This lab will be available until December $6$th$,11$:$59$ PM CST

See Canvas for more details. The Taylor series expansion of the natural logarithm function, $ln(x),$ is given by

$ln(x)=(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}-\frac{(x-1{)}^4}{4}+\frac{(x-1{)}^5}{5}$dots

This expansion is valid for values of $x$ where $xxln(x)0.$ Write a program named $approximatin{g}_{\text{I}}n.py$ that takes $as$ input from the user $a$

value for $x$ and a value for tolerance. $If$ the value for $xis$ outside the valid range, continue $to$ take input until the user enters a valid value.

Then, have your program calculate the natural $log$ function using this expansion. Stop the summation when the next term $is$ less than the

tolerance. Finally, have your program print the exact value $ofln(x)$ using the math module, $as$ well $as$ the difference between the values.

Example output using inputs $1\mathrm{.}5$ and $0\mathrm{.}001$:

Enter a value for $x$: $1\mathrm{.}5$

Enter the tolerance: $0\mathrm{.}001$

$ln(1\mathrm{.}5)is$ approximately $0\mathrm{.}40580357142857143$

$ln(1\mathrm{.}5)is$ exactly $0\mathrm{.}4054651081081644$

The difference $is0\mathrm{.}0003384633204070453$

Example output using inputs $0,2$ and $0\mathrm{.}0001$ :

Enter a value for $x$: $0$

Out $of$ range! Try again: $2$

Enter the tolerance: $0\mathrm{.}0001$

$ln(2\mathrm{.}0)is$ approximately $0\mathrm{.}6930971830599583$

$ln(2\mathrm{.}0)is$ exactly $0\mathrm{.}6931471805599453$

The difference $is4\mathrm{.}999749998702008e-05$