Find the Taylor series centered at x = -1 for the function
f(x) = x(e^x)
f(x) = x e^x = -1/e
f'(x) = (x+1) e^x = 0/e
f"(x) = (x+2) e^x = 1/e
f3(x) = (x+3) e^x = 2/e
...
x e^x = -1/e + (x+1)^2/(2! e) + (x+1)^3/(3!*2e) + (x+1)^4/(3!*4e) ...
= 1/e (-1 + (x^1)^2/2 + (x+1)^3/3 + (x+1)^4/8 + ...)
To find the Taylor series for the given function centered at x = -1, we need to compute the derivatives of the function and evaluate them at x = -1.
Let's start by finding the first few derivatives of f(x). We'll need these to determine the coefficients in the Taylor series.
f(x) = x(e^x)
First derivative:
f'(x) = (e^x) + x(e^x)
Second derivative:
f''(x) = (e^x) + (e^x) + x(e^x) = 2(e^x) + x(e^x)
Third derivative:
f'''(x) = 2(e^x) + 2(e^x) + x(e^x) = 4(e^x) + x(e^x)
Fourth derivative:
f''''(x) = 4(e^x) + 4(e^x) + x(e^x) = 8(e^x) + x(e^x)
Now, let's evaluate these derivatives at x = -1 to find the coefficients of the Taylor series:
f(-1) = (-1)(e^(-1)) = -e^(-1)
f'(-1) = (e^(-1)) + (-1)(e^(-1)) = 0
f''(-1) = 2(e^(-1)) + (-1)(e^(-1)) = e^(-1)
f'''(-1) = 4(e^(-1)) + (-1)(e^(-1)) = 3e^(-1)
f''''(-1) = 8(e^(-1)) + (-1)(e^(-1)) = 7e^(-1)
Now, we can write the Taylor series centered at x = -1 using these coefficients:
f(x) = f(-1) + f'(-1)(x - (-1)) + (f''(-1)/2!)(x - (-1))^2 + (f'''(-1)/3!)(x - (-1))^3 + (f''''(-1)/4!)(x - (-1))^4 + ...
Simplifying,
f(x) = -e^(-1) + 0(x + 1) + (e^(-1)/2!)(x + 1)^2 + (3e^(-1)/3!)(x + 1)^3 + (7e^(-1)/4!)(x + 1)^4 + ...
This is the Taylor series centered at x = -1 for the function f(x) = x(e^x).
To find the Taylor series centered at x = -1 for the function f(x) = x(e^x), we can use the definition of the Taylor series expansion. The general formula for the Taylor series expansion of a function f(x) centered at x = a is:
f(x) = f(a) + f'(a)(x - a)/1! + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ...
To find the coefficients of the Taylor series expansion, we need to calculate the derivatives of the function f(x) and evaluate them at x = -1.
Step 1: Find the first few derivatives of f(x):
f'(x) = (d/dx)[x(e^x)] = e^x + x(e^x)
f''(x) = (d^2/dx^2)[x(e^x)] = e^x + e^x + x(e^x)
f'''(x) = (d^3/dx^3)[x(e^x)] = e^x + e^x + e^x + x(e^x)
...
Step 2: Evaluate the derivatives at x = -1:
f'(-1) = e^(-1) - e^(-1) = 0
f''(-1) = e^(-1) + e^(-1) - e^(-1) = e^(-1)
f'''(-1) = e^(-1) + e^(-1) + e^(-1) - e^(-1) = 2e^(-1)
...
Step 3: Write out the Taylor series expansion using the calculated derivatives:
f(x) = f(-1) + f'(-1)(x + 1)/1! + f''(-1)(x + 1)^2/2! + f'''(-1)(x + 1)^3/3! + ...
Since f(-1) is not given, it can be calculated by substituting x = -1:
f(-1) = -1(e^-1) = -e^-1
Putting all the terms together, the Taylor series centered at x = -1 for the function f(x) = x(e^x) is:
f(x) = -e^-1 + 0(x + 1)/1! + e^-1(x + 1)^2/2! + 2e^-1(x + 1)^3/3! + ...