# How do you find a power series converging to #f(x)=e^(x/2)# and determine the radius of convergence?

with radius of convergence

Since:

we have that:

Using the ratio test:

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To find a power series converging to ( f(x) = e^{x/2} ) and determine the radius of convergence, we can use the Taylor series expansion of ( e^x ) centered at ( x = 0 ). Then, we'll substitute ( x/2 ) for ( x ). The Taylor series expansion of ( e^x ) is:

[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} ]

Now, substituting ( x/2 ) for ( x ), we get:

[ e^{x/2} = \sum_{n=0}^{\infty} \frac{(x/2)^n}{n!} ]

This simplifies to:

[ e^{x/2} = \sum_{n=0}^{\infty} \frac{x^n}{2^n \cdot n!} ]

This is the power series representation of ( f(x) = e^{x/2} ). To determine the radius of convergence, we'll use the ratio test. Let ( a_n = \frac{x^n}{2^n \cdot n!} ). Applying the ratio test:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1}}{2^{n+1} \cdot (n+1)!} \cdot \frac{2^n \cdot n!}{x^n} \right| ]

[ = \lim_{n \to \infty} \left| \frac{x}{2(n+1)} \right| ]

[ = \lim_{n \to \infty} \frac{|x|}{2(n+1)} ]

This limit converges to zero for all ( x ), which means the radius of convergence is infinite, or in other words, the power series converges for all real numbers ( x ).

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