Interesting Free Math Books

[gatorman]

I was looking through the title list on the Gutenberg project and found quite a good number of math texts and treatises. This little gem on number theory from 1914 caught my interest:

http://www.gutenberg.org/files/13693/13693-pdf.pdf

and I started reading it. The first section is on the basic properties of numbers and operations on them, and there were a few problems to solve at the end. I thought they would be very simple (and to some of you they probably are very simple) but instead I found them to be interesting, and you might find them fun to solve as well. Here are a couple of examples:

Show Sum(n=1,N) n = N(N+1)/2
and
Show Sum(n=1,N) 2n-1= N^2

(If you have problems with them, pm me and I'll give you my solutions, which probably wouldn't satisfy a real mathematician, but are persuasive to me)

But I digress, because the real purpose of this post is to provide a spot for people to post links to books on math they think are interesting and might help build/refresh math skills and to post interesting problems which would do the same. I know this topic might be viewed as being a little nerdy, but there is no denying that math skills are one of the keys to success in the science/engineering world and, moreover, are generally useful. More than that though, math is just interesting and fun (or at least can be if properly approached). So, post your favorte links or interesting problems here.

gatorman

[nisiprius]

Show Sum(n=1,N) 2n-1= N^2

1 + 3 + 5 + 7 + 9 + 11

= a six-by-six square.

QED

[gatorman]

Show Sum(n=1,N) 2n-1= N^2

1 + 3 + 5 + 7 + 9 + 11

= a six-by-six square.

QED

I hadn't looked at it from a geometrical perspective, but you are absolutely correct! A+
gatorman

[Tycoon]

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]

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]

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.

[gatorman]

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.

Here's another one you can get for $.01 (plus exhorbitant S&H) and might enjoy, a pretty interesting book on the concept of infinity:

http://www.amazon.com/Mystery-Aleph-Mat ... 0743422996

At the end, Aczel gets into the concept of trans-infinite numbers and their properties. It all makes sense as you are reading it, but its one of those concepts you have to go back and look at more than once, and even then not easy to wrap up into a nice, neat bundle.

You have to be careful with the concept of infinity, it drove Georg Cantor insane.
gatorman

[gatorman]

Another place you can get a lot of free math is the Khan Academy, really a great resource. Here is a link:

http://www.khanacademy.org/

gatorman

[Jacotus]

The two examples in the OP are equivalent.

[gatorman]

The two examples in the OP are equivalent.

Yes, but the fun part is in showing it.

[Bungo]

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]

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]

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]

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.

[gatorman]

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.

As usual, Wikipedia has an article:
http://en.wikipedia.org/wiki/Numerical_ ... _of_%CF%80

[gatorman]

Show Sum(n=1,N) 2n-1= N^2

1 + 3 + 5 + 7 + 9 + 11

= a six-by-six square.

QED

Another way to look at it:
Sum(n=1,4) 2n-1 = (2x1-1)+(2x2-1)+(2x3-1)+(2x4-1)
so, remembering N=4, the above can be rewritten:
Sum(n=1,N)2n-1 = (2x(N-3)-1)+(2x(N-2)-1)+(2x(N-1)-1)+(2xN-1)
expanding that out:
Sum(n=1,N)2n-1 = (2N-6-1)+(2N-4-1)+(2N-2-1)+(2N-1)
= (2N-7)+(2N-5)+(2N-3)+(2N-1)
note that N appears N times in the above (and would so appear no matter what number was chosen as N) so if one collected terms, 8N is really 2x4xN, but since 4=N is really 2N^2, so the above can be rewritten:
Sum(n=1,N)2n-1 = 2N^2- (7+5+3+1)
note that the term on the right side of the equation in parenthesis is equal to Sum(n=1,4)2n-1 or when 4=N, Sum(n=1,N)2n-1, so the above can be rewritten:
Sum(n=1,N)2n-1 = 2N^2-Sum(n=1,N)2n-1
2Sum(n=1,N)2n-1 =2N^2
Sum(n=1,N)2n-1 =N^2
and as Nisiprius says, QED.
So, that's the way I got to the answer. A real mathematician would probably criticize my rigor, but it works for me (although I like Nisiprius's method better).
gatorman

[gatorman]

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.

[Jacotus]

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.

If one is willing to accept that Sum(n=1,N) n^m is a polynomial in N of degree m+1, which should be at least plausible based on your m=1 case, then there is a straightforward algebraic way to derive these relations, at least for relatively small m. Having an algebraic method is nice when our geometric intuition may not be strong enough. That's the case here: nisiprius's example works wonderfully in two dimensions, but I have a hard time visualizing a four-dimensional hypercube.

Example for m=2.
Assume

Code: Select all

Sum(n=1,N) n^2 = a + bN + cN^2 + dN^3

We have four unknown coefficients, a, b, c, and d. If our formula is true for all integers N>=0, then we can get true equations by evaluating it at N=0,1,2,3. Evaluating the left side and the right side of the equation at these N gives four linear relations for the four coefficients. These equations are

Code: Select all

0 = a
1 = a + b + c + d
5 = a + 2b + 4c + 8d
14 = a + 3b + 9c + 27c

Linear equations such as these can be solved without too much difficulty by inverting a matrix. Or you can try solving it by hand by solving the second equation for b in terms of c and d, then substituting into the third equation, etc. In any event, you can verify by direct substitution that

Code: Select all

a=0, b=1/6, c=3/6, d=2/6

is the solution.

Doing this for m=3 is left as an exercise.

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.

[gatorman]

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.

I haven't read it, but it looks interesting. Calculus of Friendship, ~$8 on the Kindle or ~$10 in paperback, 4.5 stars. Here is a link:

http://www.amazon.com/Calculus-Friendsh ... 0691150389

[gatorman]

You might find this paper to be of interest: The Unreasonable Effectiveness of Mathematics in the Natural Sciences. Here is the link:

http://www.dartmouth.edu/~matc/MathDram ... igner.html

[nisiprius]

Show Sum(n=1,N) n = N(N+1)/2

For that one:



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]

Show Sum(n=1,N) n = N(N+1)/2

For that one:



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.

I had to look at that one for a minute or two, after I did, I believe you have earned another A+.

[gatorman]

Show Sum(n=1,N) n = N(N+1)/2

For that one:



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

[Bungo]

Show Sum(n=1,N) n = N(N+1)/2

For that one:



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

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.

[nisiprius]

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.

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).

[muddyglass]

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.

gauss' boyhood trick is a lovely approach, which is of course equivalent to the blue/green picture nisiprius drew.

[Bungo]

The induction proof is also quite simple. We want to prove that

1 + 2 + 3 + ... + N = N(N+1)/2

For N = 1, the formula asserts that 1 = 1(1+1)/2, which is certainly true.

If we now assume it is true for N-1, i.e., that

1 + 2 + 3 + ... + (N-1) = (N-1)N/2, then adding N to both sides of this equation gives us

1 + 2 + 3 + ... + N = (N-1)N/2 + N

Simplifying the right hand side, we get

((N-1)N + 2N)/2

which is the same as

(N-1+2)N/2

or simply

(N+1)N/2

Thus we have shown that the formula is valid for N=1, and that if it is valid for any given N-1, it is also valid for N. This implies that the formula is true for all positive integer values of N.

[Bungo]

gauss' boyhood trick is a lovely approach, which is of course equivalent to the blue/green picture nisiprius drew.

Absolutely. It's a beautiful proof whether viewed geometrically or algebraically.

[nisiprius]

I wasn't sure what "Gauss' boyhood trick" was, Googled, and found this rather amazing web page:
Tellings of the Gauss anecdote. "Transcribed below are 109 tellings of the story about Carl Friedrich Gauss's boyhood discovery of the "trick" for summing an arithmetic progression..."

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."

[gatorman]

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
1^2+3^2+5^2+ . . . +(2n-1)^2
1^3+3^3+5^3+ . . . +(2n-1)^3

If I understood what he was doing, Jacotus may have already provided the answer for the first series.

gatorman

[Bungo]

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?

[Epsilon Delta]

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?

Von Neumann alleged that he solved the problem by figuring out how far the fly travels before he reaches the first train and reverses direction, how far it travels before the second reversal, etc. and summing the results to get the answer. I.e. he did it the hard way.

Generally there are many ways of solving a problem. There's a trade off between looking for an easy way to perform a calculation and actually doing the work of whatever method you choose. Making the trade off wisely is part of being good at math.

Of course the trade off will depend in part on how good you are at calculation. When I first met the train problem my first thought was to calculate the intersections, and sum the series, but I quickly realized that a) it would take a while and b) I would probably get it wrong anyway. So I cast around for a different method (the total time method). Presumably Von Neumann had the same first thought I did but knew he could do the algebra quickly and correctly, so he did.

[Bungo]

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

1^2 = 1
2^2 = 4 = 1 + 3
3^2 = 9 = 1 + 3 + 5
4^2 = 16 = 1 + 3 + 5 + 7
and so on. (Here, we have used the first picture provided by nisiprius.)

So, 1^2 + 2^2 + 3^2 + ... + N^2 = (N)(1) + (N-1)(3) + (N-2)(5) + ... + (1)(2N-1)

Writing this in a more compact notation, we have

sum(k=1 to N) k^2 = sum(k = 1 to N) (N+1-k)(2k-1)

The right hand side can be rewritten as

sum(k=1 to N) (N+1)(2k-1) - sum(k = 1 to N) k(2k-1)

which equals

2(N+1) sum(k = 1 to N) k
- (N+1) sum (k = 1 to N) 1
- 2 sum(k = 1 to N) k^2
+ sum(k = 1 to N) k

Now can apply our previous formula, namely sum (k = 1 to N) k = N(N+1)/2, to lines 1 and 4. Line 2 is simply -(N+1)(N). And line 3 is (-2) times the sum of squares (i.e. the left hand side). Summarizing, the right hand side becomes

2(N+1)(N)(N+1)/2
- (N+1)(N)
- 2 sum(k = 1 to N) k^2
+ N(N+1)/2

and this all equals the left hand side, which was sum(k = 1 to N) k^2. Adding two times this to both sides, we get

3 sum(k = 1 to N) k^2 = 2(N+1)(N)(N+1)/2 - (N+1)(N) + (N+1)(N)/2 = (N)(N+1)^2 - (N)(N+1)(1/2)
Dividing both sides by 3, we finally obtain

sum(k = 1 to N) k^2 = (1/3)(N)(N+1)^2 - (1/6)(N)(N+1)

This can be further simplified to

(1/6)N(N+1)(2(N+1) - 1)
= (1/6)(N)(N+1)(2N+1)

OK, that was pretty ugly. I know there are geometric proofs for this, too, involving adding a bunch of pyramids together in clever ways. I will look forward to pretty 3d pictures from nisiprius!

[555]

"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]

[Jacotus]

"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
[removed image]

[RobInCT]

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

[gatorman]

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

Wow, that's a great resource! Thank you!
gatorman

[gatorman]

"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

You just need a set of 4th dimensional spectacles, on sale this week at The Cafe at the End of the Universe, or online at their website, special offer code:ArthurDent.

[555]

"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

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]

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.

[gatorman]

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.

Jacotus- Would this work for you?

http://www.math.union.edu/~dpvc/jsMath/

gatorman

[gatorman]

I think it is fairly easy to prove this one:
Sum(n=1,N) n^3=(N(N+1)/2)^2

is true by induction as well using the approach excellently explained by Bungo.

Sum(n=1,(N-1))n^3=1^3+2^3+3^3+...+(N-1)^3= (((N-1)((N-1)+1))/2)^2
Working with just the right side of the equation:

(((N-1)((N-1)+1))/2)^2 can be simplified to:
(((N-1)N)/2)^2 and further simplified to
(N^2/2-N/2)^2 which can be expanded to
N^4/4-N^3/2+N^2/4

now adding N^3 to both sides of the original equation yields:

Sum(n=1,N)n^3= 1^3+2^3+3^3+. . . +N^3= N^4/4-N^3/2+N^2/4+N^3
= N^4/4+n^3/2+n^2/4, which is the same thing we get if we expand

(N(N+1)/2)^2, so the formula appears to be correct by induction.

That said, I haven't yet worked out the "show" part to my satisfaction, but I think that Bungo gave a hint as to how to do it in his demonstration of how to obtain the expression for Sum(i=1,N)n^2, so I'm trying to work off of that, in between bowl games.

I wish someone would pay me as much as I make at my day job to do this, so much fun.
gatorman