For math geeks out there, I have something to share.
I learned something new in Mathematics. Not really new, but I found it quite new. 😄
What do e, i and PI have in common?
To begin the discussion, as we probably all know,
e is the base of the natural logarithm
i is the number which when squared gives -1
pi is the number that is the ratio of the circumference to the diameter of a circle.
What do they have in common? Not really much, but when we put them together, they make -1, as in the following equation:
e^(i pi) = -1
It is a surprising result, because the left-hand side is a complex number involving the imaginary component, and the right-hand side is a pure real number. When the author of this was young [let's call him Mathmate], a school-friend of his showed him this equation at a time when he did not even understand what an imaginary number was. Gabriel, now a Ph.D. in physics, works in oncology to help people who live with cancer. Brilliant people remain brilliant.
Back to mathematics, how is this possible that so many favourite numbers are linked together by this single simple equation? The answer is really quite simple if you read on.
If you have learned about Taylor’s expansions, you will know that the three following mathematical functions can be represented by an infinite series, meaning that if you take enough terms to the point that the remaining terms are small enough, you get the answer to the function.
e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...
sin x = x - x^3/3! + x^5/5! - x^7/7! + ....
cos x = 1 - x^2/2! + x^4/4! - x^5/5! + ...
For those who have not worked with infinite series before we will show an example:
sin(45 degrees)=sin(pi/4 radians)
=pi/4 - (pi/4)^3/3! + (pi/4)^5/5! - (pi/4)^7/7! + (pi/4)^9/9! - ...
=0.78540 - 0.08075 + 0.00249 - 0.00004 + 0.00000 -....
=0.70710
The accurate answer is 0.7071067812... So now you get the idea about infinite series! I hope, too 😄
In the exponential equation, let’s put x=i pi, then it reduces to:
e^(i pi) = 1 + i pi + (i pi)^2/2! + (i pi)^3/3! + (i pi)^4/4! + (i pi)^5/5! + ...
By regrouping terms, substituting i^2=-1, i^4=1, and factoring out i, we obtain:
e^(i pi)= 1 - (pi)^2/2! + (pi)^4/4! -... _+i ( pi - (pi)^3/3! + (pi)^5/5! - ...)
= cos (pi) + i sin(pi)
= -1 + i . 0
=-1
😎 Cool, isn't it? 😄
The Magic Sum
Study the matrix below.
You'll see that the numbers in left diagonal will give you a total sum of 111.
This is the same with the numbers at the right diagonal.
Check:
Left: 1+8+15+22+29+36 = 111.
Right: 6+11+16+21+26+31 = 111.
Obviously, 111 here is our Magic Sum.
And now here's the trick.
I will let you choose 6 numbers out of the given 6x6 matrix.
And I'll show you that the numbers you'll choose will give you a total sum of 111.
But first, you have to give me your first choice, before we go on with the next step.
I'll wait.
😉
Originally posted by JuryIt's so blatent how that trick works 😛
[b]The Magic SumStudy the matrix below.
You'll see that the numbers in left diagonal will give you a total sum of 111.
This is the same with the numbers at the right diagonal.Check:
Left: 1+8+15+22+29+36 = 111.
Right: 6+11+16+21+26+31 = 111.Obviously, 111 here is our Magic Sum.
And now here's the trick.
I will let you choose 6 numbers out of the given 6x6 matrix.
And I'll show you that the numbers you'll choose will give you a total sum of 111.But first, you have to give me your first choice, before we go on with the next step.
I'll wait.
😉 [/B]