let f be the function f(x) = e^(2x^2)
find the first four nonzero terms and the general term of the power series for f(x) about x = 0
i tried doing this problem but finding the derivatives gets way too complicated
can anyone help??
HuH?
THe power series is
eu= Σxu/n! from n=1 to inf
So put in u (2x^2) and compute for n=1, ...4.
so ik this 14 years and 1 day late butttttt um this isn't even a real problem
Sure, I can help you with finding the power series for the given function. Instead of finding derivatives directly, we can use the Taylor series expansion to derive the power series representation.
The Taylor series expansion for a function about x = 0 is given by:
f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ...
To find the first four nonzero terms of the power series for f(x) = e^(2x^2), we need to calculate the function value and its first, second, and third derivatives at x = 0.
Step 1: Calculate f(0)
f(0) = e^(2(0)^2) = e^0 = 1
Step 2: Calculate f'(x)
f'(x) = d/dx[e^(2x^2)]
To find the derivative, we can use the chain rule:
f'(x) = 4x*e^(2x^2)
Step 3: Calculate f''(x)
f''(x) = d/dx[4x*e^(2x^2)]
Apply the product rule:
f''(x) = 4e^(2x^2) + 16x^2e^(2x^2)
Step 4: Calculate f'''(x)
f'''(x) = d/dx[4e^(2x^2) + 16x^2e^(2x^2)]
Again apply the product rule:
f'''(x) = 32xe^(2x^2) + 64x^3e^(2x^2) + 8xe^(2x^2)
Now we have the necessary terms to expand the power series:
f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ...
= 1 + 0x + (4e^(0)x^2)/2! + (32xe^(0) + 64x^3e^(0) + 8xe^(0))/3! + ...
= 1 + 2x^2 + 32x + 64x^3 + 4x + ...
The first four nonzero terms of the power series for f(x) about x = 0 are:
1 + 2x^2 + 32x + 64x^3
The general term of the power series can be written as:
aₙxⁿ, where n is the term number starting from 0.
Therefore, the general term of the power series for f(x) about x = 0 is:
aₙ = aₙ₋₁ * (2n) / ((n+1)(n+2)), for n > 0.
And a₀ = f(0) = 1.
So, the power series can be written as:
f(x) = 1 + 2x^2 + ∑[aₙxⁿ], where n starts from 3 and continues.
To find the power series representation of the function f(x) = e^(2x^2) about x = 0, we can use the Taylor series expansion.
Step 1: Find the first four nonzero terms.
To do this, we need to calculate the derivatives of f(x) with respect to x at x = 0. However, as you mentioned, finding the derivatives of f(x) directly can be quite complicated. But, there is an alternative method to simplify the process.
We can use the chain rule and the power rule for derivatives to express the higher-order derivatives of f(x) in terms of the derivative of e^u, where u = 2x^2. The derivative of e^u with respect to u is simply e^u.
Let's calculate the derivatives:
f'(x) = d/dx (e^(2x^2))
= d/du (e^u) * du/dx
= e^(2x^2) * 4x
= 4x * e^(2x^2)
f''(x) = d^2/dx^2 (e^(2x^2))
= d/dx (4x * e^(2x^2))
= 4 * e^(2x^2) + 8x * e^(2x^2)
= (4 + 8x) * e^(2x^2)
f'''(x) = d^3/dx^3 (e^(2x^2))
= d/dx ((4 + 8x) * e^(2x^2))
= (8 + 16x) * e^(2x^2) + (4 + 8x) * (4x * e^(2x^2))
= (8 + 16x + 16x^2) * e^(2x^2)
f''''(x) = d^4/dx^4 (e^(2x^2))
= d/dx ((8 + 16x + 16x^2) * e^(2x^2))
= (16 + 32x + 32x^2) * e^(2x^2) + (8 + 16x + 16x^2) * (4x * e^(2x^2))
= (16 + 32x + 32x^2 + 64x^3) * e^(2x^2)
Now, we can evaluate these derivatives at x = 0 to get the first four nonzero terms:
f(0) = e^(2(0)^2) = e^0 = 1
f'(0) = 4(0)e^(2(0)^2) = 0
f''(0) = (4 + 8(0))e^(2(0)^2) = 4
f'''(0) = (8 + 16(0) + 16(0)^2)e^(2(0)^2) = 8
Therefore, the first four nonzero terms are 1, 0, 4, and 8.
Step 2: Find the general term of the power series.
Now that we have the first four nonzero terms, we can write the general term of the power series using the formula for Taylor series expansions:
f(x) ≈ f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ...
Substituting the values we found:
f(x) ≈ 1 + 0x + (4x^2)/2! + (8x^3)/3! + ...
Simplifying gives:
f(x) ≈ 1 + 2x^2 + (4x^3)/3 + ...
Therefore, the general term of the power series for f(x) about x = 0 is (4x^n)/(3n!) for n ≥ 2.