# How do you evaluate #ln 5#?

##### 1 Answer

You can use Newton's method to help a little.

#### Explanation:

If you can calculate

#f'(x) = e^x#

Starting with an approximation

#a_(i+1) = a_i - (f(x))/(f'(x))=a_i - (e^x-5)/e^x=a_i - 1 + 5/e^x#

As for

#e^x = sum_(k=0)^oo x^k/(k!)#

I would be interested to know how well this method works using a truncated version of this series for

I may dig a little deeper...

Using

#ln 5 ~~ 1.6094454239#

Using

#ln 5 ~~ 1.6094389965#

Compare those with the proper value:

#ln 5 ~~ 1.6094379124#

If you do want to evaluate:

#sum_(k=0)^10 x^k/(k!)#

then it may be easier to use the form:

#sum_(k=0)^10 x^k/(k!) = 1+x(1+x/2(1+x/3(1+x/4(1+x/5(1+#

#x/6(1+x/7(1+x/8(1+x/9(1+x/10)))))))))#