If `|x|lt1` and `y=x-(x^(2))/(2)+(x^(3))/(3)-(x^(4))/(4)+...`, then x =

Options

A. `y+(y^(2))/(2!)+(y^(3))/(3!)+…oo`
B. `y-(y^(2))/(2)+(y^(3))/(3)-….oo`
C. `y-(y^(2))/(2!)+(y^(3))/(3!)+…oo`
D. `y-(y^(2))/(2!)+(y^(3))/(3!)- ….. oo`

Answer

To solve the problem, we start with the given series for \( y \):

\[ y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

This series is an infinite series that can be recognized as the Taylor series expansion for \( \log(1+x) \) when \( |x| < 1 \). Therefore, we can write:

\[ y = \log(1+x) \]

### Step 1: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides:

\[ e^y = 1 + x \]

### Step 2: Solve for \( x \)
Now, we can rearrange this equation to solve for \( x \):

\[ x = e^y - 1 \]

### Step 3: Expand \( e^y \)
Next, we can use the Taylor series expansion for \( e^y \):

\[ e^y = 1 + \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots \]

### Step 4: Substitute into the equation for \( x \)
Substituting this expansion back into our equation for \( x \):

\[ x = \left(1 + \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots\right) - 1 \]

### Step 5: Simplify
The \( 1 \) cancels out, so we have:

\[ x = \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots \]

This is the series expansion for \( e^y - 1 \), which is valid for \( |x| < 1 \).

### Final Result
Thus, we have derived that:

\[ x = y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots \]

This means that \( x \) can be expressed in terms of \( y \) as the sum of the series.