41 st amatyc annual conference themed session t2b leibnitz’s method and other methods for summing...
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41st AMATYC Annual Conference
Themed Session T2B
Leibnitz’s Method and Other Methods for Summing the Powers of Positive Integers
Dr. Siham AlfredGermanna Community College
November 19, 2015New Orleans, Louisiana
∑ 𝑖❑ ∑ 𝑖2 ∑ 𝑖3 ∑ 𝑖4 ∑ 𝑖5
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Abstract
• Students are introduced to sums of powers of positive integers when finding areas under curves to motivate the introduction of integration.
• They begin to approximate areas under curves by using right-Left hand side rectangles, midpoint or trapezoid methods. All these calculations are done for small values of n.
• When they proceed to calculate the area as a limit of a Riemann Sum, they encounter the formulas for the sums of powers of positive integers.
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Abstract
• Or they might encounter these formulas in a course on Mathematical Induction.
• But where do these formulas come from and how are they computed remains a mystery to the students.
• In this presentation primarily Leibnitz’s Method and some modifications his method for calculating sums of positive integer powers will be explored.
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On Finite Sums of Powers of Positive Integers
OUTLINE
I. The Sum of the First n Positive Integers
II. The Leibnitz Algebraic Method
A. Alternative Algebraic Method
B. Calculus-Based Method
C. Looking at the General Case
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Some Powers of Positive Integers
1n
i
n (1)
1
( 1)
2
n
i
n ni
= 21 1
2 2n n (2)
2
1
( 1)(2 1)
6
n
i
n n ni
= 3 21 1 1
3 2 6n n n (3)
2 2
3
1
( 1)
4
n
i
n ni
= 4 3 21 1 1
4 2 4n n n (4)
2
4 ( 1)(2 1)(3 3 1)
30
n
i
n n n n ni
= 5 4 31 1 1 1
5 2 3 30n n n n (5)
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Mathematicians who computed the sums of powers of positive integers
See Janet Beery’s valuable article in [2].
Beery traces the fascinating and rich history of the computation of the sums of powers of positive integers from Pythagoras to Bernoulli.
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Mathematicians who computed the sums of powers of positive integers
Pythagoras (570-500 B.C.)Greece, Italy
Thomas Harriot (1560-1621) England
Archimedes (287-212 B.C.), Greece, Italy
Johann Faulhaber (1580-1635) Germany
Aryabhata (b. A.D. 478)Northern India
Pierre de Fermat (1601-1665) France
Abu-Bakr Al-Kharaji (d. 1019) Baghdad
Blaise Pascal (1623-1662)France
Ibn Al Haytham (965-1039)Egypt
Jakob Bernoulli (1654-1705) Switzerland
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I. The Sum of the First n Positive Integers
1 1 + 2 = 3 1 + 2 + 3 = 6 1 + 2 + 3 + 4 = 10
1
( 1)
2
n
i
n ni
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The Pythagorean Method of Computing the Sum of n Positive Integers
Twice a triangular number is an oblong number:
Generalize this result to get the sum of the first n positive integers
2(1 2 3 ... ) ( 1)n n n ( 1)
(1 2 3 ... )2
n nn
2(1 2 3 4 5) 5 6
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Leibnitz Experimented with Finite Differences
In [3] on page 31, Leibnitz computes finite differences on the set of squares below 0 1 4 9 16 25 1 3 5 7 9 and notes that 25 - 0 = 1 + 3 + 5 + 7 + 9 He notes that this sum: first term minus the last term of the original sequence is the sum of the first differences no matter where one starts.
For example: 25 – 1 = 24 = 3+5+7+9
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Leibnitz Experimented with Finite Differences Then he proceeds to study 2nd, third and fourth finite differences on the set below which he called combinatory numbers. In this array a preceding series, whether horizontal or vertical, is the first difference of the series immediately following it. In addition, each series either horizontal or vertical contains the sum of the series immediately preceding it
1 1 1 1 1 1
1 2 3 4 5 6
1 3 6 10 15 21
1 4 10 20 35 56
1 5 15 35 70 126
1 6 21 56 126 252
1 7 28 84 210 462
In this array a preceding series whether horizontal or vertical is the first difference of the series immediately following it.
Each series either horizontal or vertical contains the sum of the series immediately preceding it.
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He argued in [3] page 51 that if x is the general term of a series of natural numbers,
then the general term of the series of squares is 2x and the next square in the series would
be 2( 1)x . Then the general term for the finite difference between the two successive
squares is 2 2( 1) 2 1x x x , which an odd number.
Similarly the general term for the finite difference of successive cubes is
3 3 2( 1) 3 3 1x x x x .
Then Leibnitz applied the same approach he used on the two successive terms of squares and cubes and considered computing the difference between two successive terms of the entire series as will be shown next.
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II. The Leibnitz Algebraic Method
2
1
( 1)(2 1)
6
n
i
n n ni
3 22
1
( 1)(2 1)
6 3 2 6
n
i
n n n n n ni
The sum of the first n integer squares =
As a polynomial, it will be shown that the above sum can also be written as
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II. The Leibnitz Algebraic Method
Leibnitz made the assumption that the sum of squares of the first n integers is a polynomial of degree n+1. So by assumption the sum of the first n squares is a cubic polynomial
Since S(0) = 0 it follows that d = 0
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II. The Leibnitz Algebraic Method
It is clear that 2 2( 1) ( ) ( 1) 2 1S x S x x x x
The left hand side can be written as:
3 3 2 2 2[( 1) ] [( 1) ] [( 1) ] 2 1a x x b x x c x x d d x x
𝑆 (𝑥+1)=02+12+22+32 ...+𝑥2+ (𝑥+1 )2Consider
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II. The Leibnitz Algebraic Method
Expanding the Left hand side and collecting like terms gives
2 23 (3 2 ) 2 1ax a b x a b c x x
1 1 1,3 2 6a b and c
Equate coefficients of like terms
to get
3 2 ( 1)(2 1)1 1 1( ) 3 2 6 6
x x xS x x x x
as required to prove
3 1, 3 2 2 1a a b and a b c
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For the sum of thefirst n integers
Use S(x) = ax2 + bx + cS(0) = 0, so c = 0
S(x+1) – S(x) = (x+1) = a[(x+1)2 – x2 ] +b[(x+1) – x] + c –c = 2ax + a + b = x + 1, a = ½ , b = ½
For the sum of thefirst n cubes
UseS(x) = ax4 + bx3 + cx2 + dxS(0) = 0, so e = 0
S(x+1) – S(x) = (x+1)3
= a[(x+1)4– x4 ] +b[(x+1)3 – x3]
+ c[(x+1)2 – x2 ] + d[(x+1) – x] = (x+1)3
= 4ax3 + (6a+3b)x2 + (4a+3b+2c)x +a+b+c = (x+1)3 = x3 + 3x2 + 3x + 1
4a =1, 6a + 3b = 3, 4a + 3b + 2c = 3a = ¼ , b = ½ , c = ¼ , d = 0
For the sum
for a positive integer k
Use S(x), a polynomialof degree k+1
S(x+1) – S(x) = (x+1)k
1
( 1)
2
n
i
n ni
2 23
1
( 1)
4
n
i
n nfor i
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A.Alternative Algebraic Method
3 3 2 2 2[( 1) ] [( 1) ] [( 1) ] 2 1a x x b x x c x x d d x x
Since we have three unknowns we need three equations to solve the system. When x = 0 the above equation becomes 1a b c
7 3 4a b c
19 5 9a b c
Note: Leibnitz Method requires expanding the binomial powers of (x+1). An alternative method would be equating two polynomials which requires equating coefficients of like terms.
When x = 1 the above equation becomes
When x = 2 the above equation becomes
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A.Alternative Algebraic Method
When solving the system the same solution is obtained
1 1 1,3 2 6a b and c
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B.Calculus Based Method
3 3 2 2 2[( 1) ] [( 1) ] [( 1) ] 2 1a x x b x x c x x d d x x 2 2[3( 1) 3 ] [2( 1) 2 ] 2 2a x x b x x x
(1)(2)
(2) is an equation in a and b, the c term vanishes. Note that the term in (2)
[2( 1) 2 ] 2 2 2 2b x x bx b bx b
By the time the students get to Riemann sums and integration they have already covered the derivative, they can differentiate both sides of the equation in Example 1
So equation (2) becomes 2 2[3( 1) 3 ] 2 2 2a x x b x
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B.Calculus Based Method
Differentiate equation (2) again will eliminate the constant term 2b to get
(6 1) 6 ) 2a x x
When x = 0 in equations (1), (2) and (3) the system can be easily solved by back substitution
(3)
1
3 2 2
1 1 16 2, , ,3 2 6
a b c
a b
a So a b and c
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C.Proving the General Case
To prove the general case that the sum of positive integer powers of k
( ) 0 1 2 3 ...k k k k kS x x 1 1
1 1 1...k k kk k k oa x a x a x a x a , is a polynomial of
degree k+1 with 1
1
1kak
and 0 0a , assume that
( ) 0 1 2 3 ....k k k k kS x x = 1 21 2 1( ) ...m m m
m m m oP x a x a x a x a x a an arbitrary
thm degree polynomial. It is required to show that 1m k , the leading coefficient of ( )P x
1
1mak
and the constant term 0 0a .
Since (0) 0S , it follows that 0 0a . 1 1
1 1( 1) ( ) ( 1) [( 1) ] [( 1) ] ... [( 1) ]k m m m mm mS x S x x a x x a x x a x x
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C.Proving the General Case
It follows that the leading term of ( 1)kx is kx = 1( ) mmm a x the leading term of the polynomial on the
left. Therefore 1k m and 1m k as required to prove.
Moreover, 1 ( ) mm a and 1 1
1mam k
namely that the leading coefficient of the sought after
polynomial of degree k+1 is the reciprocal of the degree of the polynomial as required to prove.
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Leibnitz’s Method and Other Methods of Summing Up the Powers of Integers
REFERENCES
1. J.A, Suzuki, “Introducing the Sums of Powers”, The College Mathematics Journal, Vol. 35, No. 4, September 2004, pp. 303-305
2. J. Beery http://www.maa.org/press/periodicals/convergence/sums-of-powers-of-positive-integers
3. G.W. Leibnitz (Translated by J.M. Child), The Early Mathematical Manuscript of Leibnitz, Open Court, 1920, pp. 31-33, 51