The Maclaurin series for a function f(x) is given by the formula:
f(x) = f(0) + f'(0)x + rac{f”(0)}{2!}x^2 + rac{f”'(0)}{3!}x^3 + rac{f””(0)}{4!}x^4 + …
To find the Maclaurin series for f(x) = cos(x²), we first need to compute the derivatives of the function at x = 0.
1. **Finding f(0):**
f(0) = cos(0²) = cos(0) = 1
2. **Finding the first derivative:**
f'(x) = -2x * sin(x²)
f'(0) = -2(0)(sin(0)) = 0
3. **Finding the second derivative:**
Using the product rule on f'(x):
f”(x) = -2(sin(x²)) – 4x² * cos(x²)
f”(0) = -2(sin(0)) – 4(0)² * cos(0) = 0
4. **Finding the third derivative:**
Calculating f”'(x) gets a bit involved, but we can derive it:
f”(x) = -2sin(x²) – 4x²cos(x²), thus:
f”'(x) = -4xcos(x²) – 8xsin(x²) – 8x²sin(x²) + 8x³cos(x²) = -4xcos(x²) + …
After performing this calculation, we find that:
f”'(0) = 0
5. **Finding the fourth derivative:**
Repeating the process leads to:
f””(0) = -8
So, now we can plug these values into the Maclaurin series:
f(x) = 1 + 0 rac{x}{1!} + 0 rac{x^2}{2!} – rac{8}{4!}x^4 + … = 1 – rac{8}{24}x^4 + … = 1 – rac{1}{3}x^4 + …
This series converges for all x. Now, we can use this to compute f(4):
f(4) = 1 – rac{1}{3}(4)^4 = 1 – rac{1}{3}(256) = 1 – rac{256}{3} = 1 – 85.33 = -84.33.
Therefore, the final value is:
f(4) = -84.33