Question

Find a uniformly convergent sequence of differentiable functions fₙ: (0, 1) -> R

such that the sequence f₁',f₂',f₃',.... does not converge.

Answer 1

Consider
`\(f_n(x) = (x^n)/(n)\)`on (0, 1). It converges uniformly, but
`\(f_n'(x) = x^(n-1)\)`does not converge pointwise.

Consider the sequence of functions defined on the open interval (0, 1):

`\[ f_n(x) = (x^n)/(n) \]`

Each function
`\( f_n(x) \)` is differentiable on (0, 1) since it is a power function. To show uniform convergence, we can use the Weierstrass M-test. Note that
`\( |f_n(x)| \leq (1)/(n) \)` for all
`\( x \in (0, 1) \)` and n in the natural numbers.

Now, let's consider the derivatives
`\( f_n'(x) \)`. The derivative of
`\( f_n(x) \)` with respect to x is given by:

`\[ f_n'(x) = (nx^(n-1))/(n) = x^(n-1) \]`

As n approaches infinity,
`\( f_n'(x) \)` does not converge for \
`( x \in (0, 1) \)` because it becomes unbounded as n increases.

In conclusion, the sequence
`\( f_n(x) = (x^n)/(n) \)`is a uniformly convergent sequence of differentiable functions on (0, 1) since
`\( |f_n(x)| \leq (1)/(n) \) for all \( x \)`in the interval. However, the sequence of derivatives
`\( f_n'(x) = x^(n-1) \)` does not converge for
`\( x \in (0, 1) \)`, illustrating that the convergence of derivatives is not guaranteed even if the original functions converge uniformly.