Question

Determine whether the statement is true or false. If f(x) = 2x - x^2 + 1/3x^3 - .. converges for all x, then f''' (0) = 2.

a. True
b. False

Answer 1

The statement that
`\( f'''(0) = 2 \)` is true given that the series converges for all
`\( x \)` and the pattern of the derivatives continues as implied.

To determine whether the statement is true, we need to find the third derivative of
`\( f(x) \)` and evaluate it at
`\( x = 0 \)`. The given function is:

`\[ f(x) = 2x - x^2 + (1)/(3)x^3 - \ldots \]`

We assume the series continues in such a way that it converges for all
`\( x \)`, implying it's a power series. The power series expansion of \( f(x) \) about
`\( x = 0 \)` is:

`\[ f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots \]`

Where the coefficients
`\( a_n \)` are given by:

`\[ a_n = (f^((n))(0))/(n!) \]`

The third derivative of
`\( f(x) \)`, evaluated at
`\( x = 0 \)`, would be the coefficient
`\( a_3 \)` times
`\( 3! \)` (factorial of 3):

`\[ f'''(0) = a_3 \cdot 3! \]`

From the given function, we can see that the coefficient of
`\( x^3 \)` is
`\( (1)/(3) \)`. Therefore,
`\( a_3 = (1)/(3) \)` because for the term
`\( a_3x^3 \)` in the power series, \( a_3 \) must be the coefficient of
`\( x^3 \)`.

So,

`\[ f'''(0) = a_3 \cdot 3! \]`

`\[ f'''(0) = (1)/(3) \cdot 3! \]`

`\[ f'''(0) = (1)/(3) \cdot 3 \cdot 2 \cdot 1 \]`

`\[ f'''(0) = 2 \]`

Therefore, the statement that
`\( f'''(0) = 2 \)` is true given that the series converges for all
`\( x \)` and the pattern of the derivatives continues as implied.