Find the Maclaurin series of xcos(x^2).
Oh, you want me to don my mathematician's hat, do you? Well, alright then! Let's get into the mathematical circus!
The Maclaurin series expansion is like the clown car of mathematics. It's a fancy way of expressing a function as an infinite sum of simpler functions. In this case, we're dealing with xcos(x^2).
To find the Maclaurin series of this function, we need to calculate its derivatives and evaluate them at x = 0.
First, let's start with the function itself, f(x) = xcos(x^2).
The zeroth derivative of f(x) is just f(x) itself, so f(0) = 0.
Next, we calculate the first derivative. The derivative of f(x) with respect to x is f'(x) = cos(x^2) - 2x^2sin(x^2). Evaluating this at x = 0 gives us f'(0) = 1.
Moving on to the second derivative, we differentiate f'(x) = cos(x^2) - 2x^2sin(x^2). The second derivative f''(x) = -4xsin(x^2) - 4x^3cos(x^2). Evaluating this at x = 0 yields f''(0) = 0.
For the third derivative, we differentiate f''(x) = -4xsin(x^2) - 4x^3cos(x^2). The third derivative f'''(x) = -4sin(x^2) - 4(3x^2cos(x^2) - xsin(x^2)). Evaluating this at x = 0 gives f'''(0) = -4.
Now, let's calculate the fourth derivative. We differentiate f'''(x) = -4sin(x^2) - 4(3x^2cos(x^2) - xsin(x^2)). The fourth derivative f''''(x) = -4(4xsin(x^2) + 6xcos(x^2) - 12x^3sin(x^2) + 3sin(x^2) - 4x^3cos(x^2)). Evaluating this at x = 0 yields f''''(0) = 0.
From here on, we can see that the pattern repeats. The odd-numbered derivatives evaluated at x = 0 are all 0, and the even-numbered derivatives evaluated at x = 0 alternate between being 1 and 0.
Therefore, the Maclaurin series of f(x) = xcos(x^2) is:
f(x) = 0 + 1x + 0x^2 - 4x^3 + 0x^4 + ...
You can continue this pattern to find as many terms as you need.
And there you have it, the mathematical circus trick is complete! Enjoy the infinite sum of laughter and functions!
To find the Maclaurin series of xcos(x^2), we will use the power series expansion of the cosine function.
The power series expansion of cos(x) is given by:
cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...
To find the Maclaurin series of xcos(x^2), we substitute x^2 into the power series expansion of cos(x).
cos(x^2) = 1 - ((x^2)^2)/2! + ((x^2)^4)/4! - ((x^2)^6)/6! + ...
= 1 - (x^4)/2! + (x^8)/4! - (x^12)/6! + ...
Now, we substitute x with x^2 in each term and multiply by x.
xcos(x^2) = x - (x^5)/2! + (x^9)/4! - (x^13)/6! + ...
The Maclaurin series of xcos(x^2) is given by:
xcos(x^2) = ∑ ((-1)^n * (x^(4n+1))/(2n)!) for n = 0 to infinity
To find the Maclaurin series of ƒ(x) = xcos(x^2), we can use the power series expansion of cos(x).
Step 1: Evaluate ƒ(0) and its derivatives at x = 0.
- ƒ(0) = 0cos(0^2) = 0
- ƒ'(x) = cos(x^2) - 2x^2sin(x^2)
- ƒ'(0) = cos(0^2) - 2(0)^2sin(0^2) = 1
Step 2: Find the general formula for the nth derivative at x = 0.
- The pattern we see is that the derivative of cos x^2 contains terms of the form x^(2n) and its coefficients are related to the corresponding derivatives of cos(x).
- So, the general formula for the nth derivative is: ƒ^(n)(x) = cos(x^2) - 2nx^2sin(x^2)
Step 3: Write the Maclaurin series for ƒ(x).
- The Maclaurin series is given by: ƒ(x) = f(0) + f'(0)(x - 0) + f''(0)(x - 0)^2/2! + f'''(0)(x - 0)^3/3! + ...
- Substituting the values from Step 1 and the general formula from Step 2 into the series, we get:
ƒ(x) = 0 + 1(x - 0) - 2(0)x^2sin(0^2)(x - 0)^2/2! + 0(x - 0)^3/3! + ...
Simplifying, we have:
ƒ(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
Therefore, the Maclaurin series of ƒ(x) = xcos(x^2) is given by:
x - x^3/3! + x^5/5! - x^7/7! + ...
well, you know that
cos(x) = 1 - x^2/2! + x^4/4! - ...
so,
cos(x^2) = 1 - x^4/2! + x^8/4! - ...
Now just multiply that by x.
See
http://www.wolframalpha.com/input/?i=x+cos(x%5E2)