Summing the Matrix
Alice and Bob saw this matrix with infinite rows and columns:
⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ 1 − 1 0 0 0 ⋮ 0 1 − 1 0 0 ⋮ 0 0 1 − 1 0 ⋮ 0 0 0 1 − 1 ⋮ 0 0 0 0 1 ⋮ ⋯ ⋯ ⋯ ⋯ ⋯ ⋱ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ .
They wanted to sum up all the elements in it.
Alice:
"The sum of each column is 0, so the sum of all the numbers must be 0."
Bob:
"The sum of all the rows except the first one is 0. The first row adds up to 1. Hence, the sum of all the numbers is 1."
Who is correct?
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7 solutions
Moderator note:
In order to handle Grandi's Series as a "sum", a more exact idea of what it means to sum an infinite series is necessary. In both commonly accepted definitions (Cesàro's and Abel's) the sum turns out to be 2 1 .
The Cesàro sum, for instance, is the average of all of an infinite series's partial sums. The sequence of partial sums is 1 , 1 − 1 = 0 , 1 − 1 + 1 = 1 , 1 − 1 + 1 − 1 = 0 , 1 , 0 , 1 , 0 , …
( 1 ) / 1 ( 1 + 0 ) / 2 ( 1 + 0 + 1 ) / 3 ( 1 + 0 + 1 + 0 ) / 4 ( 1 + 0 + 1 + 0 + 1 ) / 5 ( 1 + 0 + 1 + 0 + 1 + 0 ) / 6 = 1 = 2 1 = 3 2 = 2 1 = 5 3 = 2 1
The terms alternate between 2 1 and a fractional series 3 2 , 5 3 , 7 4 , … that converges to 2 1 , so the Cesàro sum is 2 1 .
Question should state that the matrix is infinite in size. For a finite size matrix, Bob is correct.
So why is the correct answer "None of them"?
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Technically they're both right in a way, since the sum would just continue to oscillate between 1 and 0. However, since the value never stops or converges to a single number (If it did, it would be considered 1/2), we consider it divergent and therefore making it indecisive.
I am not sure, but maybe because the series is divergent. Both are trying to find out the sum, but the sum cannot be found. (P.S- I am very new to the concept of infinite series, so forgive me if I said something foolish)
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Yes; the series is considered divergent; as a particular value cannot be determined. Grandi himself thought the answer, if it converged, would be 1/2.
This problem shows why terms have to be well defined. The sum of a series is defined to be the limit of its partial sums. But what are the partial sums here? How is this series indexed? It's not made clear. Alice and Bob both translate this problem into a double summation problem, but doing so changes the nature of the question dramatically. If the terms of the series are not converging in magnitude to zero, then there's no hope of the partial sums converging to anything.
The array above isn't even Grandi's sum, not strictly speaking. For any integer I can choose an ordering of the indices to reach said integer as a partial sum, and never get further than 1 away from it in any subsequent partial sums. This is what happens when you have an infinite set with no prescribed order given to it.
If the matrix is nxn than Bob is correct? Are we supposed to assume the matrix is not square (nxm)??
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If the matrix is of any finite size, Bob is correct. But we are assuming that the matrix is infinitely long and wide.
When I saw the problem, I assumed the matrix was intended to be finite and square. If it is finite but not necessarily square, the correct answer is "It depends on the dimensions of the matrix." And the phrasing "they saw the matrix and wanted to sum up all the elements in it" is not suggestive of an infinite object.
If we consider the partial sum S[k] to be the value for a k by k matrix, then S[k] = 1 for every k, so I don't see how 0 is an accumulation point.
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So it's important to show exactly which limit we are taking. To get an answer of "neither", we need to be considering the partial sum as S[j,k] the value for a j by k matrix. In that case, non-square matrices are in the series too, and then 0 is indeed an accumulation point. The problem needs to be rephrased to clarify this.
Bob is right. with any finite number of rows / columns (n), S=0. Hence when n --> infinite, then S remains = 0. Conterargument for Alice is this: The sum of last column is 1, hence S=0
To say that the Cesaro or the Abel "sum" of the infinite series 1 -1 +1 -1 ... equals one half is to assign a value to the series. This value is called a Cesaro or an Abel sum. Neither of these gives the classical result for this problem. Normally one says that the sum simply does not exist because the series does not converge.
The word sum can have different interpretations under different circumstances. Usually one accepts that the sum of two numbers is the total when they are added. Thus a sum is most fundamentally defined when two numbers are being added.
It is easy to generalize this principle to say that a sum of finitely many numbers can be calculated, grouping them such that only two values are being added at a time, but continuing the process until all numbers have been totaled.
When infinitely many numbers are being added, a careful reading of mathematics results in the introduction of the term "infinite series." When dealing with infinite series, a value is assigned to the infinite series if the sequence of partial sums converges to a value. Sometimes this assigned value is referred to as the "sum". But the more precise interpretation is to say that a number is assigned to the infinite series. In this way mathematicians acknowledge that while a "sum" in the classical sense does not make sense, the infinite series has a number associated with it that behaves much like a sum behaves a finite series of numbers is totaled.
Similarly, the Cesaro and Abel "sums" are values assigned to an infinite series, given a set of circumstances that must be met. They are not sums in the classical sense of the term.
If this process of calling an assigned number of an infinite series its "sum" is continued, results such as 1+2+3+4+...=-1/12 can result, as seen in the numberphile video at https://www.youtube.com/watch?v=w-I6XTVZXww.
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oops: ...with it that behaves much like a sum behaves when a finite series of numbers is totaled.
The sum of row/column elements is an oscillating quantity. Since you are dealing with infinities, you cannot be sure if each column "ends" with( in a crude sense; you can't really 'end' infinity) 1 or -1. Same goes for rows.
Case 1: Considering the above given matrix is Square Let B=The above given matrix. Let A=lower triangle in the above given figure. Then * I + A = B ,where I= Identity Matrix * Now we know that (By Logic) the * Identity Matrix contains the maximum number of 1's as it is the longest diagonal of the matrix. * Now consider Matrix A, This matrix contains exactly ( n − 1 ) 1's multiplied by (-1) Thus the sum of all numbers will be equal to 1 .
( Case 2: Considering the above matrix of R C, where * R<C . * Let B=The above given matrix. Let A=lower triangle in the above given figure. Then * Z + A = B ,where Z=Central triangle with diagonal elements as "1" . Now, By trial and error we can establish the relation that the Sum of all integers(i.e. Sum of all elements of Z + Sum of all elements of A) is controlled by the number of columns in the given problem. Relation is: If Number of columns > or = Number of Rows, then sum will always be 1. Otherwise Sum will be zero. Thus, **Sum of all elements will be 1.
( Case 3: Considering the above matrix of R C, where * R>C . * Let B=The above given matrix. Let A=lower triangle in the above given figure. Then * Z + A = B ,where Z=Central triangle with diagonal elements as "1" . Thus, by using the deduction from Case 2 above that sum is controlled by no. of columns. Hence, **Sum of all elements will be 0.
Each Bob and Alice is correct conditionally. Hence neither of them are correct according to the options given.
Hmmm! But Bob wanted to start the first row with only a +1. At the end of this matrix you can clearly see the next row will have a -1 only. This negates his first row +1 and so creates a set from +1 to 0. No matter how long it goes on for a plus one will come up then a set of zeros.
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If we go by either way of summing using columns or rows, there is always one extra +1 in either ways which will always give the sum as 1 and not zero.
It is evident from the first row itself and last two columns.
Not sufficient information or there is not definite data pattern, to say who is correct.
What sort of a data pattern would you like?
This is the sum:
S = 1 − 1 + 1 − 1 + 1 − 1 + . . .
One minus the sum:
1 − S = 1 − ( 1 − 1 + 1 − 1 + 1 − 1 + . . . ) = 1 − 1 + 1 − 1 + 1 − 1 + 1 − . . . = S
→ 1 − S = S
→ 1 = 2 S
→ S = 2 1
→ S = 0 , S = 1
Why should S =1/2 be true? Or better yet, why are both Alice and Bob wrong?
When summing up infinite matrices, by definition you take the limit of the sequence s n where s n is the sum of the numbers in an n by n square that starts in the top left corner of the matrix.Neither of them do this, so they are both wrong.
The matrix isn't mentioned to be a square anywhere.
So taking the limit (n to infinity) of a 2n*n matrix is equally valid
Both Alice and Bob considered only a single scenario but there are two possibilities i.e 0 (considering Alice's view when the last row ends up with -1) and 1 (considering Bob's view when the last row ends up with 1). That's why none of them is correct.
Are you saying that both 0 and 1 are possible values of the sum of all these numbers inside the matrix?
The answer- INDTERMINANT - surely
Grandi's Series is known to be divergent, as although it has two accumulation points (1 and 0), it doesn't approach a single fixed value at infinity.
Alice grouped Grandi's series as follows
S = ( 1 − 1 ) + ( 1 − 1 ) + ( 1 − 1 ) + ( 1 − 1 ) + . . . = 0
While this should have been correct, Bob performed the Eilenberg-Mazur Swindle and came up with
S = 1 + ( − 1 + 1 ) + ( − 1 + 1 ) + ( − 1 + 1 ) + . . . = 1
which was clearly from the same series.