not only provides an approximation method for sine but also directly relates to trigonometric identities such as sin(?) = 0. Evaluating the series at x = ? confirms the identity by summing the infinite series to yield zero.

Taylor Series Expansion

A Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This concept is fundamental in analysis and allows complex functions to be approximated by simpler polynomial expressions, enabling both theoretical insights and practical computational techniques.

Transcript

00:01 We want to use the taylor series to verify that the series from k equals 0 to infinity of negative 1 to the k times pi to the 2k plus 1 over 2k plus 1 factorial is zero.

00:45 First, i want to consider sign x.

00:48 And the reason i'm thinking sign is because i have the power series for sign memorized, and i know that it's an alternating series and that i'll have x to the 2k plus 1 over the 2k plus 1 factorial in the denominator.

01:03 I remember that in the deep recesses of my brain.

01:07 But just in case you don't remember that, let's go through the steps to show that.

01:12 If f of x is sine x, then the derivative is cosine x.

01:20 Then our second derivative is negative sine x.

01:30 Then our third derivative is negative cosine x.

01:43 Now when we plug in 0, f at 0 is the sign of 0, which is you is 0.

02:00 Then f prime at 0 is the cosine of 0, which is 1.

02:15 The second derivative at 0 is negative sign of 0, which is 0.

02:32 And the third derivative at 0 is, let's see, derivative of cosine, so this would be negative cosine of 0, which is negative 1.

03:01 And then we would start repeating.

03:04 Notice the fourth derivative would be the derivative of negative cosine, which would be sign, so we would start going back through the same four functions.

03:13 So now what we know is that sine of x is, is equals f of 0 divided by 0 factorial plus f prime of 0 0 divided by 1 factorial and this would be times x plus the second derivative at 0 this is divided by 2 factorial that would be times x squared plus the third derivative at 0, that's divided by 3 factorial, so it's times x cubed.

04:19 Then we've got the fourth derivative at 0, divided by 4 factorial, and so that's times x to the fourth.

04:37 The pattern continues.

04:41 Now f of zero is zero.

04:45 So that term will be zero.

04:47 The second derivative of f at zero is zero.

04:51 So that term will be zero and this term will be zero.

04:56 So we have zero plus x plus zero.

05:05 Now the third derivative is is negative 1.

05:10 So now we'll have minus x cubed over 3 factorial.

05:22 And then the fourth derivative is 0.

05:25 And then the fifth derivative we're repeating now.

05:31 We would have a positive x to the fifth over 5 factorial.