Show Sum(n=1,N) 2n-1= N^2
nisiprius wrote:Show Sum(n=1,N) 2n-1= N^2
1 + 3 + 5 + 7 + 9 + 11
= a six-by-six square.
QED
Tycoon wrote:If you have access to a library any of Ian Stewart's books are worth reading. The Magical Maze: Seeing the World through Mathematical Eyes might interest you.
gatorman wrote:Tycoon wrote:If you have access to a library any of Ian Stewart's books are worth reading. The Magical Maze: Seeing the World through Mathematical Eyes might interest you.
http://www.amazon.com/The-Magical-Maze- ... tical+maze
rated 4.5 stars on Amazon (3 reviews, 2-5s and 1-3), ~$26 in hardcover, but only $9.99 on Kindle. I was given a Kindle as a Christmas gift, so this may be my first purchase! Thanks1
Abe wrote:gatorman wrote:Tycoon wrote:If you have access to a library any of Ian Stewart's books are worth reading. The Magical Maze: Seeing the World through Mathematical Eyes might interest you.
http://www.amazon.com/The-Magical-Maze- ... tical+maze
rated 4.5 stars on Amazon (3 reviews, 2-5s and 1-3), ~$26 in hardcover, but only $9.99 on Kindle. I was given a Kindle as a Christmas gift, so this may be my first purchase! Thanks1
I just ordered one on Amazon in used like new condition, paperback for .01c, of course shipping was $3.99.
Jacotus wrote:The two examples in the OP are equivalent.
Bungo wrote:Those are nice, useful formulas that are easy to prove, for example using the geometric argument given by nisiprius, or a simple induction argument.
Here are some much deeper results which I find fascinating because of the unexpected pi (i.e., the area of the unit circle, 3.14159...) showing up in the answer:
(2/1)*(2/3)*(4/3)*(4/5)*(6/5)*(6/7)*(8/7)*(8/9)*..... = pi/2
where ... means that the pattern continues ad infinitum.
(1/1) + (1/4) + (1/9) + (1/16) + (1/25) + (1/36) + .... = pi^2 / 6
[i.e. the sum of the reciprocals of all the squares]
gatorman wrote:So, when folks compute pi to a large number of decimal places, do they do it by using series like the two you set down? Or, is there a more efficient way to do it?
Bungo wrote:gatorman wrote:So, when folks compute pi to a large number of decimal places, do they do it by using series like the two you set down? Or, is there a more efficient way to do it?
The earliest method, attributed to Archimedes, was to approximate the area of a circle by inscribing and circumscribing polygons. For example, he used a 96-sided polygons to estimate pi to the equivalent of 3 or 4 decimal digits. Calculating the area of such a polygon is conceptually simple as you can decompose it into 96 triangles of equal area, so the problem reduces to finding the area of one of the triangles. In theory this is easy because of the (1/2)*base*height formula, but the trick is working out the base and the height, a nontrivial thing to do without calculators as it involves clever trigonometric formulas (some of which appear to have been devised by Archimedes) and the ability to calculate square roots.
In theory, you can use formulas (infinite series) like the ones I mentioned above, but they converge rather slowly, meaning you have to add many thousands of terms in order to get a few significant digits of accuracy. There are infinite series which converge much more rapidly but which have much more complicated formulas. Here is a rather ugly looking one which has been used to calculate the first 10 trillion digits of pi:
http://en.wikipedia.org/wiki/Chudnovsky_algorithm
There are also other classes of algorithms that have been used with good success, but I don't know much about the details.
nisiprius wrote:Show Sum(n=1,N) 2n-1= N^2
1 + 3 + 5 + 7 + 9 + 11
= a six-by-six square.
QED
gatorman wrote:Here's another one to play with:
Show Sum(n=1,N) n^3=(N(N+1)/2)^2
I haven't tried this one yet, so can't offer a solution.
Sum(n=1,N) n^2 = a + bN + cN^2 + dN^3
0 = a
1 = a + b + c + d
5 = a + 2b + 4c + 8d
14 = a + 3b + 9c + 27c
a=0, b=1/6, c=3/6, d=2/6
Jacotus wrote:By the way, have you read Steven Strogatz's book The Calculus of Friendship, or read any of his New York Times columns? I suspect you would enjoy them.
For that one:gatorman wrote:Show Sum(n=1,N) n = N(N+1)/2
nisiprius wrote:For that one:gatorman wrote:Show Sum(n=1,N) n = N(N+1)/2
1 + 2 + 3 + 4 + 5 + 6 in blue fills half of a 6 x 7 rectangle. 1 + 2 + 3 + 4 + 5 + 6 in green fills the other half.
nisiprius wrote:For that one:gatorman wrote:Show Sum(n=1,N) n = N(N+1)/2
1 + 2 + 3 + 4 + 5 + 6 in blue fills half of a 6 x 7 rectangle. 1 + 2 + 3 + 4 + 5 + 6 in green fills the other half.
gatorman wrote:nisiprius wrote:For that one:gatorman wrote:Show Sum(n=1,N) n = N(N+1)/2
1 + 2 + 3 + 4 + 5 + 6 in blue fills half of a 6 x 7 rectangle. 1 + 2 + 3 + 4 + 5 + 6 in green fills the other half.
Another way to look at it:
Sum(n=1,4)n = 1+2+3+4, which can be rewritten, remembering N=4 as:
Sum(n=1,N)n= (N-3)+(N-2)+(N-1)+(N)
regrouping yields:
Sum(n=1,N)n= 4N-(3+2+1), now the first term, since 4=N, is just N^2, (this is good, we need an N^2 term) so we can rewrite as:
Sum(n=1,N)n= N^2-(3+2+1), where the second term is equal to Sum(n=1,(N-1))n, which is not quite where we need to be, but, if we recognize that
Sum(n=1,(N-1))n = Sum(n=1,N)n-N, we can substitute back in to get:
Sum(n=1,N)n = N^2-(Sum(n=1,N)n-N)
Sum(n=1,N)n = N^2-Sum(n=1,N)n+N
2Sum(n=1,N)n = N^2+N
Sum(n=1,N)n = N(N+1)/2
Again, probably not rigorous enough for the math wizards, but it works for me.
gatorman
In my picture, the N = 6, the first row is 1 + 6, the second row is 2 + 5, the third row is 3 + 4 ... making 6 copies of (6 + 1).Bungo wrote:nisiprius wrote:
Another way to see it is to add (1 + 2 + ... + N) to itself, with the second copy written in reverse order:
1 + 2 + 3 + ... + (N-1) + N +
N + (N-1) + (N-2) + ... + 2 + 1
= (N+1) + (N+1) + (N+1) + ... + (N+1) + (N+1)
where there are N copies of (N+1), so the sum is N(N+1).
Since this is the sum of two copies of (1 + 2 + ... + N), that means the sum of one copy must be N(N+1)/2.
Bungo wrote:Another way to see it is to add (1 + 2 + ... + N) to itself, with the second copy written in reverse order:
1 + 2 + 3 + ... + (N-1) + N +
N + (N-1) + (N-2) + ... + 2 + 1
= (N+1) + (N+1) + (N+1) + ... + (N+1) + (N+1)
where there are N copies of (N+1), so the sum is N(N+1).
Since this is the sum of two copies of (1 + 2 + ... + N), that means the sum of one copy must be N(N+1)/2.
muddyglass wrote:gauss' boyhood trick is a lovely approach, which is of course equivalent to the blue/green picture nisiprius drew.
nisiprius wrote:In the same vein, there is the story of John von Neumann and the fly problem. The problem itself varies in detail in every telling, but e.g. two trains 100 miles apart approach each other. One travels 60 mph, the other 40 mph. A fly traveling 75 mph (!) flies from train A until it reaches train B, then turns around and flies back to train A, back and forth over successively shorter distances, until the trains collide and the fly is crushed between them. In total, how many miles does the fly fly? The story is that Von Neumann instantly said "75 miles," and the disappointed questioner said, "Oh, you got it. I thought you would try to sum the infinite series." Von Neumann supposedly said "But I did sum the infinite series."
Bungo wrote:nisiprius wrote:In the same vein, there is the story of John von Neumann and the fly problem. The problem itself varies in detail in every telling, but e.g. two trains 100 miles apart approach each other. One travels 60 mph, the other 40 mph. A fly traveling 75 mph (!) flies from train A until it reaches train B, then turns around and flies back to train A, back and forth over successively shorter distances, until the trains collide and the fly is crushed between them. In total, how many miles does the fly fly? The story is that Von Neumann instantly said "75 miles," and the disappointed questioner said, "Oh, you got it. I thought you would try to sum the infinite series." Von Neumann supposedly said "But I did sum the infinite series."
Hmm, so the relative speed of the trains is 100mph, and they are 100 miles apart, so they will collide in exactly one hour. The fly is traveling at 75mph, and irrespective of what path it takes between the trains, it will be squashed in one hour, so it will have traveled 75 miles. I don't see the need for an infinite series, but maybe that's what Von Neumann's remark is saying: the series was summed implicitly regardless of how you solve the problem?
gatorman wrote:Here are some more from the same introductory problem set to nibble on:
Find the sum of the following series:
1^2+2^2+3^2+ . . . +n^2
Bungo wrote:"I will look forward to pretty 3d pictures from nisiprius!"
555 wrote:Bungo wrote:"I will look forward to pretty 3d pictures from nisiprius!"
I will look forward to pretty 4d pictures to show
[\sum_{k=1}^{n} k]^2=\sum_{k=1}^{n} k^3
or longhand
[1+2+3+...+n]^2=[1^3+2^3+3^3+...+n^3]
RobInCT wrote:Not a book, but Academic Earth has plenty of free math lectures on various subjects from some of the best professors/universities in the country online, and some include free downloadable course materials.
http://www.academicearth.org/subjects/math
Jacotus wrote:555 wrote:Bungo wrote:"I will look forward to pretty 3d pictures from nisiprius!"
I will look forward to pretty 4d pictures to show
[\sum_{k=1}^{n} k]^2=\sum_{k=1}^{n} k^3
or longhand
[1+2+3+...+n]^2=[1^3+2^3+3^3+...+n^3]
Test
Jacotus wrote:555 wrote:Bungo wrote:"I will look forward to pretty 3d pictures from nisiprius!"
I will look forward to pretty 4d pictures to show
[\sum_{k=1}^{n} k]^2=\sum_{k=1}^{n} k^3
or longhand
[1+2+3+...+n]^2=[1^3+2^3+3^3+...+n^3]
Test
555 wrote:According to one of the mods
viewtopic.php?f=3&t=6&p=1234798&hilit=codecogs#p1234798
we're not allowed to use that.
Jacotus wrote:555 wrote:According to one of the mods
viewtopic.php?f=3&t=6&p=1234798&hilit=codecogs#p1234798
we're not allowed to use that.
Oh, that's unfortunate. Thanks for pointing that out. I removed the image from my post.
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