Natural Logs
Key Questions
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The Answer is 1.
# ln(e)# is the same thing as#log_e(e)# Because
#e^1 = e, log_e(e) = 1# 
Kevin B.
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1
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Mar 24 2015
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Answer:
You can approximate
#ln x# by approximating#int_1^x 1/t dt# using Riemann sums with the trapezoidal rule or better with Simpson's rule.Explanation:
For example, to approximate
#ln(7)# , split the interval#[1, 7]# into a number of strips of equal width, and sum the areas of the trapezoids with vertices:#(x_n, 0)# ,#(x_n, 1/x_n)# ,#(x_(n+1), 0)# ,#(x_(n+1), 1/(x_(n+1)))# If we use strips of width
#1# , then we get six trapezoids with average heights:#3/4# ,#5/12# ,#7/24# ,#9/40# ,#11/60# ,#13/84# .If you add these, you find:
#3/4+5/12+7/24+9/40+11/60+13/84 ~~ 2.02# If we use strips of width
#1/2# , then we get twelve trapezoids with average heights:#5/6# ,#7/12# ,#9/20# ,...,#25/156# ,#27/182# Then the total area is:
#1/2 xx (5/6+7/12+9/20+...+25/156+27/182) ~~ 1.97# Actually
#ln(7) ~~ 1.94591# , so these are not particularly accurate approximations.Simpson's rule approximates the area under a curve using a quadratic approximation. For a given
#h > 0# , the area under the curve of#f(x)# between#x_0# and#x_0 + 2h# is given by:#h/3(f(x_0) + 4f(x_0+h) + f(x_0+2h))# Try improving the approximation using Simpson's rule, using
#h = 1/2# then we get six approximate areas to sum:#h/3(25/6+73/30+145/84+241/180+361/330+505/546)~~1.947# That's somewhat better.
George C.
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3
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Nov 4 2015
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#ln(x)# is the same as saying#log_e(x)# .#e# is the base of the natural log.
Kevin B.
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1
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Apr 5 2015