Math 152 ‐ Lecture Notes # 1 ‐ David Maslanka
Approximating ln 2 with Riemann Sums
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To approximate the number ln 2, we consider upper and lower Riemann sums. By definition, 2
⌠
⎮ 1
ln 2 = ⎮
dt .
⎮
⎮ t
⌡
1
1
In Maple, we define f ( t ) = . t
> f:=t->1/t;
f := t →
1
t
1
Upon partitioning the interval I = [ 1 , 2 ] into n subintervals, each of length , we may form L ( n ) , the left hand Riemann sum for f over I :
n
> L:=n->sum(f((1+k/n))/n,k=0..n-1);
n−1
L := n →
∑
k=0
⎛
k⎞
f⎜⎜ 1 + ⎟⎟
⎝
n⎠
n
L ( n ) = A1( n ) + A2( n ) + A3 ( n ) + . . . + An ( n ).
Note that L ( 1 ) > L ( 2 ) > L ( 3 ) > . . . > ln 2 since f is decreasing on I .
Now we consider the right hand Riemann sum R ( n ) for f on I , assuming a uniform partition as before.
> R:=n->sum(f((1+k/n))/n,k=1..n);
n
R := n →
∑
⎛
k⎞
f⎜⎜ 1 + ⎟⎟
⎝
n⎠
k=1
n
R ( n ) = a1( n ) + a2 ( n ) + a3( n ) + . . . + an( n ) Note that R ( 1 ) < R ( 2 ) < R ( 3 ) < . . . < ln 2 since f is decreasing on I .
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Claim: The value of ln 2 is 0.69 rounded to the hundredths place.
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To see this, consider the following computation:
> n:=1:
> while evalf((L(n)-R(n))>0.003) do
>
n:=n+1;
> end do:
> n;
167
Now observe that
> L[167]=evalf(L(167));R[167]=evalf(R(167));
L167 = 0.6946464276
R167 = 0.6916524156
and since R ( n ) < ln 2 < L ( n ) for any positive integer n, we are certain that .6916524156 < ln 2 < .6946464276 .
Therefore, we conclude that ln 2 = 0.69 accurately rounded to the hundredths place.