Unbelievable math problem

[KingTermite]

Originally Posted by grsamf

Do you go to department stores at Christmas and tell the lttle kids there is no Santa, too?

No Santa? What?

Grsamf....how could YOU!! My dreams are burst.

Sammy and I waited 5 hours last Christmas to sit on his lap and ask for new video cards for Christmas.

[PCBruiser]

Originally Posted by kONGO

5^(3+5i) = (5^3)*(5^5i) first of all.

Then write 5 = e^(ln(5)) and you get 125*((e^(ln(5))^5i) which is 125*e^(5i*ln(5)) which is 125*e^(i*5*ln(5)).

Then remember Euler's e^ix = cos(x) + i*sin(x) and you get: 125*(cos(5*ln(5)) + i*sin(5*ln(5))).

The a = e^(ln(a)) is the key here. It's a nice little trick to pull, I think.

Very nice demonstration of Complex Algebra, kONGO.

[Centurion]

Thanks KONGO. I like your method of doing it, but was looking for something else I think. Basically I want to know how to do it without using Euler's identity. I was looking for a way to determine it for any number in general, and then plug in the number 'e' and see how it reduces to e^ix = cos(x) + i*sin(x). Any ideas?

[Underwater Mike]

Originally Posted by kONGO

5^(3+5i) = (5^3)*(5^5i) first of all.

Then write 5 = e^(ln(5)) and you get 125*((e^(ln(5))^5i) which is 125*e^(5i*ln(5)) which is 125*e^(i*5*ln(5)).

Then remember Euler's e^ix = cos(x) + i*sin(x) and you get: 125*(cos(5*ln(5)) + i*sin(5*ln(5))).

The a = e^(ln(a)) is the key here. It's a nice little trick to pull, I think.

If I have five donuts, and I give three to Officer Bob...

[k0NG0]

Originally Posted by Centurion

Thanks KONGO. I like your method of doing it, but was looking for something else I think. Basically I want to know how to do it without using Euler's identity. I was looking for a way to determine it for any number in general, and then plug in the number 'e' and see how it reduces to e^ix = cos(x) + i*sin(x). Any ideas?

Well, for a^(b+ci) you do get the end result a^b(cos(c*ln(a)) + i*sin(c*ln(a))) always.

If you're talking about (a+bi)^(c+di) you can deduct a similar formula.

Any real number with a purely imaginary exponent will result in a complex number with the absolute value of 1. This is entirely due to Euler -- raising a number with an imaginary part only is in fact a rotation in the complex plane, you rotate a vector with length 1 around the unit circle.

But all of it revolves around Euler's formula that e^ix = cos(x) + i*sin(x). I don't think you can get around that.

[stiwho]

Im stuck in using Taylor Series w/ Differential Equations. Anyone wanna give me a hand!

[stiwho]

Originally Posted by kONGO

Well, for a^(b+ci) you do get the end result a^b(cos(c*ln(a)) + i*sin(c*ln(a))) always.

If you're talking about (a+bi)^(c+di) you can deduct a similar formula.

Any real number with a purely imaginary exponent will result in a complex number with the absolute value of 1. This is entirely due to Euler -- raising a number with an imaginary part only is in fact a rotation in the complex plane, you rotate a vector with length 1 around the unit circle.

But all of it revolves around Euler's formula that e^ix = cos(x) + i*sin(x). I don't think you can get around that.

Aren you only referencing the homogenuos solution what about the particular solution?

[sandman55]

Originally Posted by Bofinn

Here is a math trick so unbelievable that it will stump you.

1. Grab a calculator. (you won't be able to do this one in your head)

2. Key in the first three digits of your phone number (NOT the area code)

3. Multiply by 80

4. Add 1

5. Multiply by 250

6. Add the last 4 digits of your phone number

7. Add the last 4 digits of your phone number again.

8. Subtract 250

9. Divide number by 2

Do you recognize the answer?

Cool huh........

Have a Great Day!!! -

We have 8 digits too, so I substituted in line 2 "the first 3 digits" for "the first four digits" and was quite pleased with my self till I read further and read PCB's responce
now that's really analizing it

Good one PCB

[HiJon89]

Originally Posted by PCBruiser

Well, not so magical at all. Let's look at the math:

Assume y = first three digits of your phone number, and x = last 4 digits. The formula then becomes:

250(80y+1)+2x-250
_________________ = 10,000y+x.
2

The factor of 10,000 just moves the position of y four places to the left (producing y0000), and then you just add x. Since you have supplied both y and x, and since phone numbers are always 7 digits, this will always produce your phone number. In essence, there's nothing magical or unbelieveable about it at all, just simple math.

H@X

[k0NG0]

Originally Posted by stiwho

Aren you only referencing the homogenuos solution what about the particular solution?

No. It's not a differential equation, it's a simple complex algebraic equation. A, b, c and d are really constants, not variables. This because you plug in a constant real number to get the end value. Neither of a, b, c or d vary with e.g. time.

In my years with math I've only come across homogenous and non-homogenous/particular solutions in diff. equations.

[Bofinn]

another math problem solved:

[Fraoch]

Interesting solution. Despite a degree in Engineering, math was never my strong suit (science was).

I thought maybe we could be playing around with self-checking numbers here. Most account numbers, credit card numbers etc. are self-checking. Ever wonder why these numbers are 16 digits+? (Andy Rooney did several years ago on 60 Minutes, but he never bothered investigating).

It's so that you can't make a number up. The digits combine together in a simple equation defined by the account owner to self-check. Kinda like "the sum of the first 3 digits should be the 7th digit, subtract the 8th digit from the 9th digit to be the 10th digit" or something similar.

Andy Rooney, I remember, was dumb enough to look at his gas bill and say "It's 28 digits, for crying out loud! There are fewer people than that on Earth! Why can't it be something simple, like 72 or 4?" Another instance of reporting without research.

I thought that perhaps phone numbers were self-checking as well, but as PCB shows it doesn't matter, which is why it can be used for any phone number in the world.

What I find interesting about phone numbers in North America is that the area code and 3-digit prefix (first 3 digits of the 7-digit number) are never reused. For example, my current prefix is 680. No area code in North America uses that number. My prefix used to be 869. Checking here, "St Kitts & Nevis" use that - they are not on the continent but are Carribean islands IIRC. Same with all the numbers of my clients. Check it out! Now that takes some coordination - all the prefix exchange numbers in the continent are checked against the area code list. I believe this is done to prevent misdialed long-distance calls from completing inside the area code - the area code does not point to a 3-digit prefix and no connection is made.

[bjbgraphics]

I do not have a high level of math comprehension but I can usually appreciate a good problem/solution... though I don't know why this problem is so "unbelievable" or impressive.

From a visual standpoint, you see that in step 2 you put in the first three digits of your phone number, and in step 6&7 you input the last four digits of your phone number... and lo and behold at the end of the calculation these numbers are displayed together displaying your full 7 digit phone number.

It doesn't seem particularly amazing to me that this problem is able to come up with your exact phone number after you already typed it in. <shrugs> Perhaps it is more impressive from a purely mathematical standpoint.

BJB

[polonyc2]

Originally Posted by BJB

It doesn't seem particularly amazing to me that this problem is able to come up with your exact phone number after you already typed it in.

good point...it would have been really impressive if it was able to come up with my correct phone number without me typing it into the problem in the first place

[polonyc2]

Originally Posted by PCBruiser

Yes, I did, but that's what a degree in Math from MIT does for you.

pretty impressive ...kONGO seems pretty good with Math himself, maybe we ought to have a ABX Math Mod face off challenge

hey PCB are you really John Forbes Nash Jr. in disguise?

http://nobelprize.org/economics/laur...h-autobio.html

I was looking through the Nobel Prize website and they said that Nash received the Nobel Prize in 1994 for ""for his pioneering analysis of equilibria in the theory of non-cooperative games"...they must have been doing their testing on an early beta of Half Life 2