What do you think?

anonimnystefy · 2012-11-04 11:40:09

I am not able to draw any generalized conclusions, but I think I sorta get the point.

bobbym · 2012-11-04 11:45:41

Then you are not getting the point yet.

2 Corinthians 6:17 wrote:

Go out from amongst them and be ye separate.

Same point! But what does it mean when we are talking about problem solving in mathematics.

anonimnystefy · 2012-11-04 11:48:37

Learn new tehniques for solving problems?

bobbym · 2012-11-04 11:53:48

Nope, learning different ones from what everyone is taught and learns. Back engineering from solution to method to question is one tool. We do not work front to back as they do. We work back to middle, front to middle, middle to front, any way that solves the problem.

phrontister · 2012-11-05 17:38:28

Hi Bobby,

M's 'RootReduce' on 1/Sqrt[1600 - x^2] + 1/Sqrt[2500 - x^2] == 1/10 gives the polynomial
x^8 - 7800 x^6 + 22350000 x^4 - 27838000000 x^2 + 12728100000000 = 0.

'N(Solve' on both equations gives
A. 37.355085334132530453574972651850703481315230858701, but the polynomial has an additional root:
B. 39.279615487303641329301362495489316846168383844149

Loading A & B back into the polynomial gives the following:
A. -3.303*10^-50
B. 9.0*10^-52

They're much closer to each other than when loading A & B back into the fraction equation, which gives these results:
A. 0.10000000000000000000000000000000000000000000000000000000000003843105
B. 0.16464704210781912709568407854860421540174931326784531936682258297947

In the words of Julius Sumner Miller: "Why is it so?"

Also, why can't I get W|A to give me the polynomial with RootReduce[Solve[1/Sqrt[1600 - x^2] + 1/Sqrt[2500 - x^2] == 1/10, x]]? It ignores the RootReduce function and instead displays a graph, three alternate forms of the original equation, and the root (positive and negative).

Don't worry about spending too much time sorting this out for me if the answer isn't simple enough for me to understand (you know what I mean).

Last edited by phrontister (2012-11-05 18:46:17)

bobbym · 2012-11-05 19:59:12

Hi phrontister;

Sometimes algebraic manipulations of the kind you did introduce spurious roots.

Alpha is not the complete Mathematica package. It behaves differently.

phrontister · 2012-11-06 00:10:57

Thanks, Bobby. I'll leave that there then.

bobbym · 2012-11-06 02:57:04

Hi phrontister;

No problem keep on experimenting with your ideas!

bobbym · 2012-12-06 19:41:08

New problem:

WDYT #200:

Inspired by the other thread C creates another similar problem.

What is the area of the circle that has two perpendicular chords with lengths as
shown in the diagram.

A says) A great problem C. I have the answer!

B says) Bah humbug!

C says) Why the negative attitude B? Is my problem too tough for ya?

D says) Yes, B is a math wimp. I solved it!

E says) Double humbug, B is right. Those solutions are kaboobly doo.

A says) Why!

E says) Maybe because...

anonimnystefy · 2012-12-06 21:45:23

Hi

bobbym · 2012-12-06 21:46:43

Hi;

anonimnystefy · 2012-12-06 21:53:30

Hi

bobbym · 2012-12-06 21:57:26

Hi;

anonimnystefy · 2012-12-06 22:02:16

Hi

bobbym · 2012-12-06 22:03:55

Hi;

anonimnystefy · 2012-12-06 22:15:41

Yes, I have. And I solved it.

bobbym · 2012-12-06 22:39:36

Hi;

I does look like you have done it.

anonimnystefy · 2012-12-09 05:24:00

Hi bobbym

I hope you don't mind me posting this. If you do, feel free to remove this post.

Here's a new problem:

You have a square of side 5 divided into 25 unit squares. In the central square a 0 is placed.
Question: In how many ways can the square be filled with non-negative numbers so that the absolute difference of numbers in two adjacent squares is never greater than 1? What is the answer if the large square has a side of 11 instead?

Last edited by anonimnystefy (2012-12-09 05:37:08)

bobbym · 2012-12-09 22:33:52

Hi anonimnystefy;

anonimnystefy · 2012-12-09 22:36:38

Let me guess, that was a wild guess to see if I have the answer yet or not.

I a now wondering why I posted it without having the solution...

bobbym · 2012-12-09 22:38:48

That is not a wild guess but what I think is the answer unless my arithmetic is bad.
Some people hate broccoli, some people hate liberals, most people hate bobbym, everyone hates math, I hate hand calculations!

anonimnystefy · 2012-12-12 06:36:19

It's not true that everybody hates math!

bobbym · 2012-12-12 08:41:18

Maybe but they all hate broccoli and bobbym.

What do you think of my solution and do you require the 11 x 11?

anonimnystefy · 2012-12-12 10:27:35

Not everyone hates bobbym, either!

I am still unable to verify your answer. Sorry.

Maybe I should delete the problem and post it when I have the answer...

bobbym · 2012-12-12 10:30:05

Hi;

You could leave it. Where does the problem come from?

Math Is Fun Forum

#1926 2012-11-04 11:40:09

Re: What do you think?

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