Calculus HW Question

Question

How would I find the Taylor Series of f(x) = 7/x^2 at a=1 and expand to 4 terms.

Answer 1

First, we know that the general formula for the Taylor series at a is the following:

f(x) = ∑ (the sum) from n=0 to n=∞ of f[n](a) * (x-a)^n / n!,

where f[n](a) is the nth derivative of f(x) at x = a, so now we have:

f(x) = f[0](a) * (x-a)^0 / 0! + f[1](a) * (x-a)^1 / 1! + f[2](a) * (x-a)^2 / 2!

+ f[3](a) * (x-a)^3 / 3! + ... + f[n](a) * (x-a)^n / n! + ... (and so on, all of the way up to ∞)

Now, given that f(x) = 7 / x^2, we need to find each of the following:

f[0](a=1) = f(1), which is = 7 / 1^2 = 7;

f[1](a=1) = f'(1) = d/dx [7 / x^2] (at x=1) = -14 / x^3 (at x=1) = -14 / 1^3 = -14;

f[2](a=1) = f''(1) = d/dx [-14 / x^3] (at x=1) = 42 / x^4 (at x=1) = 42 / 1^4 = 42;

f[3](a=1) = f'''(1) = d/dx [42 / x^4] (at x=1) = -168 / x^5 (at x=1) = -168 / 1^5 = -168;

...

f[n](a=1) = d/dx f[n-1](x) (at x=1) = 7*(-1)^n * (n+1)! / x^(n+2) (at x=1) = 7*(-1)^n * (n+1)!

... (and so on, all of the way up to ∞), so for our function we have:

f(x) = 7 / x^2 = ∑ (the sum) from n=0 to n=∞ of 7*(-1)^n * (n+1)! * (x-1)^n / n!,

= ∑ (the sum) from n=0 to n=∞ of 7 * (-1)^n * (n+1) * (x-1)^n

= 7 * (-1)^0 * (0+1) * (x-1)^0 + 7 * (-1)^1 * (1+1) * (x-1)^1 + 7*(-1)^2 * (2+1)*(x-1)^2

+ 7 * (-1)^3 * (3+1) * (x-1)^3 + ... + 7 * (-1)^n * (n+1) * (x-1)^n

+ ... (and so on, all of the way up to ∞)

So, finally, we have:

f(x) = 7 / x^2 = 7 - 14 * (x-1) + 21 * (x-1)^2 - 28 * (x-1)^3 + ... +

7 * (-1)^n * (n+1) * (x-1)^n + ... (and so on, all of the way up to ∞)

Answer 2

A Taylor Series approximating f(x) near the point "a" is:

f(x)_{near a} ≈ f(a) + f '(a)(x - a) + f ''(a)(x - a)²/2! + f '''(a)(x - a)³/3! + . . .

Since a = 1, this becomes:

f(x)_{near 1} ≈ f(1) + f '(1)(x - 1) + f ''(1)(x - 1)²/2! + f '''(1)(x - 1)³/3! + . . .

Since f(x) = 7/x², f(1) = 7

f '(x) = (7x^-2)' = -14x^-3 so f '(1) = -14

f ''(x) = (-14x^-3)' = 42x^-4 so f ''(1) = 42

f '''(x) = (42x^-4)' = -168x^-5 so f '''(1) = -168

So (for 4 terms):

f(x)_{near 1} ≈ 7 + -14(x - 1) + 42(x - 1)²/2! - 168(x - 1)³/3!

f(x)_{near 1} ≈ 7 + -14x + 14 + 42(x² - 2x + 1)/2 - 168(x³ - 3x² + 3x - 1)/6

f(x)_{near 1} ≈ 21 - 14x + 21(x² - 2x + 1) - 28(x³ - 3x² + 3x - 1)

f(x)_{near 1} ≈ 21 - 14x + 21x² - 42x + 21 - 28x³ + 84x² - 84x + 28

f(x)_{near 1} ≈ -28x³ + 105x² - 140x + 70

It's actually interesting to plot the f(x) and the approximation of f(x) and look how closely they match near x = 1.

Answer 3

First find f'(x), f"(x), and f'''(x), then plug in 1 for x into f(x), f'(x), f"(x), and f'''(x)..

The sum of the first 4 terms is f(1) + f'(1)(x-1) + [f"(1)/2!](x-1)² + [f'''(1)/3!](x-1)³