Express the following functions as power series in ascending powers of [tex]\( x \)[/tex] up to the fourth term:

1. [tex]\((1+x)^{-3} = 1 - 3x + 6x^2 - 10x^3 + \ldots\)[/tex]

2. [tex]\((1-2x)^5 = 1 - 10x + 60x^2 - 280x^3 + \ldots\)[/tex]

3. [tex]\(\sqrt{4-x} = 2 - \frac{1}{4}x - \frac{1}{64}x^2 - \frac{1}{512}x^3 + \ldots\)[/tex]



Answer :

Sure, let's express the given functions as power series in ascending powers of [tex]\( x \)[/tex], as far as the fourth term. Here are the detailed expansions for each function:

### 1. Power Series for [tex]\(\sqrt{1+y}\)[/tex]:

The power series expansion for [tex]\(\sqrt{1+y}\)[/tex] is:

[tex]\[ \sqrt{1+y} = 1 + \frac{1}{2} y - \frac{1}{8} y^2 + \frac{1}{16} y^3 + \ldots \][/tex]

### 2. Power Series for [tex]\((1+x)^{-3}\)[/tex]:

The power series expansion for [tex]\((1+x)^{-3}\)[/tex] is:

[tex]\[ (1+x)^{-3} = 1 - 3x + 6x^2 + 10x^3 + \ldots \][/tex]

### 3. Power Series for [tex]\((1-2x)^5\)[/tex]:

The power series expansion for [tex]\((1-2x)^5\)[/tex] is:

[tex]\[ (1-2x)^5 = 1 + 10x + 60x^2 + 280x^3 + \ldots \][/tex]

### 4. Power Series for [tex]\(\sqrt{4-x}\)[/tex]:

The power series expansion for [tex]\(\sqrt{4-x}\)[/tex] is:

[tex]\[ \sqrt{4-x} = 2 - \frac{1}{4}x - \frac{1}{64}x^2 - \frac{1}{512}x^3 + \ldots \][/tex]

These power series expansions provide the necessary expressions up to the fourth term. If there are any specific questions about how to use these expansions or other related queries, feel free to ask!