Unutterably Silly Quiz

That series, the harmonic series, diverges. Can't put the symbols to make this simple but the area under the curve of the function 1/x from x=1 to x=infinite, or the integral from x=1 to x = infinite of 1/x dx is less than the sum of the rectangles of height 1/n and width 1 from n=1 to n=infinite. The mentioned integral diverges and therefore so does the series.

Oh and your first series is that which arises from Xeno's Achilles and the tortoise 'paradox.' It took the development of the concept of limits to reveal that there was no paradox.

[pdurrant]

Originally Posted by Hamlet53

That series, the harmonic series, diverges. Can't put the symbols to make this simple but the area under the curve of the function 1/x from x=1 to x=infinite, or the integral from x=1 to x = infinite of 1/x dx is less than the sum of the rectangles of height 1/n and width 1 from n=1 to n=infinite. The mentioned integral diverges and therefore so does the series.

You're right that it diverges, but your proof isn't very elegant.

Here's the elegant proof.

Divide the series into groups as follows:

1 + (1/2 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/14 + 1/14 + 1/15 + 1/16) + ....

Replace each term in each group with the smallest term in the group. Note that this reduces the sum of the series.

1 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + .....

Note that each and every group sums to 1/2

1 + (1/2) + (1/2) + (1/2) + .....

Which clearly diverges to infinity. Our original sequence has a larger sum than this, and so also diverges to infinity.

Q.E.D.

Hamlet53 gets to set the next question.

[Bilbo1967]

Originally Posted by pdurrant

You're right that it diverges, but your proof isn't very elegant.

Here's the elegant proof.

Divide the series into groups as follows:

1 + (1/2 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/14 + 1/14 + 1/15 + 1/16) + ....

Replace each term in each group with the smallest term in the group. Note that this reduces the sum of the series.

1 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + .....
Note that each and every group sums to 1/2

1 + (1/2) + (1/2) + (1/2) + .....

Which clearly diverges to infinity. Our original sequence has a larger sum than this, and so also diverges to infinity.

Q.E.D.

Hamlet53 gets to set the next question.

Sorry, why do you replace with the smallest term in each group? From my decidedly uninformed standpoint it just smacks of cheating.

[Daithi]

It diverges to infinity. My proof, which wouldn't pass for a proof in a math class, is that the denominators contain all the primes up to infinity, and if you have all the primes it's going to infinity.

Originally Posted by pdurrant

You're right that it diverges, but your proof isn't very elegant.

Here's the elegant proof.

Divide the series into groups as follows:

1 + (1/2 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/14 + 1/14 + 1/15 + 1/16) + ....

Replace each term in each group with the smallest term in the group. Note that this reduces the sum of the series.

1 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + .....

Note that each and every group sums to 1/2

1 + (1/2) + (1/2) + (1/2) + .....

Which clearly diverges to infinity. Our original sequence has a larger sum than this, and so also diverges to infinity.

Q.E.D.

Hamlet53 gets to set the next question.

I take exception to your not elegant characterization. Perhaps if I could have put in the standard symbols with a graph to illustrate? But mine is just the standard integral test for infinite series.

Originally Posted by Bilbo1967

Sorry, why do you replace with the smallest term in each group? From my decidedly uninformed standpoint it just smacks of cheating.

Actually I think there is a mistake in Pdurrant's proof.


1+1/2+[1/3 +1/4] +[1/5+1/6+1/7+1/8] + . . . > 1 + 1/2 + (1/4 +1/4) + (1/8+1/8+1/8+1/8) + . . . = 1 +1/2 +1/2+1/2+ . .. diverges.

The point is that each group in the [] in the original series is larger than the group in the () in the second series; 1/3+1/4 > 1/4 +1/4.

So with my mind on mathematics and trying to match the speed this quiz has been proceeding at . ..

Leonardo of Pisa is generally given credit with spreading the Hindu-Arabic number system (0,1,2,3,4,5,6,7,8,9,10,11 . . . 100, 101, . . .) to Europe. The introduction of these numerals and the numeral 0 to serve as a place holder for powers of 10 was a major improvement over the old Roman (I,II,III,IV,V,VII,VIII,IX,X,XI, . . . C, CI, . ...) system.

So quick and simple two part question. By what name is Leonardo of Pisa more commonly known by today? What in mathematics is his name commonly associated with?

[Bilbo1967]

Originally Posted by Hamlet53

So with my mind on mathematics and trying to match the speed this quiz has been proceeding at . ..

Leonardo of Pisa is generally given credit with spreading the Hindu-Arabic number system (0,1,2,3,4,5,6,7,8,9,10,11 . . . 100, 101, . . .) to Europe. The introduction of these numerals and the numeral 0 to serve as a place holder for powers of 10 was a major improvement over the old Roman (I,II,III,IV,V,VII,VIII,IX,X,XI, . . . C, CI, . ...) system.

So quick and simple two part question. By what name is Leonardo of Pisa more commonly known by today? What in mathematics is his name commonly associated with?

Oooh, I think I know this one.

Fibonacci. Most famous for his eponymous sequence, which is, without Googling; 1,1,2,3,5,8,13,21 etc...?

[Stitchawl]

Originally Posted by Hamlet53

So quick and simple two part question. By what name is Leonardo of Pisa more commonly known by today? What in mathematics is his name commonly associated with?

I know! I know! Leonardo Di Caprio!! And the numbers are associated with the number of girlfriends he's amassed! Unless Bilbo is right...



Stitchawl

Originally Posted by Bilbo1967

Oooh, I think I know this one.

Fibonacci. Most famous for his eponymous sequence, which is, without Googling; 1,1,2,3,5,8,13,21 etc...?

Correct. A very easy sequence to remember as well since from the third element in the sequence on the element is just the sum of the two previous elements. I bet you also know that as N increases without bound the ratio of the Nth term to the (N-1)th term becomes the Golden Ratio = 1.6180339887. . .. The sequence and that ratio appear in amazing places in nature, entire books have been written on the subject.

The next quiz is yours. Maybe not math this time?

Originally Posted by Stitchawl

I know! I know! Leonardo Di Caprio!! And the numbers are associated with the number of girlfriends he's amassed! Unless Bilbo is right...


Stitchawl

No, but that is clever!

[SneakySnake]

my brain hurts!

[pdurrant]

Originally Posted by Hamlet53

Actually I think there is a mistake in Pdurrant's proof.


1+1/2+[1/3 +1/4] +[1/5+1/6+1/7+1/8] + . . . > 1 + 1/2 + (1/4 +1/4) + (1/8+1/8+1/8+1/8) + . . . = 1 +1/2 +1/2+1/2+ . .. diverges.

The point is that each group in the [] in the original series is larger than the group in the () in the second series; 1/3+1/4 > 1/4 +1/4.

That's rather the point. We've shown that an infinite series of terms, each of which is the same or smaller than the term in our original series, diverges to infinity. Since our original series must have a larger sum than the series that diverges to infinity, it also diverges to infinity.

[Stitchawl]

Numbers are for people who can't deal with reality...

Give us a question about something real... like parapsychology or theology!


Stitchawl

[Billi]

I knew you would love the last quiz questions!!!

[Stitchawl]

Originally Posted by Billi

I knew you would love the last quiz questions!!!

The very notion that folks can actually hold so many numbers in their minds at the same time without blowing a fuse completely boggles my brain! I realize that I suffer from a numbers learning disorder (Dyscalculia) so I don't mind so much that I can't, but it just SOOOOO amazes me every time I witness real math people at play! I'd be satisfied to just be able to dial a telephone number correctly on the first try!


Stitchawl