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.