Integrals - Series
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Integrals & Series n Ù0 Sin HxL â x Π
Suppose we wanted to find the area under the graph of Sinn HxL for any value of n. Well, you would start by setting up the above integral. (I preformed a very similar calculation with the function Log@xDn , and arrived at a very similar result.) The problem with this integral is that it's tough to solve for all values of n, so instead I looked only at odd values, for n = 2 k + 1. 2 k+1 @xD â x = à sinHxL Sin2 k @xD â x = à sinHxL I1 - Cos2 @xDM â x à Sin Π
Π
0
0
Π
k
0
And, making the substitution u = Cos@xD , d u = -Sin@xD d x, with uH0L = 1, uHΠL = 0, we obtain 2 à I1 - u2 M â u 1
k
0
Which, using the binomial formula and evaluating the integral (it's a simple polynomial), can be expanded and nested into the summation: 2â k
i=0
H-1Li K
k O i
2i+1
Which is very nice. Now, there is something useful we can do with this formula: we know that as n ® ¥, the integral should go to zero. That's a result of the fact that the sine function has the range H0, 1L on the interval H0, ΠL, and raising a number between zero and one to a large power makes it go to zero; thus, as n gets large most of the points on the graph will be very close to zero. Below are a few graphs to illustrate this:
Out[22]=
2
Integrals - Series.nb
1.0
0.8
0.6
Out[22]=
:
,
0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
1.0
0.8
0.6
, 0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
,
Integrals - Series.nb
1.0
0.8
0.6
, 0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
,
3
4
Integrals - Series.nb
1.0
0.8
0.6
, 0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
1.0
0.8
0.6
, 0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
1.0
0.8
0.6
>
0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
So, we have the final result that Lim 2 â k
k®¥
i=0
H-1Li K
k O i
2i+1
=0
The sum is actually a cool one. Below are the first few summations for small k and their numerical values. The series converges really slowly because, as you can see from the graph, Sin2 k+1 @xD still keeps a not insignificant area as k gets large. In the second table, notice that the numerators are the k th row of Pascals Triangle while the denominators are simply increasing odd numbers. (With an alternator stuck in there.)
Integrals - Series.nb
The sum is actually a cool one. Below are the first few summations for small k and their numerical values. The series converges really slowly because, as you can see from the graph, Sin2 k+1 @xD still keeps a not insignificant area as k gets large. In the second table, notice that the numerators are the k th row of Pascals Triangle while the denominators are simply increasing odd numbers. (With an alternator stuck in there.) Out[36]=
4 16 :82, 2.0000<, : , 1.3333>, : , 1.0667>, 3 15 32 256 512 2048 : , 0.91429>, : , 0.81270>, : , 0.73882>, : , 0.68198>> 35 315 693 3003 :2 H1L, 2 1 5 2 1-
1
2 , 2 1-
3
3
10 +
3
10 -
5
1 , 2 15
5 +
7
3
+
3
1 -
9
3 + 5 6
7
+ 3
4 , 2 1-
15
, 2 111
1 -
3 20
5
6 + 5
15 +
7
1 +
7
6 -
9
4 1 +
11
, 9 ...>
13
n x Ù-¥ x ã â x 0
This one is a little quicker and clearner, though it still relies on n being an integer. After doing the calculation I noticed that this integral is quite similar (almost identical) to the one that defines the gamma function; I wonder if the calculation I end up with is at all useful, or if it has already been done. (It has probably already been done.) We proceed with an integration by parts; multiple integration by parts. But the pattern appears quickly. à
0
-¥
xn ã x â x = Bxn ã x - à n xn-1 ã x â xF
0 -¥
= Bxn ã x - n xn-1 ã x + à nHn - 1L xn-2 ã x â xF
0 -¥
And you can now probably see the pattern. Aã Ix - n x x
n
n-1
+ nHn - 1L x
n-2
- ... + H-1L
= Bã â n
n
0 n !ME-¥
x
i=0
H-1Li n ! xn-i Hn - iL !
F
0
-¥
And, noticing that for x = -¥ all the terms go to zero, and for x = 0 all the terms are zero except where i = n, we have the final value of the integral: H-1Ln n !
The value of this integral is alternating because if n is even, then xn ® ¥ as x ® -¥ , whereas if n is odd, xn ® -¥ as x ® -¥. Below are a few graphs for increasing integral values of n.
Out[85]=
5
6
Integrals - Series.nb
-14
-12
-10
-8
-6
-4
-2 -0.05 -0.10
Out[85]=
:
-0.15
, -0.20 -0.25 -0.30 -0.35
0.5
0.4
0.3
, 0.2
0.1
-14
-12
-10
-8
-6
-4
-2
-14
-12
-10
-8
-6
-4
-2 -0.2 -0.4 -0.6
,
-0.8 -1.0 -1.2
,
Integrals - Series.nb
4
3
, 2
1
-14
-12
-10
-8
-6
-4
-2
-14
-12
-10
-8
-6
-4
-2
-5
-10
,
-15
-20
100
80
60
40
20
-14
-12
-10
-8
-6
-4
-2
>
7
Suppose we wanted to find the area under the graph of Sinn HxL for any value of n. Well, you would start by setting up the above integral. (I preformed a very similar calculation with the function Log@xDn , and arrived at a very similar result.) The problem with this integral is that it's tough to solve for all values of n, so instead I looked only at odd values, for n = 2 k + 1. 2 k+1 @xD â x = à sinHxL Sin2 k @xD â x = à sinHxL I1 - Cos2 @xDM â x à Sin Π
Π
0
0
Π
k
0
And, making the substitution u = Cos@xD , d u = -Sin@xD d x, with uH0L = 1, uHΠL = 0, we obtain 2 à I1 - u2 M â u 1
k
0
Which, using the binomial formula and evaluating the integral (it's a simple polynomial), can be expanded and nested into the summation: 2â k
i=0
H-1Li K
k O i
2i+1
Which is very nice. Now, there is something useful we can do with this formula: we know that as n ® ¥, the integral should go to zero. That's a result of the fact that the sine function has the range H0, 1L on the interval H0, ΠL, and raising a number between zero and one to a large power makes it go to zero; thus, as n gets large most of the points on the graph will be very close to zero. Below are a few graphs to illustrate this:
Out[22]=
2
Integrals - Series.nb
1.0
0.8
0.6
Out[22]=
:
,
0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
1.0
0.8
0.6
, 0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
,
Integrals - Series.nb
1.0
0.8
0.6
, 0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
,
3
4
Integrals - Series.nb
1.0
0.8
0.6
, 0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
1.0
0.8
0.6
, 0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
1.0
0.8
0.6
>
0.4
0.2
0.0
0.5
1.0
1.5
2.0
2.5
3.0
So, we have the final result that Lim 2 â k
k®¥
i=0
H-1Li K
k O i
2i+1
=0
The sum is actually a cool one. Below are the first few summations for small k and their numerical values. The series converges really slowly because, as you can see from the graph, Sin2 k+1 @xD still keeps a not insignificant area as k gets large. In the second table, notice that the numerators are the k th row of Pascals Triangle while the denominators are simply increasing odd numbers. (With an alternator stuck in there.)
Integrals - Series.nb
The sum is actually a cool one. Below are the first few summations for small k and their numerical values. The series converges really slowly because, as you can see from the graph, Sin2 k+1 @xD still keeps a not insignificant area as k gets large. In the second table, notice that the numerators are the k th row of Pascals Triangle while the denominators are simply increasing odd numbers. (With an alternator stuck in there.) Out[36]=
4 16 :82, 2.0000<, : , 1.3333>, : , 1.0667>, 3 15 32 256 512 2048 : , 0.91429>, : , 0.81270>, : , 0.73882>, : , 0.68198>> 35 315 693 3003 :2 H1L, 2 1 5 2 1-
1
2 , 2 1-
3
3
10 +
3
10 -
5
1 , 2 15
5 +
7
3
+
3
1 -
9
3 + 5 6
7
+ 3
4 , 2 1-
15
, 2 111
1 -
3 20
5
6 + 5
15 +
7
1 +
7
6 -
9
4 1 +
11
, 9 ...>
13
n x Ù-¥ x ã â x 0
This one is a little quicker and clearner, though it still relies on n being an integer. After doing the calculation I noticed that this integral is quite similar (almost identical) to the one that defines the gamma function; I wonder if the calculation I end up with is at all useful, or if it has already been done. (It has probably already been done.) We proceed with an integration by parts; multiple integration by parts. But the pattern appears quickly. à
0
-¥
xn ã x â x = Bxn ã x - à n xn-1 ã x â xF
0 -¥
= Bxn ã x - n xn-1 ã x + à nHn - 1L xn-2 ã x â xF
0 -¥
And you can now probably see the pattern. Aã Ix - n x x
n
n-1
+ nHn - 1L x
n-2
- ... + H-1L
= Bã â n
n
0 n !ME-¥
x
i=0
H-1Li n ! xn-i Hn - iL !
F
0
-¥
And, noticing that for x = -¥ all the terms go to zero, and for x = 0 all the terms are zero except where i = n, we have the final value of the integral: H-1Ln n !
The value of this integral is alternating because if n is even, then xn ® ¥ as x ® -¥ , whereas if n is odd, xn ® -¥ as x ® -¥. Below are a few graphs for increasing integral values of n.
Out[85]=
5
6
Integrals - Series.nb
-14
-12
-10
-8
-6
-4
-2 -0.05 -0.10
Out[85]=
:
-0.15
, -0.20 -0.25 -0.30 -0.35
0.5
0.4
0.3
, 0.2
0.1
-14
-12
-10
-8
-6
-4
-2
-14
-12
-10
-8
-6
-4
-2 -0.2 -0.4 -0.6
,
-0.8 -1.0 -1.2
,
Integrals - Series.nb
4
3
, 2
1
-14
-12
-10
-8
-6
-4
-2
-14
-12
-10
-8
-6
-4
-2
-5
-10
,
-15
-20
100
80
60
40
20
-14
-12
-10
-8
-6
-4
-2
>
7
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