In order to differentiate a exponential function such as g(x) = ax we invented the natural exponential function
f(x) = ex
where e = 2.71828182845905... For the exponential function f(x) = ex we have f'(x) = ex and
f(0) = 1 and f(1) = e The natural log of this function is ln(f(x)) = ln( ex) = x
Let L(x) = ln(x) then L'(x) = 1/x . We are not proving the above as it can be easily obtained from any textbook on Calculus.
We can also obtain the ln of any number by noting the following
| ln(c) = lim [( ch - 1)/h] where h tends to 0 |
Let f(t) = ct where c = ez Then f ' (t) = zezt=zct=
lim [(ct+h - ct)/h] where h tends to 0
So z = ln(c) = lim [( ch - 1)/h] where h tends to 0 . Now lets test for ln(2) . The calculator gives a value of 0.6931471805599453094...
Let h = 10-6 Then ln(2) = 0.693147420.. which is quite near to the real value. So now we have a method of determining ln values.
Now lets try to invent a method along the same line to determine e
| e = lim ( 1/x + 1)x = lim ( (x + 1)/x))x where x tends to oo |
We have that ln(c) = lim [( ch - 1)/h] where h tends to 0 . Then h.ln(c) = ( ch - 1) where h tends to 0
So h.ln(c) +1 = ch where h tends to 0. Therefore c = ( h.ln(c) +1)(1/h) where h tends to zero. Now let c = e and we have
e = (h + 1)(1/h) where h tends to zero. Note as h tends to zero that 1/h tends to oo. So let 1/h = x where x tends to oo
Then we have that e = lim ( 1/x + 1)x = lim ( (x + 1)/x))x where x tends to oo
Lets test this taking x = 1000000. Then e is approximately 2.7182805693 which is quite good.
In the same way we get
| ea = lim ( a/x + 1)x = lim ( (x + a)/x))x where x tends to oo |
if we let c = ea
So we have c = ( h.ln(c) +1)(1/h) where h tends to zero. Therefore ea= ( h.ln(ea) +1)(1/h) where h tends to zero.
Thus ea= ( h.a +1)(1/h) where h tends to zero. Letting h = 1/x we get that
ea= ( a/x +1)(x) where x tends to infinity.
Or writing it differently we have ea= ( [x+a]/x )(x) where x tends to infinity.
We test this using a = 2 and x = 1000000. The pocket calculator gives e2 = 7.3890560989.......
Using our formula we have e2= ( 2/1000000 +1)(1000000) = 7.3890413....... which is quite close
Now lets look at the relationship between complex numbers and e
| z = a + ib = r(cos(x) + i sin(x)) = r e i x |
Lets deduce the last part using differentiation and integration. So z = a + ib = r(cos(x) + i sin(x)).
Therefore dz/dx = r(-sin(x) + i cos(x)) and thus dz/dx = i.r(cos(x) + i sin(x)) = i z
So dz/z = i dx and we therefore have that ∫ (1/z) dz = "∫ i dx that results in ln(z) + k1 = i x + k2
So ln(z) = i x + k3 At x = 0 we have z = r , so ln(r) = k3
So z = e( i x + k3) = e( i x + ln(r)) = r e i x
After proving the above we can now do some examples.
We know that e π i = -1 and thus ln(-1) = π i where π = 3.1415926.. and i is the square root of -1
We can easily confirm that e π i = -1 or approaches -1 using our formula ea= ( a/x +1)(x) where x tends to infinity.
So let a = π i and let x = 10000000 Then we have e π i = ( π i /10000000 +1)(10000000)
So e π i = (1 + 3.1415x10-7i)10000000 = 1 angle 180 degrees = -1
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