Question

Find the Maclaurin polynomials of orders n = 0, 1, 2, 3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation.
1/1+x

Answer 1

Answer:

Σ(-1)^kx^k for k = 0 to n

Step-by-step explanation:

The nth Maclaurin polynomials for f to be

Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......

The given function is.

f(x) = 1/(1+x)

Differentiate four times with respect to x

f(x) = 1/(1+x)

f'(x) = -1/(1+x)²

f''(x) = 2/(1+x)³

f'''(x) = -6/(1+x)⁴

f''''(x) = 24/(1+x)^5

To calculate with a coefficient of 1

f(0) = 1

f'(0) = -1

f''(0) = 2

f'''(0) = -6

f''''(0) = 24

Findinf Pn(x) for n = 0 to 4.

Po(x) = 1

P1(x) = 1 - x

P2(x) = 1 - x + x²

P3(x) = 1 - x+ x² - x³

P4(x) = 1 - x+ x² - x³+ x⁴

Hence, the nth Maclaurin polynomials is

1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n

= Σ(-1)^kx^k for k = 0 to n

Answer 2

Maclaurin polynomials of orders n = 0, 1, 2, 3,4 will be

1-x+x²-x³+x⁴.......+(-1)ⁿxⁿ.

How do Maclaurin Polynomials look like?

The Maclaurin Polynomials is of type:

Pₙ(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +f''''x⁴/4! +. ......

The given function is.

f(x) = 1/(1+x)

Differentiating till the fourth derivative with respect to x:

f(x) = 1/(1+x)

f'(x) = -1/(1+x)²

f''(x) = 2/(1+x)³

f'''(x) = -6/(1+x)⁴

f''''(x) = 24/(1+x)^5

So, f(0) = 1

f'(0) = -1

f''(0) = 2

f'''(0) = -6

f''''(0) = 24

P₀(x) = 1

P₁(x) = 1 - x

P₂(x) = 1 - x + x²

So, the Maclaurin polynomials will be:

1-x+x²-x³+x⁴.......+(-1)ⁿxⁿ

Therefore, Maclaurin polynomials of orders n = 0, 1, 2, 3,4 will be

1-x+x²-x³+x⁴.......+(-1)ⁿxⁿ.

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