Question

In Exercises 35 and \(36,\) use a power series and the fact that \(i^{2}=-1\) to verify the formula. $$ g(x)=\frac{1}{2 i}\left(e^{i x}-e^{-i x}\right)=\sin x $$

Short answer

The formula \( g(x)=\frac{1}{2 i}\left(e^{i x}-e^{-i x}\right)=\sin x \) is verified by expanding the expressions of \( e^{ i x } \) and \( e^{-i x } \) using power series, substituting them into \( g(x) \), simplifying it and comparing the obtained expression with the power series representation of \( \sin x \).

Step-by-step solution

Expand the expressions using the power series

From the formula for power series, we can write the expressions \( e^{ i x } \) and \( e^{-i x } \) as follows: \( e^{ i x } = \sum _{ n=0 }^{ \infty }{ \frac { (i x)^{ n } }{ n! } } \) and \( e^{-i x } = \sum _{ n=0 }^{ \infty }{ \frac { (-i x)^{ n } }{ n! } }\).

Simplify the expressions using the property \( i^{ 2 }=-1 \)

Taking into account that \( i^{ 2 }=-1 \), we can simplify the above expressions. For \( n=4k \), \( i^{ n }=i^{ 4k }=(i^{ 2 })^{ 2k }=(-1)^{ 2k }=1 \), for \( n=4k+1 \), \( i^{ n }=i^{ 4k+1 }=i(i^{ 2 })^{ 2k }=(-1)^{ k }i \), for \( n=4k+2 \), \( i^{ n }=i^{ 4k+2 }=(i^{ 2 })^{ 2k+1 }=(-1)^{ 2k+1 }=-1 \), for \( n=4k+3 \), \( i^{ n }=i^{ 4k+3 }=-i(i^{ 2 })^{ 2k }=(-1)^{ k }(-i) \).Thus, we update the expression for \( e^{ i x } \) and \( e^{-i x } \) taking into account the above simplifications.

Combine the series and simplify

We substitute expressions of \( e^{ i x } \) and \( e^{-i x } \) into \( g(x)=\frac{1}{2 i}\left(e^{i x}-e^{-i x}\right) \) to get the series form of \( g(x) \). As we go through the calculation, we will find that only the terms for 4k+1 and 4k+3 remain, and these are precisely the terms in the series for \( \sin x \). In other words, \( g(x)=\sin x \).

Verify the expression

We can verify the expression by comparing the series form of \( g(x) \) obtained in the previous step with the power series representation of \( \sin x \), \( \sin x = \sum _{ n=0 }^{ \infty }{ (-1)^{ n }\frac { x^{ 2n+1 } }{ (2n+1)! } }\). We can see that the two expressions match, hence, verifying the given formula.