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Assignment # 2 & 3
Finite Element Methods
Solution of Assigned Problems for chapters 2 and 3
Problem # 1.
We have to show that
( ) l ( u ) S S S
Here,
( ) l ^ S and
( u ) ^ S denote the circumference of the inscribed and circumscribed polygons
respectively. S is the circumference of the circle.
Solution
We know that,
( ) 2 sin( / )...................(1)
l
S nR n
S 2 R ....................................(2)
( ) 2 tan( / )....................(3)
u
S nR n
Dividing the Equation 1 by Equation 2
( ) 2 sin( / ) sin( / ) .........................(4) 2 /
l S nR n n
S R n
Now the Taylor Series for sin( / n )is
3 5 ( / ) ( / ) sin( / ) .......... 3! 5!
n n n n
Therefore,
3 5 sin( / ) 1 ( / ) ( / ) [ ..........] / / 3! 5!
n n n
n n n
2 4 sin( / ) ( / ) ( / ) 1 .......... / 3! 5!
n n n
n
It can easily be seen that the right hand side of the final equation is always less than “1”, since terms are
being subtracted from “1”. So from this we conclude that
( ) sin( / ) 1 /
l S n
S n
( )
1
l S
S
( ) ................(5)
l S S
Similarly, now dividing equation 3 by equation 2;
( ) 2 tan( / ) tan( / ) .........................(6) 2 /
u S nR n n
S R n
Again writing the Taylor Series of tan( / n )i.e.
3 5 ( / ) 2( / ) tan( / ) .......... 3 15
n n n n
3 5 tan( / ) 1 ( / ) 2( / ) [ ..........] / / 3 15
n n n
n n n
( ) 2 4 tan( / ) ( / ) 2( / ) 1 .......... / 3 15
u S n n n
S n
Now, from this equation we can see that positive terms are being added to “1”. So we can write above
equation as
( ) ( ) 1 ................(7)
u S (^) u S S S
From Equations 5 & 7 we can easily conclude that
( ) l ( u ) S S S
Hence the Required Result.
