Geometry, Algebra, Graphing, and Calculus are very useful for analyzing relationships and finding equations to express those relations.
Since exponential growth/decay is 'everywhere' in living systems ... and chemistry and physics and finance even,
so ... it's VERY GOOD to understand exponential curves and (the inverse) logarithm curves.
Some interesting facts ... (you will soon know) ...
Just like pi has the constant value (of non repeating decimals) 3.14159...
(the ratio of circumference to diameter for any/every circle)
So ... e has the constant value of 2.718281828... (also with NON-repeating decimals)
e is a VERY SPECIAL number ...
e is related to pi even ...
Euler showed that e^(i*pi) + 1 = 0
Below, you will see more re. e^(i*x), but especially more about ...
algorithms to find the value of the 'real constant e', to as many decimal places of significance as needed or desired,
and how e, e^x and ln(x), i.e. log to the base e ... and the whole of 'Calculus' ... are THE ....
real bridge ...to the algebra that uses 'real' numbers, from an algebra that only uses 'rationals'.
*** But see also the link above ... provided at the bottom of the first page, now copied here ***
An other perspective from MIT ... 'The Big Picture of Calculus'http://www.youtube.com/watch?v=UcWsDwg1XwMWhere
calculus also LEADS us to derive a pair of related functions,
i.e. how to derive ANY one of the pairs of functions below ... from the other.
i.e if ...
(1.1) df(x) / dx = F(x)
then ...
(2.1) Integral( F(x) ) = f(x) + C, where C is some constant
AND ... if ...
(1.2) f(x) + C = Integral( F(x) ), where C is some constant
then ...
(2.2) d( Integral( F(x) ) ) / dx = F(x) = df(x) / dx
i.e. the calculus may also be understood as ...
How to find these related pairs of functions ...
i.e. how to find the matching related function in the pair below, having been given the other:
i.e.
1) How to derive the derivative function from any (differentiable) function
2) How to derive the integral function from any function (that is integrable)
Heads up!
You will see that:
1) d( e^x + C ) / dx = d( e^x ) = e^x
and ...
2) Integral( e^x ) = e^x + C, where C is some real constant
Also that ...
3) d( ln(x) + C ) = 1/x, where ln(x) means ... take the logarithm of any real number x to the base e
and ...
4) Integral( 1/x ) = ln(x) + C
Also that ...
ln( e ) = 1
So if ...
y = e^x
then ...
ln(y) = ln( e^x ) = x * ln( e ) = x
(but this is really just the definition of ln, i.e. x is the logarithm <i.e. the exponent> to the base e)
(i.e. x is the exponent of e, or another way to say it ... x is the power e is raised to)
Now ... recall from your work with multiplying numbers, that if y = b*b*b = b^3, then y^(1/3) = b
But how do we evaluate b^r, where b > 0 and r, (especially r), are ANY real numbers, and not just integers or rational numbers?
To the rescue ... Please hold your applause ... Please welcome: e ... And log to the base e
Since, b^r = e^( ln(b^r) ) = e^( r*ln(b) )
(because ln( b^r ) = r*ln(b) = ln( e^( r*ln(b) )
Once we learn how to calculate e^x, for any 'real' x,
And learn to take the natural log of any 'real' b > 0,
we then can always evaluate b^r, for all 'reals' r and b, with b != 0 (b not equal to 0)
So 'e' ... IS such a big 'BIG-E' ... Can you start to see?
So with all this gala introduction ...
HOW do we find the road that leads straight to e ? ? ?
Note! In the following, that y' (y primed) symbolizes the rate of change in y
1. for y' = y (type relationships ... where the rate of change of something depends on the (amount of that) something present instant by instant ... )
y = A*e^x is the ONLY solution (set)
Recall: e^x means 'e to the exponent x', A*e means 'A times e'
So ... you see here ... 'e' ... is a 'real' and also a 'constant value' (similar to pi)
We discovered 'e' here, by finding here the form of equations that satisfy, (that are solutions to), the problem of y' = y
If we take the particular solution case of A = 1, then the constant value e = 2.718281828... falls out, when we then also take the case of x = 1
On a following page, I will show how the constant e may be found out to as many places of decimals as desired ... i.e. how to derive from just y' = y ... the same series expansions, as given below on this page, that (below) comes from the limit, as n grows arbitrarily large, for the value of (1 + 1/n)^n ... and also, the series expansion from the limit, for arbitrarily large n, of (1 + x/n)^n
Heads Up!!!
You will see that these amazing infinite series expressions are both the same:
e^x =
1 + x/1! + x^2/2! + x^3/3! + x^4/4! + ...
And the limit, as n grows arbitrarily large of (1+x/n)^n =
1 + x/1! + x^2/2! + x^3/3! + x^4/4! + ...
Recall:
'Things/Expressions' that are equal to the same 'alternate thing/expression' ... are equal to each other. As per if x = w and also y = w, then it follows that x = y
(Recall also:
n! means 'n factorial' and n! = 1*2*3* .... *n
Can you see that n! becomes ever much larger as n grows larger?)
Note also, if y' = y, then y = y' = y'' = y''' = ... and for however many repetitions of the prime in y primed ... as in for example y = y''''''''''
To prove that our 'infinite power series solution' ...
y = f(x) = e^x = 1 + x/1! + x^2/2! + x^3/3! + ....
WORKS for y' = y and y'' = y, etc...
All you need to know, that's 'new', is ...
0. If y = f(x) = c, you already know
the slope (of an horizontal line) is 0
So here, y' = 0
0. If f(x) = m*x + b, then f'(x) = m
you already know that ...
when finding the slope of a straight line ...
you just get the 'real' constant m,
and the 'real' constant 'b' ... is dropped
i.e. the 'y intercept' has NO effect on the slope
1. How to find g'(x) when ...
g(x) is some polynomial in powers of x like ...
g(x) = a*x^n + b*x^m with x, a, b real and integers n,m >0
On a following page we prove for this above g(x) ...
g'(x) = a*n*x^(n-1) + b*m*x^(m-1)
See if you can now prove this 'power series expansion' works?
i.e. SEE IF ...
for y = 1 + x/1! + x^2/2! + x^3/3! + ....
does y' = 1 + x/1! + x^2/2! + x^3/3! + ....
You will learn that this amazing relationship of exponential growth to the base e ...
means that the amount is continuously growing at an instant by instant (new) rate
exactly proportional to the amount present at each instant ...
i.e. y' = k*y implies y = A*e^(k*t) implies that A is the amount present when time t = 0
Note also (exponential) decay is described by ...
y' = -k*y implies y = A*e^(-k*t) and thus A is the amount present at the start of the decay measurement when time t = 0
To verify the above two, you will NEED to 'know' ...
If y = e^(k*x) then y' = k *e^(k*x)
See if you can show this by substituting k*x for x
everywhere .... in the 'power series'
then find y' AND SEE IF y' = k*y
A second way to find e^x comes from ...
firstly defining ln(x) ...
If ln(x) is defined as the area under the curve of 1/t,
for t from 1 to x, for x > 0
we can then see that the inverse function is e^x
If you Google 'y = e^x' ... for example ...
http://math.usask.ca/emr/examples/expon2.htmlhttp://www.youtube.com/watch?v=Yo-UN392NDchttp://www.youtube.com/watch?v=IcZIeuOzAB0http://www.math.utah.edu/~wortman/1050-text-ef.pdfAnd then 'reflect' that curve y = e^x in the (straight) line y = x
that 'reflected curve' is the graph of 'the inverse function'
y = ln(x) ...
And vice-versa ... If you reflect y = ln(x) in the line y =x, then you obtain the graph of y = e^x
Note: ln(x) is pronounced 'lawn x'
or ... 'log to the base e of x'
or ... 'natural logarithm of x'
Also recall that if you have a graph of y = f(x) for some x, with a <= x <= b and a, b and x all 'real', then you can also read, from (range of) that graph, the values of x, for differing values of y (thus an inverse relationship EXISTS). Thus one CAN find x for differing y, as well as y for differing x, just from the graph!!!
http://en.m.wikipedia.org/wiki/Domain_of_a_functionNote: the area under the curve of y = 1/x from 1 to e is 1
i.e. ln(e) = 1 (Just as expected ... for e = 2.71828... !!!)
http://arcsecond.wordpress.com/2011/12/17/why-is-the-integral-of-1x-equal-to-the-natural-logarithm-of-x/(But more about the proof, to follow, re. this above 'area' definition of ln(x) being the same, i.e. evaluating to the same, as the the inverse definition that ln(x) is the inverse of e^x, for x != 0, for all real x)
(Computer program link inserted ... here it is now ...
http://developers-heaven.net/forum/index.php?topic=2602.msg2993#msg2993re. demo to find e by finding the value x that makes the area under the curve of 1/t from 1 to x sum to a total area of value = 1, an Archimedes type method of repeated finner approximations to e so as to make area, i.e. ln(e) ever nearer to 1 )
Note: if y = e^x
then ln(y) = x
And e^(ln(y)) = y (take ln of both sides to verify)
So note well: e is the base of natural logs
Recall e = 2.718281828... and on and on ...
with non repeating decimals
Recall e^1 = 1 + 1/1 + 1/2! + 1/3! + 1/4! + ...
Note n! reads as n factorial
meaning 1*2*3*4*5* ... *n
And n! becomes very large very quickly as n grows larger
e^x = 1 + x/1! + x^2/2! + x^3/3! + x^4/4! + x^5/5! + ...
The above series expansions for e and e^x can be used to find as many decimal places of accuracy as you may need for any calculation using e or e^x
(Link to be inserted ... here it is now ...)
http://developers-heaven.net/forum/index.php?topic=2602.msg2994#msg2994 re. computer program to find e^x to up to a 200,000 decimal places.)
Our third (the first historically) approach to finding ... e .... (and to finding expressions to evaluate e)
arises from compounded growth ... as in compounded interest in financial dealings or bacteria growth.
If 1/n represents the fraction of growth in each compounding period ... and if we compound the growth for n periods ... then as n grows very large ... at the end of all n compounding bits, the original amount will have grown (nearly) 2.71828... = (nearly) e times the original amount.
Note: Compared to simple (linear or arithmetic sequence) progression, where after n periods of adding in a (1/n)th part each period, the sum of original 1 + n x 1/n = (1 + n/n) i.e. 2 times the original amount ... we see rather, that in compounded growth ... comparable to the progression of terms in a 'geometric sequence' ... that after n compounding terms, that the new term then in the geometric sequence is now (1+1/n)^n times the original, (which is always greater than 2, but less than 2.71828... as we see from below.)
If you keep calculating (1 + 1/n)^n ...
(use your calculator and also find binomial expansion)
as n gets to be larger and larger ...
you can see that the limiting value of (1 + 1/n)^n =
1 + 1/1! + 1/2! + 1/3! + ... = 2.718281828... = e
One can use the binomial expansion of (1+x/n)^n ...
(recall coefficients are as per Pascal's Triangle) ...
then let n grow arbitrarily large ...
and see how that series approaches the very same series as for e^x
1 + x/1! + x^2/2! + x^3/3! + x^4/4! + x^5/5! + ...
(All the above talk about 'limits' means we need to be looking into the formal definition and properties of limits.)
http://www.mecca.org/~halfacre/MATH/limits.htmlhttp://www.rose-hulman.edu/FC/lc/tutorial/limit-continuity.php3(More to follow ...)
Now a very useful feature of semi-log graphing paper ...
If you are doing an experiment that involves EXPONENTIAL BEHAVIOUR ...
NOTE that if y = A*e^(k*t)
Then ln(y) = ln( A*e^(k*t) ) = ln(A) + k*t
This is the same form as the graph of a straight line with slope k and y intercept ln(A)
So ... IF the scale is logarithmic on the y axis, one can just read the y intercept to find ln(A) and then
taking the anti-log, find A ... and k is just the slope of the straight line.
2. for y'' = -y type relationships, where (the rate of change of (the rate of change)) is related to the 'thing' in a 'negative feedback' like way ...
The only ('real') solutions are (of form) ...
y = a*sin(x) and ...
y = b*cos(x)
Note: we have yet to define cos(x) or sin(x)
Now, this is true IF ...
(sin(x))'' = -sin(x) and ...
(cos(x))'' = -cos(x)
And that is true IF ...
(sin(x))' = cos(x) and ...
(cos(x))' = -sin(x)
Note: a 'complex plane' solution to y'' = -y
could take the form ...
z(x) = A*e^(i*x), where i is sqrt(-1) ...
which, when you substitute in i*x for x,
in the power series expansion of e^x ...
THIS leads to the 'power series definitions':
cos(x) = 1 - x^2/2! + x^4/4! - ...
sin(x) = x - x^3/3! + x^5/5! - ...
And then, we can see that z'' = -z when...
z = A*e^(i*x) = A*(cos(x) - i*sin(x))
So now we can evaluate these 'circular/periodic' functions for any 'real' x
Can you begin to see how e, e^x, e^(i*x) with i = sqrt(-1) ...
and the 'whole calculus' .... IS ...
THE BRIDGEto extend the one dimension real number line,
to be extended now to the 2D (complex) plane ? ? ?
So the general (real 1D) solution to this y'' = -y relationship would have the form:
y = a*sin(x) + b*cos(x)
(i.e. 'linear multiples' of an 'orthogonal basis'
Note: we can see that cos(x) = sin(x+pi/2)
So ... cos(x) 'LEADS' sin(x) by 90 deg's, check the graphs
Note: cos(x) reaches the 'peak value' of 1, 90 deg's before sin(x)
So we also have: cos(x-pi/2) = sin(x) for all 'real' x )
http://m.dummies.com/how-to/content/comparing-cosine-and-sine-functions-in-a-graph.html http://www.electronics-tutorials.ws/accircuits/AC-inductance.html Or for z'' = -z, the general 2D solution would take the form ...
z = A*e*(i*x) = A*(cos(x) - i*sin(x))
Note also:
If on an 'xy plane rectangular coordinate system', if you graph an unit circle with center at the origin, so that x^2 + y^2 = 1 describes the relationship between the allowed values of x and y ...
(The radius r is always 1 here, since it is an 'unit circle')
If t is the angle from the line y = 0, i.e. the x-axis, to the radius line r = 1, then we can define new 'functions of the circle' / 'circular functions' such that:
y = sin(t)
read as y is sine of angle t
x = cos(t)
read as x = cosine of angle t
Recall an unit circle has a circumference of 2*pi, (calculated via Archimedes 'limit' method), so the angle t, measure cc from the x-axis, is just that fraction (its ratio) of the circumference of an unit circle, i.e. the arc length of that sector of the unit circle divided by 2*pi ...
THIS defines how we measure A PLANE angle in 'radians' !!!
Then here (in our 'unit circle, centered at the origin on the 'plane') ...
r^2 = 1^2 = 1 = cos^2(t) + sin^2(t)
Since Pythagoras showed us that the 'square on the hypotenuse is the sum of the squares on the other two sides of the right angle' ...
And this yields the equation for an unit circle for angle t, radius r = 1, in polar coordinates, r, t, (here r is constant = 1)
Note here ...
We are relating circles and their (related) triangles and using the property of similar triangles having equal corresponding angles and also similar triangles having ALL their sides grow proportionately ... if a triangle or circle is expanded ... Recall C = 2*pi * r
Implies ... d(C)/dr = d(pi*r)/dr = 2*pi ( where here we use notation dC/dr instead of notation C' )
which agrees with the constant rate of growth that we know exists here.
The rate that C grows with r here ... is the constant 2*pi, thus C = 2*pi*r
Note also: r^2 = (r*cos(t))^2 + (r*sin(t))^2
We have yet to show that both the above two approaches, to defining cos(x) and sin(x), give us functions that evaluate to the same consistent values for all 'real' x
One way to do this, would be to show that for both definitions ...
(cos(x))' =-sin(x) AND
(sin(x))' = cos(x) AND then, for say x = 0 ...
THAT BOTH definitions of each of cos(x) and sin(x)
EVALUATE to the same values.
(We could further verify with say ...
x in { 0, pi/6, pi/3, pi/2, pi, 3*pi/2, 2*pi }
3. Note below (the place of) special ln(x) ...
Below using d(y) to mean y' (i.e. y primed)
in relationships like y = x^n
(read as ... y = x to exponent n)
d( x^n) = n*x^(n-1)
...
d( x^3) = 3*x^2
d( x^2) = 2*x^1 = 2*x
d( x^1) = 1*x^0 = 1
d( x^0) = 0 **zero slope**
d( ln(x)) = x^(-1) **NOTE**d(1/x^1) = -x^(-2)
d(1/x^2) = -2* x^(-3)
d(1/x^3) = -3*x^(-4)
...
Above edited/typos-fixed for (neg exponents) on 2015-01-03
https://www.whitman.edu/mathematics/calculus/calculus_03_Rules_for_Finding_Derivatives.pdfhttp://www.zweigmedia.com/RealWorld/proofs/powerruleproof.htmlhttp://de2de.synechism.org/c3/sec33.pdfhttp://www.math.hmc.edu/calculus/tutorials/limit_definition/http://www.math.uakron.edu/~dpstory/tutorial/c1/c1d_tp.pdfhttp://betterexplained.com/articles/calculus-building-intuition-for-the-derivative/I plan to show how to 'derive' the above relationships ... in a following page.
(Edit: added new appendix showing a derivation of the 'chain rule' on 2015-01-04 and thus this next link)
http://developers-heaven.net/forum/index.php/topic,2601.msg3139.html#msg3139Note also: for x^n, this can (only) be differentiated only n+1 times ... and then we have 'zero'
Or n times ... and then we have a constant.
Where as ln(x), 1/x, 1/x^2, ... can be repeatedly differentiated as often as you wish!!!
Recall y = m*x + b is form of equation that has for it's graph a straight line with slope m and y intercept b
So here y' is m (for the equation y = m*x + b)
(The slope, just like we would expect.)
Also note that the 'b' is just a 'constant' value and so (the term) d(b)/dx = 0 and 'drops out' here in the expression for y'
4. For unit circle in (complex) plane...
Recall i = (-1)^(1/2) read as i = square root of (-1)
z = cos(t) + i*sin(t) = e^(it)
z' = i*cos(t) - sin(t) = i*e^(it)
z'' = -z
Note 1st dif... rotates all cc pi/2
Note 2nd dif... rotates all cc another pi/2 ...
z'''' = z implies ...
After 4 rotations cc of pi/2 ... Then back to starting condition
If z is a velocity vector
Then z' is an acceleration vector at RIGHT angles
What does all this mean for circular motion, rotations of 'solid bodies', (conservation of) angular momentum ... regarding ', '', ''', and back to start position via '''' (i.e. the 4th derivative) ? ? ?
Foot note for 2 above. cos(t) and sin(t) are 90 degrees apart!!!
Note you may contact the author via ...
http://secureservices.cadwzavitz@secureservices.cahttp://sites.google.com/site/andeveryeyeshallseehim/P.S. Following is a link to PDF file of a 'Classic' textbook on Calculus
It contains loads of worked examples and insights into the history of science as it was 100 years ago. Note: then 1 billion in the UK was a million millions or 10^12
The American system today has 10^9 as a billion and 10^12 as a trillion
http://www.oxforddictionaries.com/words/how-many-is-a-billionEnjoy ...
http://djm.cc/library/Calculus_Made_Easy_Thompson.pdfOr a shorter online PDF updated just last month
http://www.whitman.edu/mathematics/calculus/An other rich resource ...
http://www.sosmath.com/calculus/calculus.html
