expand-1-x-1-using-maclaurins-theorem-and-talyors-formula-

Question Number 87245 by redmiiuser last updated on 03/Apr/20

e x p a n d

{(1 + x)}^{- 1}

u s i n g m a c l a u r i n s

t h e o r e m a n d t a l y o r s

f o r m u l a

Commented by jagoll last updated on 03/Apr/20

f (0) = 1

f^{'} (x) = - {(1 + x)}^{- 2} \Rightarrow f^{'} (0) = - 1

f^{″} (x) = 2 {(1 + x)}^{- 3} \Rightarrow f^{″} (0) = 2

f^{‴} (x) = - 6 {(1 + x)}^{- 4} \Rightarrow f^{(3)} (0) = - 6

f^{(4)} (x) = 24 {(1 + x)}^{- 4} \Rightarrow f^{(4)} (0) = 24

f (x) = 1 - x + x^{2} - x^{3} + x^{4} + \dots

= \underset{n = 1}{\sum^{\infty}} {(- x)}^{n - 1}

Commented by jagoll last updated on 03/Apr/20

x \neq - 1

Commented by redmiiuser last updated on 03/Apr/20

o k m i s t e r i w a n t t o a s k

a r e t h e r e a n y c o n d i t i o n s

f o r t h e a b o v e e x p a n s i o n .

Commented by jagoll last updated on 03/Apr/20

i think no sir

Commented by redmiiuser last updated on 03/Apr/20

A r e y o u 100 % s u r e .

Commented by Ar Brandon last updated on 03/Apr/20

∣ x ∣< 1

Commented by redmiiuser last updated on 03/Apr/20

b u t w h y n o t ∣ x ∣> 1

Commented by redmiiuser last updated on 03/Apr/20

m r j a g o l l w h y \underset{n = 1}{\sum^{n}} * *

Commented by jagoll last updated on 03/Apr/20

o yes it typo

Commented by redmiiuser last updated on 03/Apr/20

t h e n w h a t a r e t h e l i m i t s

o f s u m m a t i o n .

Commented by redmiiuser last updated on 03/Apr/20

c a n a n y o n e c o m m e n t

Commented by Joel578 last updated on 03/Apr/20

\frac{1}{1 + x} = 1 - x + x^{2} - x^{3} + \dots

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{n}

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{n} will converge if ∣ x ∣< 1, otherwise diverge

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