Let f(x)=∑n=0^∞ ((-1)^n x^2 n+1)/((2 n+1) !) and g(x)=∑n=0^∞ ((-1)^n x^2 n)/((2 n) !) (a) Find t

Let f(x)=∑n=0^∞ ((-1)^n x^2 n+1)/((2 n+1) !) and g(x)=∑n=0^∞...

Key Concepts

Power Series and Interval of Convergence

A power series expresses a function as an infinite sum of terms involving powers of x. An important aspect of studying power series is determining the interval (or radius) of convergence, where the series converges to a finite value. Techniques such as the ratio test are often used to determine this interval, and for many familiar functions, such as the sine and cosine series, the series converge for all real numbers.

Term-by-Term Differentiation

Term-by-term differentiation allows one to differentiate a power series by differentiating each term individually. This method is justified within the interval of convergence of the series and is a powerful tool for relating the derivatives of functions represented by power series to other power series. This concept is essential for establishing relationships between functions where one is the derivative of the other.

Taylor Series Representations of Functions

Taylor series provide a way to represent functions as infinite sums of polynomial terms, and many elementary functions have well-known Taylor series expansions. In the context of the problem, recognizing the series for sine and cosine functions facilitates the identification of the functions defined by the given power series. Such representations are crucial for analyzing properties like convergence and differentiability.

Derivative Relationships in Trigonometric Functions

Many fundamental relationships in calculus arise from the derivative rules for trigonometric functions, such as the fact that the derivative of sine is cosine and the derivative of cosine is negative sine. These relationships can be derived directly from their Taylor series representations through term-by-term differentiation, highlighting a deep interplay between power series and classical differential calculus.

Transcript

00:02 In this problem, we are given two functions, f and g, are represented by power series.

00:11 And in the first part of this problem, we are asked to find the intervals of convergence of each function.

00:19 So let's start with f of x.

00:24 So remember, to find the interval of convergence, we will use the ratio test.

00:33 Ration test says that to find the interval of convergence, we need to calculate the limit of a n plus 1 over a n in our case a n plus 1 is equal to you just need to replace n by n plus 1 in the general term of the series and you can ignore minus 1 to the end because we are taking absolute values so for f of x a n plus 1 is equal to x the 2 times n plus 1 plus 1 remember we're replacing n by n plus 1 over 2 times n plus 1 plus 1 factorial divided by an but remember when you're dividing 2 fractions it's basically same as keeping the first fraction and multiplying by the reciprocal or the second fraction so reciprocal of x to the 2n plus 1 over 2n plus 1 factorial is 2n plus 1 factorial divided by x to the 2n plus 1 so we need to calculate this limit as n and goes to infinity.

02:01 This is equal to the limit of.

02:03 Now, after cancellations, we will get, so in the numerator, we have x to the 2n plus 2 plus 1, which is x to the 2 n plus 3.

02:14 Divided by x to 2n plus 1 gives us x squared.

02:20 In the denominator, we'll get 2n plus 3 factorial, and there still will be 2n plus 1 factorial remaining in the numerator and goes to infinity and this is equal to the limit absolute value x squared now remember 2 n plus 1 factorial is simply 1 times 2 times so on times 2 n plus 1 and 2n plus 3 factorial is 1 times 2 n plus 2 n plus 2 2 and plus 3 so it's same as 2 n plus 1 factorial but you have 2 more extra factors 2 and 2n plus 2n plus 3 so we can cancel the first 2n plus 1 term since the numerator and denominator and we will get limit of x squared over 2n plus 2 to n plus 3 now x is a fixed number and the denominator goes to infinity therefore this limit will be equal to 0 now by the ratio test if the limit is less than 1 the series converges now since the limit is equal to 0 and and since the limit is less than one for every value of x, this means that the series converges for all x.

04:01 Similarly, we can find the interval of convergence for function g of x.

04:09 We will need to calculate the limit, an plus 1 over n, n goes to infinity.

04:17 Now what is an plus 1 and an for g? let's go up.

04:24 So it's, so an plus 1 is going to be x, replace n by n plus 1.

04:30 Everywhere x to the 2n plus 2 over 2n plus 2 factorial that's a n plus 1 times reciprocal of a n it's just 2n factorial over x to the 2n and goes to infinity now after cancellations we'll get limit absolute value x squared over 2n plus 1 times 2n plus 2 now as n goes to infinity this limit is equal to 0 and 0 is always and this is always and this is series less than 1...

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