function f(x) = x⁴ - 8x² + 12 and its Taylor series approximations around x = 0. Show the 2nd-order, 4th-order, and 6th-order Taylor approximations on the same graph.

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Let's examine the Taylor series approximations of the function f(x) equals x to the fourth minus 8x squared plus 12 around x equals 0. The second-order Taylor approximation P_2(x) equals 12 minus 8x squared. The fourth-order approximation P_4(x) equals 12 minus 8x squared plus x to the fourth. The sixth-order approximation P_6(x) is identical to the fourth-order one. Notice that since our original function is a fourth-degree polynomial, the Taylor series matches the function exactly from the fourth-order approximation onward. The green curve represents both P_4(x) and P_6(x), which are identical to the original function shown in red. The blue curve shows the second-order approximation, which is a parabola that matches the function's value and curvature at x equals 0. Now let's analyze the accuracy of our Taylor approximations. For this particular function, which is a 4th-degree polynomial, the 4th-order Taylor approximation is exactly equal to the original function. However, the 2nd-order approximation P_2(x) has some error. The purple curve shows the error between the original function and the 2nd-order approximation. Notice how this error is zero at x equals zero, and increases as we move away from the center point. This is a key property of Taylor series - they're most accurate near the point of expansion. Let's visualize this by moving a point along the x-axis and observing how the error changes. As we move further from x equals zero, the vertical distance between the red and blue curves - which represents the error - grows larger. This error is exactly equal to x to the fourth, which is the term we omitted in the 2nd-order approximation. Let's derive the Taylor series for our function f(x) equals x to the fourth minus 8x squared plus 12 around x equals 0. We start by calculating the derivatives of the function. The first derivative is 4x cubed minus 16x. The second derivative is 12x squared minus 16. The third derivative is 24x. The fourth derivative is simply 24. And all higher derivatives are zero. Next, we evaluate these derivatives at x equals 0. We get f(0) equals 12, f prime of 0 equals 0, f double prime of 0 equals negative 16, f triple prime of 0 equals 0, and the fourth derivative at 0 equals 24. Using the Taylor series formula, we can now construct our approximations. The second-order approximation P_2(x) equals 12 minus 8x squared. The fourth-order approximation P_4(x) equals 12 minus 8x squared plus x to the fourth. Since all higher derivatives are zero, the sixth-order approximation is identical to the fourth-order one, which is also identical to the original function. Let's analyze the error in our Taylor approximation using Taylor's Theorem with Remainder. According to this theorem, we can express our function as the sum of the Taylor polynomial and a remainder term. For a function with continuous derivatives, the remainder term R_n(x) can be expressed using the n+1 derivative evaluated at some point ξ between 0 and x. For our second-order approximation, the remainder term R_2(x) involves the third and fourth derivatives. Since the third derivative at any point ξ is 24ξ, and the fourth derivative is constant 24, we get R_2(x) equals 4ξx cubed plus x to the fourth. For this specific function, since all derivatives beyond the fourth are zero, this remainder term simplifies to just x to the fourth when we consider the full Taylor series. This is exactly the term we omitted in our second-order approximation. Let's visualize how this remainder term changes as we move along the x-axis. Notice that as x increases, the remainder term grows rapidly because it's proportional to x to the fourth power. This explains why our second-order approximation becomes increasingly inaccurate as we move away from x equals zero. Let's summarize what we've learned about Taylor series approximations. Taylor series allow us to approximate functions using polynomials centered at a specific point. For our function f(x) equals x to the fourth minus 8x squared plus 12 expanded around x equals 0, we found that the second-order approximation P_2(x) equals 12 minus 8x squared. The fourth-order and sixth-order approximations are both equal to 12 minus 8x squared plus x to the fourth, which is exactly the original function. We discovered that the error term for the second-order approximation is precisely x to the fourth, which explains why the approximation becomes less accurate as we move away from x equals 0. This illustrates a key property of Taylor approximations: they are most accurate near the expansion point. Finally, we observed that for polynomial functions, the Taylor series becomes exactly equal to the original function when we reach the polynomial's degree. This is why our fourth-order approximation perfectly matched our fourth-degree polynomial. These principles apply to approximating more complex functions as well, where higher-order terms provide increasingly accurate approximations.