Question

The McLaurin series expansion for 1/1+x is ?​

Answer 1

Answer:

HI,ANSWER IS HERE

Step-by-step explanation:

Given:

   f(x)=11−x

It seems to me that the easiest way to find the Maclaurin Series is basically to start to write down the multiplier for (1−x) that results in a value of 1...

Write down:

   1=(1−x)(...

The first term of the multiplier will be 1, in order to get 1 when multiplied, so add that to the right hand side:

   1=(1−x)(1...

When −x is multiplied by 1 it will give us −x to cancel out. So the next term on the right hand side is x...

   1=(1−x)(1+x...

When −x is multiplied by x it will give us −x2 to cancel out. So the next term on the right hand side is x2...

   1=(1−x)(1+x+x2...

Continuing in this way, we get:

   1=(1−x)(1+x+x2+x3+x4+...)

So:

   11−x=1+x+x2+x3+x4+...=∞∑k=0xk

Note that if |x|<1 then the remainder gets smaller each time we add a term on the right hand side. Hence the Maclaurin series converges for |x|<1.

Answer 2

Given:

$\frac{1}{1+x}$

To Find:

The McLaurin series of expansion

Solution:

The McLaurin series of expansion states:

∞∑ n=0     $\frac{f^{(n)} (a)}{n!} (x-a)^{n$

• The expression  $\frac{1}{1+x}$ can also be written as $(x+1)^{-1}$.

• According to McLaurin series of expansion the number next to "x" in the bracket should be subtracted from "x".

$(x-a)^{n}$

• Hence, $(x+1)^{-1}$ can be written as $(x-(-1))^{-1}$

Now,

$f(x) = (x-(-1))^{-1} \\ f(0) = (0 - (-1))^{-1} = 1\\\\f'(x) = -(x-(-1))^{-2} \\f'(0) = 1\\\\f''(x) = 2(x-(-1))^{-3} \\f''(0) = 2\\\\f'''(x) = -3.2(x-(-1))^{-4} \\f'''(0) = -6\\\\f''''(x) = -3.2.-4(x-(-1))^{-5} \\f''''(0) = 24$

So we get,

1. the first term = 1
2. the coefficient of x = -1
3. coefficient of $x^{2}$= 2

and so on..

$\frac{1}{1+x} = 1 + \frac{-1x}{1!} + \frac{2x^{2} }{2!}+ \frac{6x^{3} }{3!} + \frac{24x^{4} }{4!} \\\\\frac{1}{1+x} = 1 -x + x^{2} - x^{3} + x^{4}$

Hence we get 1/1+x is :

∞∑ n=0  $(-1)^{n} x^{n}$

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