Mengoli had posed the following series to be evaluated:
Some great mathematicians, including Liebnitz, John Bernoulli and D’Alembert, failed to compute this infinite series. Euler established himself as the best mathematician of Europe (in fact, one of the greatest mathematicians in history) by evaluating this series initially by a not-so-rigorous method. Later on, he gave alternative and more rigorous ways of getting the same result.
Prove that the series converges and gets an upper limit. Then, try to evaluate the series.
Due Nicolas Oresine:
Consider the following infinite series:
We can re-write the preceding series as follows: , which in turn is less than
. Now, the RHS of this can be re-written as
, which is a geometric series and it is given by
Now, we can say that will converge if .
In order to prove what is asked, we start with
And, then multiply both sides by and then subtract the resulting equation from the preceding equation to get
where all the terms containing the reciprocals of the sth power of even numbers vanished.
Repeating this procedure with gives
where all terms containing the reciprocals of the sth power of multiples of 3 vanished.
By continuing this with all prime numbers, we get
, where p represents all prime numbers. Thus, we get
This is a remarkable result because the LHS is concerned with only positive integers, whereas the RHS is concerned with only primes. This result is known as the “Golden Key of Euler”.
Riemann created his famous function by extending the variable s to the entire complex plane, except with
This function is now very famous as the Riemann zeta function.
How can we apply the Golden Key of Euler to Mengoli’s question that we started with?
Ans. In the Golden Key of Euler, substitute .
Hence, we get the upper limit of the given series is 2.
Euler’s proof (1775):
The proof ran as follows:
It is a little roundabout way of arriving at the correct answer from a known result. Consider McLaurin’s series expansion of sin x:
By dividing both sides by x and then substituting on the right side, we get the following:
By taking a special value of (and, hence ), we get the following:
Note that preceding equation is not a polynomial, but an infinite series. But, Euler still treated it as a polynomial (that is why it was not accepted as a rigorous result) and observed that this “infinite” polynomial has roots equal to . Then, Euler had used the fact that the sum of the reciprocals of the roots is determined by the coefficient of the linear term (here, the y-term) when the constant is made unity. (check this as homework quiz, for a quadratic to be convinced). So, Euler had arrived at the following result:
. With , we get the following:
Another proof also attributed to Euler that uses the series expansion of sin (x) goes as follows below:
has roots given by 0, , , , …So does this polynomial that Euler reportedly constructed:
So, Euler considered the preceding equation to be equivalent to:
Then, he had equated the coefficient of in both to get the result:
Later on, Euler had provided a few more alternate and rigorous proofs of this result.
Reference: Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books.
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Hope you all enjoyed it — learning to think like Euler !! By the way, it did take a long time for even analysis to become so rigorous as it is now….You might like this observation a lot. 🙂 🙂 🙂
If a, b, c are non-negative real numbers such that , then prove that the product abc cannot exceed 1.
Solution will be posted after you attempt it…so that you can compare the two approaches.
In the Feb 23 2018 blog problem, we posed the following question:
Sum the following infinite series:
The sum can be written as:
, where .
Thus, . This is the answer.
If you think deeper, this needs some discussion about rearrangements of infinite series also. For the time, we consider it outside our scope.
Let f be a bijective function from the set to itself. Show that there is a positive integer such that for each . Here, denotes the composite function repeated M times.
The student should be familiar with the following properties of bijective functions:
a) If is a bijective function then there is a unique bijective function such that , the identity function on A. The function g is called the inverse of f and is denoted by . Thus,
c) If f and g are bijections from A to B, then so are and .
d) If f, g, h are bijective functions from A to B and , then .
Apply at left to both sides to obtain .
Coming to the problem, since A has n elements, we see that there are only finitely many (in fact, n!) bijective functions from A to A as each bijective function f gives a permutation of by taking . Since f is a bijective function from A to A, so is each of the functions in the sequence: , , , , \ldots
All these cannot be distinct since there are only finitely many bijective functions from A to A. Hence, for some two distinct positive integers , say, we must have
If , we take $, latex M=m$, to obtain the result. If , multiply both sides by , to get . We take to get the relation when .
Note this means for all .
Take any element r in the set A and consider the sequence of elements
obtained by applying f successively. Since A has only n elements there must be repetitions in the above sequence. But, when the first repetition occurs, this must be r itself, for, if the above sequence looks (for instance) like:
where the first repetition is an element c other than r, this would imply and contradicting the fact that f is a bijection. Thus, for some positive integer , we have .
This is true for each r in the set . By taking M to be the lcm of , we get
for each .
Note: If f itself is the identity function the above proof fails because each . But, in this case, we may take M to be any integer greater than or equal to 2.
Show that there exists a convex hexagon in the plane such that:
(a) all its interior angles are equal, (b) its sides are 1,2,3,4,5,6 in some order.
Let ABCDEF be an equiangular hexagon with side-lengths as 1,2,3,4,5,6 in some order. We may assume WLOG that . Let , , , , .
Since the sum of all the angles of a hexagon is equal to , it follows that each interior angle must be equal to . Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. Use vectors. If the vector is denoted by , we then have
This is because these vectors are inclined to the positive x-axis at angles respectively.
Since the sum of all these 6 vectors is , it follows that ,
and, . That is, …call this I
and ….call this II.
Since , in view of II, we have the following:
Note: the possibility that and in (i), for instance,need not be considered separately, because we can reflect the figure about and interchange these two sets.
Here, . Since , , this is not possible.
Here, =2c-2=2$. This is satisfied by
This is satisfied by .
Hence, we have essentially two different solutions: and . It may be verified that (I) and (II) are both satisfied by these sets of values.
Consider an equilateral triangle of side 9 units. Remove from the three corners equilateral triangles of sides 1 unit, 2 units, and 3 units respectively. The remaining portion is now an equiangular hexagon ABCDEF with sides 1,6,2,4,3,5 as required.
There are ten objects with total weight 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed in the two pans of a balance.
Let denote the weights of the 10 objects in decreasing order. It is given that and that . For each i, , let . (For example, , , etc.Consider the 11 numbers . Note that all these 11 numbers are non-negative and we have and for . Now, look at the remainders when these 11 numbers are divided by 10. We have 10 possible remainders and 11 numbers and hence, by the Pigeon Hole Principle at least some two of these 11 numbers have the same remainder.
For some j, has the remainder 0, that is, is multiple of 10. But, since , the only possibility is that . Thus, we get a balancing by taking the two groups to be and .
Suppose is a multiple of 10. But, then since , this forces , which, in turn, implies that all the weights are equal and equal to 2 as they add up to 20. In this case, taking any five weights in one group and the remaining in the other we again get a balancing.
For some j and k, say , we have that and have the same remainder, that is, is a multiple of 10. But, again since , we should have , that is, , and we get a balancing.
Suppose and for some j () have the same remainder, that is, is a multiple of 10. As in the previous cases, this implies that . That is, . Hence, and balance each other.
Thus, in all cases, the given 10 objects can be split into two groups that balance each other.
At each of the eight corners of a cube write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so written down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?
Let be the numbers written at the corners. Then, the final sum is given by
Because there are fourteen terms in the above sum, and each of the terms is +1 or -1, the sum will be zero only if some seven terms are +1 each and the remaining 7 terms are -1 each.
But, the product of the fourteen terms is .
Therefore, it is impossible to have an odd number of -1’s in the above sum.
We conclude that the desired arrangement is not possible.
Given the 7-element set , find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.
One possible solution collection is as follows:
, , , . , , . Note that there could be other combinations obtained by permitting the letters. WLOG, a can be associated with three pairs b,c,d; e, f, g. Now b can be associated with d, f and e, g. The possible choices left for c are only the pairs e,f and d,g. This arrangement does work.
Hope you all enjoy such type of questions. Please compare your attempts with the solutions given above.
Reference: Problem Primer for the Olympiad, C. R. Pranesachar, B J Venkatchala, C. S. Yogananda, Prism Books.
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If you are interested in knowing more about the RMO and INMO, please refer to:
Let f be an bijective function (one-one and onto) from the set to n itself. Show that there is a positive integer such that for each .
Note: denotes the composite function repeated M times.
Show that there exists a convex hexagon in the plane such that : (a) all its interior angles are equal (b) its sides are 1,2, 3, 4, 5, 6 in some order.
There are ten objects with total weight 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the two pans of a balance.
At each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so written down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that the final sum is zero?
Given the seven element set , find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.
Solutions will be put up tomorrow. Meanwhile, make a whole-hearted attempt to crack these!
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