write a power series for e to the 2x and find the interval of convergence

Since e^x = 1 + x + x^2/2! + x^3/3! + ...

e^2x = 1 + 2x + 4x^2/2! + 8x^3/3! + ...

It converges for all x.

To find the power series representation of e^(2x), we can use the Taylor series expansion for the exponential function:

e^z = 1 + z + (z^2)/2! + (z^3)/3! + (z^4)/4! + ...

Substituting z = 2x into the series, we get:

e^(2x) = 1 + (2x) + ((2x)^2)/2! + ((2x)^3)/3! + ((2x)^4)/4! + ...

Simplifying each term gives us:

e^(2x) = 1 + 2x + 4x^2/2 + 8x^3/6 + 16x^4/24 + ...

Simplifying further gives:

e^(2x) = 1 + 2x + 2x^2 + (4/3)x^3 + (2/3)x^4 + ...

Now let's determine the interval of convergence. For a power series, the interval of convergence is the range of x-values for which the series converges. We can use the ratio test to find this interval.

The ratio test states that a power series ∑(n = 0 to ∞) cnx^n converges if the following limit exists and is less than 1:

lim┬(n→∞)⁡〖|c_(n+1) x^(n+1) |/| c_n x^n 〗| < 1

Applying the ratio test to our series, we get:

lim┬(n→∞)⁡〖|(2n+2)x^(n+1) |/|(2n)x^n 〗| < 1

Simplifying the ratio and using some algebraic manipulation, we have:

lim┬(n→∞)⁡〖|2(n+1)x |/|(2n) 〗| < 1
lim┬(n→∞)⁡〖|(n+1)x |/|n 〗| < 1

Taking the absolute value and applying the limit, we get:

| x | < 1

Therefore, the interval of convergence for the power series representation of e^(2x) is (-1, 1).

So, the power series for e^(2x) is 1 + 2x + 2x^2 + (4/3)x^3 + (2/3)x^4 + ..., and it converges for x values between -1 and 1.