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Monday, May 02, 2005 - 01:31 PM
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 I was young, a school-friend of mine showed me this equation at a time when I 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!
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
Simple, easy and straight forward, as my secondary two English teacher often said.
Note: this article is an extract from http://www.mathpath.uni.cc with permission.
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 I was young, a school-friend of mine showed me this equation at a time when I 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!
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
Simple, easy and straight forward, as my secondary two English teacher often said.
Note: this article is an extract from http://www.mathpath.uni.cc with permission.
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Comments
[img]http://www.allmathematics.spiralmindsinc.com/Images/Equations/Pi_e_Link_2.gif[/img]
Cheers,
RyanJ
how(if possible!)can you put this ,(e^(i pi)=?),on a calculator,when there's no function for ( i ),I know it's imaginary!.
(stupid question?)
A friend of mine who's calculator doesn't have an explicit i button simply stored the variable i as sqrt(-1).
Additionally, if your calculator does have a the ability for complex representation, you can use Complex(0,1) as i.
For instance, the full relationship is better described as
e^(x i) = cosx + i*sinx
where x is variable.
From this you can define trig functions as differences between complex power functions, for instance
cosx = [ e^(i x) + e^(- i x) ] / 2
sinx = [ e^(i x) - e^(- i x) ] / 2i
And from these you can find, in my opinion, easier to remember derivations of trig identities.
Additionally, these show the relationship to the hyperbolic trig functions cosh, sinh, tanh, etc.
coshx = cos(i x) = [ e^x + e^(-x) ] / 2
-Ja
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