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rodikova [14]
3 years ago
13

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
Mathematics
1 answer:
Zielflug [23.3K]3 years ago
6 0

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

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Step-by-step explanation:

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Hey~freckledspots!\\----------------------

We~will~solve~for~i^{425}!

Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\

Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------

Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------

Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------

Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------

Answer:\\\large\boxed{-1+i}\\----------------------

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