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General => Ask the Author => Topic started by: David on January 06, 2013, 02:18:56 AM

Title: Beginning Calculus ...
Post by: David on January 06, 2013, 02:18:56 AM

This new thread is to help prepare students for ...

Beginning Calculus

Update:  please see this next link:

http://developers-heaven.net/forum/index.php/topic,2636.0.html

FREE homework help NOW available ...

You can contact me via:
http://sites.google.com/site/andeveryeyeshallseehim/home/he-comes
(See e-mail link near bottom of first page)
http://developers-heaven.net/forum/index.php?topic=2587.0

(Note to Computer Programming Students:
Many computer programming problems require a good grasp of Calculus.  As the pages on this topic of Beginning Calculus are developed, I hope to link to several demo computer programs that require some knowledge of Calculus.)


The goal here is:

In an as simple/straightforward way as possible, and using a 'step by step build' ... to get a good foundation/grasp of Calculus ...

so as to give a 'head-start' for serious students that need to pass Calculus and then want to use it with understanding.


We will try to make good use of many existing useful Calculus 'texts' already up on the web ...

and a top-down/bottom-up 'Modular Computer Programming' style here  ...

in our attempt to avoid the too common modern science fallacy of 'circular reasoning'.

One example of this fallacy in circular reasoning, an example that needs to be actively exposed, goes something like this:

If you ask an (evolutionary) geologist, how to know that certain (sedimentary) rocks, (rocks we now know, from lab tests, are laid down by fast moving flood water, when it slows, as per 'Noah's Flood') ... if they are of a certain age ... you may be told that 'the type of fossils in that rock tells its age.'

So off you go to the (evolutionary) paleontologist and ask to know the age of those fossils, and you are told: 'The geologist has supplied me with the date of that rock.'

So ... the (evolutionary) presumption of LONG ages is 'presumed true' from the start ... here ... a fallacy.

Euclid's proofs are a 'step by step' build that are a worthy classic in logic, used by Archimedes and Newton ... and were taught even when I did my High School Geometry, all the way up to Ontario grade 13, ending June '66.

Enjoy ...

The 'top-down' goes like this ...

We will 'name and describe' some major modules right at the start, which we need to 'call' as we progress in 'running' our 'program' ... but we also must ensure that each module (and sub-module) is 'correct' and 'well built' ...

From the 'bottom up' ...

We will ensure the we build each module and sub module from appropriate 'true primitives' of our 'machine programming language', before our program is 'finished' and we expect it to run at all, let along, correctly. 


Firstly: Pre-Calculus ...


Review of basic Math, Geometry, Algebra
   
Note: e^x means 'e to the exponent x', A*e means 'A times e'

Recall 'Bedmas'  ... so exponent is calculated first ... then multiply by constant value 'A'

( So you will 'know' how to evaluate an expression like A * e^x )

'Bedmas' is a mnemonic that means first do work inside brackets, then do exponent, then do division and/or multiplication in the order they occur, then do addition and/or subtraction in the order they occur ... All these are 'binary operations' ... Input a, b ... then doing 'binary op' ... output c


Number types:


Natural counting numbers (positive integers)


Discovery of 'zero' ... (extends natural numbers to include 0)


Discovery of negative integers ... extends 'whole' number line to 'left'
(Now we can solve any equation like y = x + b, and have y, x and b ALL be (just) integers)


Discovery of rational numbers a/b, (extends base number set to include decimal fractions, with a, b in set of all integers and b != 0)
(Now we can solve any equation like y = m*x + b and have y, m, x and b ALL be (just) 'rationals')
Recall 'trick' to find a and b in rational q= a/b = 0.123123etc...

1000 * q = 123.123123...
            q = 000.123123...

Now subtract ...
So    999 * q = 123
and            q = 123/999
Thus can find integers a and b and thus we show that any repeating decimal is a rational number. 


Note!!!

Can you see now, how we are moving from discrete counts of quanta, to a continuum?

You will want to note that between any two different rational numbers, a, b with a < b, no matter how close together they might be, we can always insert the one rational, (a+b)/2, or the 9 rationals a + k*(b-a)/10 where k takes values 1 to 9 ...  And between each of these we can keep on inserting, and so on ...

We will see that Calculus helps us to find ways to evaluate, to as many decimal places of precision as desired, quantities that are 'real' ... but may NOT be 'rational' ... i.e. NOT 'ratio' of two integers!  (We already know about some such numbers like square root of 3, for example, or pi.  You may have learned an algorithm to find the square root of any decimal fraction, to as many places of decimal precision as you like, but do you know a method, like Archimedes did, over 2200 years ago, to find pi to as much precision as desired?)

http://itech.fgcu.edu/faculty/clindsey/mhf4404/Archimedes/Archimedes.html

See link to modern computer program that uses an Updated Archimedes Method of finding pi ...

http://developers-heaven.net/forum/index.php?topic=2602.0


Discovery/extension to the 'real continuum' ... i.e. the 'complete' number line to include numbers that are non repeating decimals like pi, e, sqrt(2), etc. 

Note: you may be required to prove some 'real' ... like sqrt(2) ... is not a 'rational' on an exam.
 
An often used method here is *proof by contradiction*. 
So firstly, assume 'true' and that for some integers a, b ... sqrt(2) = a/b with NO common factors
(i.e. fraction is in 'lowest terms'), then square each side and 2*b^2 = a^2 ... which implies that a^2 is 'even' ... which implies that a is 'even' (since an odd times an odd is odd) ... So ... now can re-write a as 2*c ... So substituting '2*c' in the place of 'a' ... (2*c)^2 = 4*c^2 = 2*b^2 ... And thus ... 2*c^2 = b^2 ... Which implies that b^2 and then b are even.   

So we see 'here' that both a and b must 'then' be even.   But this contradicts that a/b was in lowest terms with NO common denominator (like 2 here).  So we conclude that sqrt(2) is NOT a rational number. 

To show that a number like e is not a rational, we could show that it is a non-repeating infinite decimal sequence ... by looking at the infinite series expansion for e below:

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ...

(Note: n! means 'n factorial' and n! = 1*2*3* .... *n
Can you see that n! becomes ever much larger as n grows larger?)

... and see that adding in the next term, in a never ending series of ever increasingly SMALL decimal bits ... ensures the next decimals in the sequence have no repeating pattern, since the new factor in the ever increasing factorial, that we next divide by, will keep adding new re-scrambled smaller bits .... 'all the way' ... out to infinity ...


Discovery/extension to 'complete' 2-Dimensional Plane via 2D vectors or z = r*e^(i*t) = r*(cos(t) + i*sin(t)) and i has the value sqrt(-1)

Above begs question(s) of what is(are) ...
e,
cos(t),
sin(t)

...


Hint:

Think of music, like a violin string vibrating, think of the the alternating current induced into an electrical circuit by a microphone near by, think of the radio wave vibrations that fill our skies, think of nerve pulse electrical activity coursing though you right now, as you read this, think of the waves on the lake as a fish jumps up and breaks through the surface ... and dives back down into the deep ...  Ah ... we are emersed in so much beautiful and intricate sine wave and exponential activity ... Don't you think?



What I am seeing myself ... and from some recent research of well seasoned Calculus Professors ... that what is especially needed as a necessary Pre-Calculus preparation ... is a good grounding in Algebra and Problem Solving with close attention to excellent skill in 'details manipulation' of even fairly complex algebra ... so as to get the correct result in the end ... and to 'know' that your answer is correct 100 % of the time, (or nearly most of the time.)

The reality to me seems ... that although the math needed in ones routine practice will soon enough become 'routine' ... the 'bar' to even get to 'practice' ... often seems 'pretty high'. 


More to follow ...


P.S.

You may be interested in what motivated Newton in his 'discovery and pursuit of the tools of calculus' ...

"The greatest Scientist of all time, Sir Isaac Newton, was also a great student of the Bible. In fact, his motivation to grind telescope lens, to better track the planets positions in the sky, to study and to discover the laws of gravity, motion and the laws of calculus that allowed him to predict, with great precision, where the planets would be in the sky, back in Bible times and into the future to our days … was … because he wanted to see, if he was part of that generation that would still be living when the Lord Jesus, the Anointed Holy One of Israel, would return.  However Newton came to realize, that that time was ahead, even for this present generation.  First Israel had to be become a nation again (in 1948), and then Jerusalem had to come back to Israel (in 1967), in order for all the trouble that was predicted to occur with all the nations being troubled over Jerusalem … and then for Jesus to return, to save his people that remain, just outside Jerusalem, at the same spot He ascended into heaven about 2000 years ago."


The above 'excerpt' was taken from ...

http://developers-heaven.net/forum/index.php?topic=2587.msg2918#msg2918


P.P.S.

An other perspective from MIT ... 'The Big Picture of Calculus'

http://www.youtube.com/watch?v=UcWsDwg1XwM


Jumping in to see a very common example of applied calculus ...

http://www.dummies.com/how-to/content/how-to-analyze-position-velocity-and-acceleration-.html

http://www.mathcentre.ac.uk/resources/uploaded/mc-web-mech1-10-2009.pdf

http://www.youtube.com/watch?v=d-_eqgj5-K8


On the next page, please *NOTE THIS VERY AMAZING AND VERY USEFUL FORMULA*:

Euler showed that e^(i*pi) + 1 = 0
(Recall that i is defined as sqrt(-1) ... thus i^2 = -1 and i^4 = 1)

Also, please note this typo error, on next page here corrected
Implies ... d(C)/dr = d(pi*r)/dr = 2*pi ( where here we use notation dC/dr instead of notation C' ) <wrong version>
Implies ... d(C)/dr = d(2*pi*r)/dr = 2*pi ( where here we use notation dC/dr instead of notation C' ) <corrected version>

Title: Re: Beginning Calculus ...
Post by: David on January 06, 2013, 02:49:58 AM

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=UcWsDwg1XwM

Where 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.html

http://www.youtube.com/watch?v=Yo-UN392NDc

http://www.youtube.com/watch?v=IcZIeuOzAB0

http://www.math.utah.edu/~wortman/1050-text-ef.pdf

And 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_function


Note: 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#msg2993

re. 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.html

http://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 BRIDGE

to 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.pdf

http://www.zweigmedia.com/RealWorld/proofs/powerruleproof.html

http://de2de.synechism.org/c3/sec33.pdf

http://www.math.hmc.edu/calculus/tutorials/limit_definition/

http://www.math.uakron.edu/~dpstory/tutorial/c1/c1d_tp.pdf

http://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#msg3139


Note 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.ca
dwzavitz@secureservices.ca
http://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-billion

Enjoy ...

http://djm.cc/library/Calculus_Made_Easy_Thompson.pdf

Or 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

Title: Re: Beginning Calculus ...
Post by: David on January 09, 2013, 03:44:36 AM

Below, we will start to fill in some of the details, hinted to above.


Firstly, we need to know how to find y' for any y of the form y = a * x^n, for any real a, x, and n an integer > 0

because ...

we are going to suppose that we can find a solution to our main problem of ... find some function, f(x), for any 'real' x, so that f'(x) = f(x)

we are going to suppose 'a solution' may take the following 'power series' form: (for any real x)

f(x) = a0 + a1*x + a2*x^2 + a3*x^3 + ... (and so on ... out to infinity) ...

with all the 'a' coefficients being 'real' ... and the increasing exponent (power) of x^n having n as a positive integer only ...

Once we have 'solved' ... and found all the coefficients ...
(if a solution exists) ...

we can then use that 'power series' expression to
obtain some increasingly significant decimal places
when we need to evaluated f(x) for some real x

IF hopefully, this infinite series, converges to some unique real value, for each real x input.

A 'result' we need to use in our solution process here,
(to find all the 'a' coefficients for the above is):

If y = a * x^n, then y' = a * n * x^(n-1)
for all n in 1, 2, 3, ... and a, x in the 'reals'

We will 'assume' this y' = a * n * x^(n-1)
result here, and prove it below, using 'limits'.   

( We already 'know' the result for the case for ...
  y = c  when  y' = 0, since the slope of an horizontal line is 0,
  and for y = m*x + b, when y' = m, since the slope of a line is m
  Note: y intercept of b for the straight line has NO effect on slope )

Now, if we add the 'initial condition' that ...
f(0) = 1
we are now ready to solve for all the 'a' coefficients in the above power series ...

If      f(x) = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 + a5*x^5 + ...
then f(0) = 1 implies a0 = 1


If      f(x) = 1 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 + a5*x^5 + ...
then f'(x) = a1 + a2*2*x + a3*3*x^2  + a4*4*x^4 + a5*5*x^4 + ...
But   f'(0) = f(0) = 1 implies that a1 = 1

If     f'(x) = 1 + a2*2*x + a3*3*x^2  + a4*4*x^3 + a5*5*x^4 + ...
then f''(x) = a2*2 + a3*3*2*x + a4*4*3*x^2 + a5*5*4*x^3 + ...
But   f''(0) = f(0) = 0 implies a2 = 1/2

If    f''(x) = a2*2 + a3*3*2*x + a4*4*3*x^2 + a5*5*4*x^3 + ...
then f'''(x) = a3*3*2 + a4*4*3*2*x + a5*5*4*3*x^2 + ...
But   f'''(0) = f(0) = 0 implies a3 = 1/3!

If    f'''(x) = a3*3*2 + a4*4*3*2*x + a5*5*4*3*x^2 + ...
Then  f''''(x) = a4*4*3*2 + a5*5*4*3*2*x + ...
But   f''''(0) = f(0) = 0 implies a4 = 1/4!

And so a5 =1/5! ... and so on ...

So a power series solution to f'(x) = f(x) is ...
f(x) = 1 + x/1!  + x^2/2! + x^3/3! + x^4/4! + ...



Now to prove...

If y = a * x^n, then y' = a * n * x^(n-1)
for all n in 1, 2, 3, ... and a, x in the 'reals'

Note: we actually need to prove that ...
If      y = a*x^n + b*x^m
then y' = a*n*x^(n-1) + b*m*x^(m-1)
( for all real a,b,x and positive integers m,n > 1 )
But we will take this in steps ...

Above, I have already supplied several links, that each cover, the fairly standard 'limit approach'.  But let's start out here from the beginning ...
to make sure we avoid all fallacy of 'circular reasoning'.

We will assume for now, (proofs to follow), several limit theorems that we will need here in our proof that:

y' = a*n*x^(n-1) + b*m*x^(m-1)

The basic approach used to prove the above is the idea that for any 'nice' f(x), (like we have here), then f'(x) is the limiting slope of secants that get nearer and nearer to the tangent to the curve, at x

Note!!! In this proof, h is some arbitrarily small rational like 1/n, for a very large n

i.e.
If y = f(x)
By definition then y' = lim h -> 0 of  ( f(x+h) - f(x) ) / h

1.  If f(x) = c, then f'(x) = 0
     This, we take as obvious,
      since the slope is everywhere 0,
      and does NO-where, depend on x

2.  If f(x) = m* x + c, the f'(x) = m
     This too, we take as obvious,
      since the slope is everywhere, the constant m,
      and does NO-where, depend on x

Note: for 1. and 2. above we do NOT use any limit theorems!

3.  If f(x) =m * x^2,
     we aim to show that f'(x) = 2 * m * x
     Let y = g(x) = x^2
     Then lim h -> 0 ( (x+h)^2 - x^2 ) / h =
      lim h -> 0 ( 2*x*h + h^2 ) / h =
      lim h -> 0 ( 2*x+ h ) = 2*x
      So whenever g(x) = x^2
      Then we have g'(x) = 2 * x
     
      And since lim x  -> a of  k * u(x) =
      k * ( lim x  -> a of u(x) )

      Then we may conclude that for f(x) = a * x^2,
      f'(x) = a * 2 * x


4.  Now taking general case of g(x) = x^n

     Then lim h -> 0 ( (x+h)^n - x^n ) / h =

     lim( x^n + n*x^(n-1)*h + n*(n-1)*x^(n-2)*h^2/2! + n*(n-1)*(n-2)*x^(n-3)*h^3)/3! + ... + h^n  - x^n ) / h =

     lim( n*x^(n-1) + n*(n-1)*x^(n-2)*h/2! + n*(n-1)*(n-2)*x^(n-3)*h^2)/3! + ... + h^(n-1) ) =

     n*x^(n-1)
     
     Now since lim x -> a of ( k * u(x) + m * v(x) ) =
 
     k*( lim x  -> a of u(x) )  + m*( lim x  -> a of v(x) )
     
     We can conclude that ...

     If      y = a*x^n + b*x^m
     then y' = a*n*x^(n-1) + b*m*x^(m-1)


So ... our power series solution 'works' as a solution to the problem of finding some f(x) so that if y = f(x) we always have y' = y

BUT ... can we also show that this power series has the desired 'exponent' behavior?

Recall .. we are looking for ...

e^a * e^b = e^(a+b)
e^(-a)      =  1/(e^a)

We can show the above two hold for our power series expression. 

And using the above, we can show that ...

(e^x)^n = e^(x*n) for all integer n

But how do we show ... (for all real r)

(e^x)^r = e^(x*r) ?

Let's try taking the ln of both sides ...

ln((e^x)^r) = r * ln(e^x) = x*r

ln(e^(x*r)) = x*r

So now ... all we need to know is that ...

ln(a^b) = a * ln(b)

Looks like we are 'stuck' in a big circle of 'circular reasoning' ...

Here's an attempt to 'unlock' the above circle ...

http://www.softmath.com/tutorials-3/relations/notes-on-ax-and-logax.html

But also this next link (that was also given on the previous page) ...

http://arcsecond.wordpress.com/2011/12/17/why-is-the-integral-of-1x-equal-to-the-natural-logarithm-of-x/

If you have studied through the above links, you may see the importance of the 'continuum' ... (and limiting values) ... in transitioning from 'all the rationals' ... to ... 'all the reals' ...


Footnote:

See also MIT introducing e^x ...

http://www.youtube.com/watch?v=oo1ZZlvT2LQ


     

Title: Re: Beginning Calculus ...
Post by: David on January 10, 2013, 10:30:38 AM

This page is reserved for the limit theorems we need to prove and talk about ...

The following looks like some good links to get started here ...

http://www.youtube.com/watch?v=MccSGWZdM_A

http://www.youtube.com/watch?v=kyE1TyblyLg

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html

http://www.sosmath.com/calculus/limcon/limcon01/limcon01.html

http://www.sosmath.com/calculus/calculus.html

Note:  We say that a function, f(x), is continuous ... at the input value 'p' ... if the value f(p)  =  lim x->p f(x)  (of course the limit needs to exist)
           We say that a function, f(x), is continuous over some input domain, if it is continuous at each value in that domain.

http://mathworld.wolfram.com/ContinuousFunction.html


Below are some 'limit theorems' you may like to do some 'web research' to see if you can (get help to) prove (if you need help):

1.  The limit of the sum of two functions is the sum of their limits.
2.  The limit of the difference of two functions is the difference of their limits.
3   The limit of a product of two functions is the product of their limits.
4.  The limit of a constant times a function is the constant times the limit of the function.
5.  The limit of a quotient of two functions is the quotient of their limits, provided the denominator does not tend to zero.

For some demo proofs, please see this next link:

http://en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules


But here are the three type of limits, we have actually used so far, that we need to prove, 'bottom up' ...

lim h -> 0   of  ( 2*x*h + h^2 ) / h  =  lim h -> 0  of  ( 2*x+ h )  =  2*x  ... see (1) above

lim x  -> a  of   k * u(x)   =    k * ( lim x  -> a  of  u(x) )   ... see (4) above

lim x -> a  of  ( k * u(x) + m * v(x) )  =    k*( lim x  -> a  of  u(x) )  + m*( lim x  -> a  of  v(x) )   ... see (1,4) above

This link ... linked from above will give considerable 'research material' ...

http://www.math.ucdavis.edu/~kouba/ProblemsList.html

for ...

Problems on the continuity of a function of one variable
Problems on the "Squeeze Principle"
Problems on the limit definition of the derivative
Problems on the chain rule
Problems on the product rule
Problems on the quotient rule
Problems on differentiation of trigonometric functions
Problems on differentiation of inverse trigonometric functions


Also ...


Problems using summation notation
Problems on the limit definition of a definite integral
Problems on u-substitution
Problems on integrating exponential functions
Problems on integrating trigonometric functions
Problems on integration by parts
Problems on integrating certain rational functions, resulting in logarithmic or inverse tangent functions
Problems on integrating certain rational functions by partial fractions
Problems on power substitution


Lots to do eh ... Enjoy! ...

More to follow ... All the above links should give a 'substantial start'.



Footnote:


The general relationship noted in 'Geometry', between the circumference and the radius of any circle could be expressed like this:

d( C(r) )/dr  =  k, where k, r, C(r) would all be 'reals', k some constant value, r the independant variable and C, the dependant value.

Thus, a general solution to the above would look like this:  (which graphs as a straight line)

C(r) = k*r + b

Because when we 'differentiate back' we get back ...

d( k*r + b )/dr  =  k


So ... how do we get C(r) = 2*pi * r as 'the solution' ?


Well, we know that when r = 0, C(r) = 0, and this implies then, that  b = 0


So now we have:

C(r) = k*r for all real r


So all we need then, is some 'experiment' to find our constant value k

Recall Archimedes 'experiment' above, noting that 'k' does NOT depend on 'r', so we can do our experiment for a circle with any 'r' ... and will get the same result for k.  (Thus, Archimedes just used a 'unit circle' to fine k/2 = pi = 3.14159... )



Now contrast this with our 'problem' of finding e ...

A general relationship, for which we sought a solution, concerned our noting that the rate of something 'growing' depended on the (instant by instant changing) quantity of that 'something' ...

Which we expressed as:

y' = k * y, where k and y are 'reals' and k is some constant value

Then our general solution looked like this:

y = A * e^(k*t) ... because y' is then  k * A* e^(k*t)  =  k * y   (Note that A is some constant real number here also.)

Now here we found that e had the constant value of e = 2.71828... = 1 + 1/1! +1/2! + 1/3! + 1/4! + ...
And noting that ln(e)  =  1, then taking the ln of both sides of above equation, we have ...

ln(y) = ln(A) + k*x  ... which we note is like the graph of a straight line, slope k, 'ln(y) intercept' here being 'ln(A)'

This suggests using semi-log graph paper on our y axis to find the slope 'k' and the 'ln(y) intercept' value of 'ln(A)',
then taking the anti-log of the intercept, we find the value for 'A', the initial amount present at the start of our experiment.

So ... we see here, for this type of relationship 'experiments' ...

A is the initial amount at the start of each experiment,
and k is the 'experimentally found' constant that we can use for all similar experiments of this identical process

then ... we can expect that the new, instant by instant amount,
for any experiment of this identical process, to be given by ...

y = A * e^(k*t)  as the time t progresses from the start of the experiment, forward



Can you see that the Calculus here ...

is about finding the type of equations that predict the future of some process that is governed by a 'simple law' ?


Example 1:  C(r) = 2*pi * r           (linear growth)
Recall this was the particular solution (to the problem concerning circles) C'(r) = k (which holds for ANY circle)

Example 2:  y(t) = A * e^(k*t)      (exponential growth)
Recall this was the general solution to the problem y' = k*y


To use Calculus to find roots, max and min of polynomial, see this next link ...

http://developers-heaven.net/forum/index.php?topic=2602.msg2997#msg2997

Title: Re: Beginning Calculus ...
Post by: David on January 22, 2013, 07:11:49 AM

I have checked out a lot of material on the web and linked to some of what looks relevant.  But it seems to me that there is a big problem of circular reasoning in the standard presentations ...


The following is a slightly edited series of e-mails to a friend:

First letter:

The problem I'm seeing is in the transition from x^n or x^(n/m) ... to x^r

and could be stated like this:

We can progress to defining x^(n/m) where x is any real, n and m are integers with m > 0
(and x and n not both zero at the same time) ...

BUT how do we get to x^r where r is any real ? ? ?

If we use x^r  = e^( ln(x^r) )   ...which is just the the definition of log

then how do we get that to be equal to e^( r * ln(x) ) for any real r  ? ? ?

when all we can really show is that the above holds for rational  r ...

(but we want r to be any real) ? ? ?


If ...

The definition x^r = e^(ln(x^r))

Is used to get to (i.e. to prove?)
the derivative of x^r (for any real r)
i.e. to prove that ...
d( x^r )/dx = r * x^(r-1)
where r is any 'real' number

The problem in transitioning
from r in rationals to r in reals ...
seems to exist then in (shifting) ...

The primary definition of ln(x) to ...
ln(x) = Area under curve of 1/t from 1 to x

Where, using above (Area under curve of 1/t) we can show ...
(for all real a, b > 0 and n, m in 1, 2, 3...)
(1)  ln(a*b) = ln(a) + ln(b)
(2)  ln(a/b) = ln(a) - ln(b) = -(ln(b/a))
(3)  ln(1/a) = -ln(a) = ln(a^(-1))
(4)  ln(a^n) = n*ln(a)
(5)  ln((1/a)^n) = -n*ln(a) = -ln(a^n) = ln(a^(-n))
(6)  ln(a^(1/n)) = 1/n * ln(a)
(7)  ln((a^n)^m) = n*m*ln(a) = ln(a^(n*m))
( 8 )  ln((a^n)^(1/m) = n/m * ln(a) = ln(a^(n/m))

See 2nd program at this link to see the above being used ...

http://developers-heaven.net/forum/index.php?topic=2602.msg2993#msg2993


But how do we show (4)..( 8 ) above ?
When n, m are not integers and
i.e. when n and/or m are real (m != 0)?

It seems we have just 'shifted the problem' of defining x^r
for real r
from the exponent here above ...
to the exponent in ln(x^r) ? ? ?

Quite a 'slight of hand' ... but still no proof!


Next letter to friend:

If we use definition of ln(x) as the
Area under curve of 1/t from 1 to x

As the primary definition of ln(x)

Then we can show all my previous
(1) to ( 8 ) re. ln(...)

For all real a, b > 0
And for m, n in 1, 2, 3...

If then ( 8 ) ...

ln( (a^n)^(1/m) ) = ln( (a^(1/m))^n ) =
n * ln(a^(1/m)) =
n/m * ln(a) =
ln( (a^(n/m) )


Is used as a 'starting point' ...

Then since rational number line is 'dense' ...

i.e. we can approach,
from above or below, any real number r
with a rational as close as we wish ...

So we can approach the value of

r * ln(a)

By using from above and below ...
closer and closer ... n/m

such that ...

as n/m -> r

then we also have  (n/m * ln(a))  ->  (r * ln(a))

Thus we have 'validated'

our definition of

x^r = e^(ln(x^r)) = e^( r * ln(x) )

Quite a 'deep' proof that really merely acknowledges the continuity of God's created Heavens ...

In the beginning, God (aleph tau) created the heavens and the earth.
(Genesis 1:1)

So we see here the origin of ...
Time (continuous)
Space (continuous)
Matter (quanta/discrete charge)

And us made from that 'discrete matter'

Baruch HaShem!



The 3rd e-mail, a reply to a friend about this ...


Hi my friend,


Thanks for taking a look and the great question. 


If you ...

Take two rational numbers a < b

Then (a+b)/2 is in-between a and b
And so is a+k(b-a)/10, k in  {1,2,3...,9}
Then in-between each of the above,
we can insert 9 more rationals ...
And so on ... And so on ... And ...

This ... is a little of what is meant when one says that the rational number line is dense.

Also ...

we can approach any real number ...
from above and from below ...

***as close as we like***

by rational numbers. 

For example, take pi ...

3 < pi < 4
3.1 < pi < 3.2
3.14 < pi < 3.15
etc...
The sequence values -> pi from each sequence, both from below and from above. 

Or take a rational like 2

1.9 < 2 < 2.1
1.99 < 2 < 2.01
1.999 < 2 < 2.001
etc...
An < 2 < Bn

As n -> inf,
then An -> 2
and  Bn -> 2

Note:

In computing, all we really have is rationals, since with limited memory to hold only so many digits, we can only approximate the real value, (for so many values we deal with that can't be expressed as integers or the ratio of two integers.)


From the definition of ln(x) as Area under curve of 1/t as t goes from 1 to x ...

We can show my previously labelled (1) to ( 8 ) theorems hold for the Area ln(x) definition i.e ln(x) = Area under curve of 1/t as t goes from 1 to x

Demo case (1) ...

Firstly take case of b > a > 1

Area under 1/t from 1 to ab
=
Area under 1/t from 1 to a
+
Area under 1/t from a to ab
=
Area under 1/t from 1 to a
+
Area under 1/u from 1 to b
(note set u = a*t)

Thus we have shown from areas under curve of 1/t ...

ln(a*b) = ln(a) + ln(b)


And recalling ( 8 ) was ...

Area under curve of 1/t
for t = 1 to t = x^(n/m)
is ... (as shown by areas under curve)

ln( x^(n/m) ) = ln( (x^n)^(1/m) )
                   = ln( (x^(1/m))^n )
                   = n*ln( x^(1/m) )
                   = (n/m)*ln(x)

But ... what we are looking for is a way to handle ...

Area under curve of 1/t
as t goes from 1 to r
(where r is any real)

What we have is ...

Area under curve of 1/t
as t goes from 1 to n/m
(where n, m are integers, m > 0)

ln( x^(n/m) ) = ln( (x^n)^(1/m) )
                   = ln( (x^(1/m))^n )
                   = n*ln( x^(1/m) )
                   = (n/m)*ln(x)

So ... if we just take closer and closer rational n/m approximations to r, (as we
only can in computing, anyways) ...

Thus we claim ...
in the limiting case of all the (n/m) 's ...

As n/m -> r

ln( x^r ) = r * ln(x)

And so our idea of replacing ...

x^r with e^( ln(x^r) ) = e^( r * ln(x) )
holds ...

And then we can do ...
d( x^r )/dx =
d(  e^( r * ln(x) ) )/dx =
(r/x) *  e^( r * ln(x) )  =
r/x * x^r   =  r * x^(r-1)

So our formula holds for all r, (excepting x, r both zero)

QED.

Thanks for your question.  I've been wanting to write something ... (more clearly?) ... about this ... for a few days now. 

The 'continuum' of time and space and electromagnetic fields that extend indefinitely ...
in all directions ...

is in distinct contrast to the 'discrete (pos and neg) quanta' of conserved charge found with each proton and electron, (which quanta of moving charge seem each to generate all these continuous electromagnetic fields.)

For a model of the electron and proton that well describes this, and seems to 'closely fit' the present observations of reality around us, please see ...

http://www.commonsensescience.org/

We are highly complex discrete creations made in the image of our (every-where present) Creator ... made out of the very discrete elements He first created ... and then we ... and all our constuent elements are dependent continuously upon the elements around us ... 

But we were also breathed upon personally by our Creator ... so as to be able to commune with Him ... in and by continuously depending upon His very and every Spirit/Breath!

Yeshua/Jesus revealed clearly that God is Spirit ... and that He is looking for those who would worship (see the supreme high value of) His every breath, seeing the eternal abundant life He has for us in each and every breath.

Yeshua/Jesus became discrete like us, but is filled with the Spirit of God 'without measure' (i.e. unbounded) ... and we are invited to be 'complete' ... (reconciled to an entirely Holy God) ... in and by Him ... simply by (continual faith) in Him i.e. faith in His Faithfulness.  (Really, when you think about it just a little, it is apparent, that ONLY God can keep His promises.  He is really the ONLY ONE who is able TO DO what He says He will do ... and when He says He will do it!!!)

So although we are discrete, God has placed 'eternity in our hearts' that we would seek after Him and find Him, for HE IS always very near to each of us .. but desires that we 'see' him with the eyes of faith ... by examining His creation ... and by researching His dealings with us, especially via the Holy Book of Books (The Holy Bible) He had inspired for us ...

beginning with Adam, down to Noah, Abraham, Moses ... all the prophets ... and finally ... the long promised Saviour and Messiah ... who was God's long promised lamb to take away the sin of this world ...

the root sin being NOT believing in God's Word ... (as per Adam and Eve in the garden). 

So ... we are invited to return to our Omnipresent, Omnipotent and Eternal Creator and experience hearing His Voice, His Word (like Abraham did) ... and following Him. 



Best regards

David


P.S.

In a fashion similar to 'e', e^x, ln(x) being the bridge from the rationals to the reals ...

Yeshua/Jesus is the bridge between us and the Eternal All Holy God. 

http://secureservices.ca
dwzavitz@secureservices.ca

http://sites.google.com/site/andeveryeyeshallseehim/

Sent from my iPhone

On 2013-01-21, at 13:15, my friend wrote:

Hello David,

I was reading the equations (1) to ( 8 ) and... until now, in every course we had as definition that the exponent is always a Natural number (as in 1,2,3,...).  But on going for the prove... I haven't been lit up for what goes where, still.  Would you please explain this line? "Then since rational number line is 'dense'"

Title: Re: Beginning Calculus ...
Post by: David on January 04, 2015, 01:49:45 PM

Appendix added 2015-01-04

A little demo of the derivation of the 'chain rule' ...


Note:

We WILL see (on next page now) that we have ...

(1)  d(f*g) = f * dg   +   g * df   (where f and g are both differentiable functions of x)

and thus ...

d( f/g ) = -f/g^2 *dg    + 1/g * df   =  (g*df - f*dg )  / g^2 .... with g(x) not zero, for all considered x


AND ... we WILL see ... using the chain rule, (for function composition), we have ...

(2)  d( f(x)^(n/m ) )  =  (n/m) * f(x)^((n/m) - 1) * d( f(x) )
for ALL differentiable f(x) ... where x is some real value ... and n/m is any rational, positive or negative ( or even zero, BUT BOTH n and x can NOT be zero at the same time ... SINCE .... (0)^(0) is NOT defined !!! )


HOWEVER ... WHAT we REALLY need  to show firstly ...  is that ... for all rational r ...and real x ... that ...
d( x^r ) = r * x^(r-1)   where r is a constant and x is the real variable raised to power r
 
To start, we can find some n/m = r, where n and m are integers, m not zero ...
( let's say m > 0 and ratio is in lowest terms, i.e. there are no common factors,
 but note this is not needed since n/m has ONE (unique) value, same as kn/(km) )

( THUS ... function composition and the 'chain rule') is what we want to use here ... since ...
  x^(n/m) can be evaluated as (x^n)^(1/m) )
   
 
 So ... we now want to show that ...
 d( x^(n/m) ) =  (n/m) x^(n/m - 1)
 
 Ok, let's re-write  x^(n/m)  as  (x^n)^(1/m)  and differentiate using function composition (and chain rule) ...
 
 d( (x^n)^(1/m) )
 
 = (1/m)*( x^n )^( 1/m - 1 ) * n*x^(n-1)
 
 = (n/m) * (x^n)^( 1/m - 1 ) * x^n / x ... (after top and bottom of last factor both times x)
 
 = (n/m)/x * (x^n)^( 1/m -1 + 1 )
 
 = (n/m)/x * (x^n)^( 1/m )
 
 = (n/m) * ( x^(n/m ) ) / x
 
 = (n/m) * x^( n/m - 1 ) ... which is what we wanted to show ... q.e.d.
 
 
 ( But, but, but  ... we have yet to show that d( x^(1/m) ) = (1/m)*x^(1/m - 1) ... where m is an integer)
 
 We could do this by using the fact that if y = x^(1/m) ... then y^m = x
 then dx/dx = 1 = m*y^(m-1) * dy/dx
 
 Now we can solve for dy/dx ...
 
 so dy/dx = (1/m) * 1 / ( y^(m-1) )  = (1/m) * y^(1-m) ... but y = x^(1/m) ... so can substitute for y back in ...
 
 so dy/dx = (1/m) * (x^(1/m))^(1-m)  = (1/m) * x^(1/m - 1)   ... which is what we wish to show ...q.e.d.
 
 
So now ... all we need to do is to establish this very important result we get from using the  'chain rule' :

d( f(x)^r ) = r * f(x)^(r-1) * d( f(x ) )

i.e.  We YET NEED to PROVE the 'chain rule' !!!

i.e. that for the case of y = f(u) ... with  u = g(x) ...  so that we then have y = f( (g(x) )  ... then:

dy/dx  =  dy/du * du/dx  =  d(f(u))/du * d(g(x))/dx


Ok ... we will suppose for now ... that we have looked at limits and that ... we understand ...

(see these next links to get a good introduction to this)

https://www.khanacademy.org/math/differential-calculus/taking-derivatives/secant-line-slope-tangent/v/slope-of-a-line-secant-to-a-curve
https://www.khanacademy.org/math/differential-calculus/taking-derivatives/derivative_intro/v/calculus-derivatives-1-new-hd-version
https://www.khanacademy.org/math/differential-calculus/taking-derivatives/derivative_intro/v/calculus-derivatives-2-new-hd-version
https://www.khanacademy.org/math/differential-calculus/taking-derivatives/derivative_intro/v/alternate-form-of-the-derivative

... what is meant by the limit definitions of the derivative for 'nice' functions ...
y = f(u) and u = g(x) and so then y = f( (g(x) ) ...
( where u = g(x) is differentiable and is u' (u primed).... and f(u) is differentiable and is f' (f primed) )

Note!  We are below using  D  to stand for operator  delta  = 'some small change in'
... and h to represent some (small) value h, THAT ALSO approaches 0, as Du->0 )

So ... since we can find some real (usually very small value) h, for any (usually very small value) Du ...
and thus can write:

Dy = y'*Du  + h*Du   ( If you see the graph in the next link below, of this, you can see this quite readily,
                                    that (total) Dy from x=a to x=b  IS  exactly the sum of these two (lesser) parts )

http://www.4shared.com/download/5HFOcfQbce/fundamental_lemma.PNG?lgfp=3000

( ... this above line, is an extremely important fundamental lemma of calculus )
( where h->0 as Du ->0 for all considered u for f(u) )

Thus ... Dy/Dx = y'*Du/Dx + h*Du/Dx ... where now, as Dx->0, also Du->0 and so also h->0 ...

And thus ... in the limit where Dx->0, also Du->0 and also h->0 and Dy->0
... we can see that: dy/dx = y'*du/dx  ( since also then, the factor h->0 ) ... and this is the chain rule we were seeking to show!

Note!  We also can express this 'chain rule' (in an easy to remember form) as:  dy/dx = (dy/du) * (du/dx)

So now, we can apply our 'chain rule' to y = u^(n/m) where u = f(x) ...
And thus ...

y'  =  (n/m)*u^((n/m) - 1) * u'  =  (n/m)*((f(x))^((n/m) - 1) * (f(x))'

... which is what we set out to show ... q.e.d.

Title: Re: Beginning Calculus ...
Post by: David on January 04, 2015, 09:20:41 PM

Ok ... now lets get a proof for (1)  ...

(1)

d(h)   

=   d(f*g)   

=   f * dg   +   g * df   ( where f and g are both differentiable functions of x and h = f*g )


(Edited/typos fixed on 2015-01-04, from an example proof up on the web at Wiki...)
http://en.wikipedia.org/wiki/Product_rule

A rigorous proof of the product rule can be given using the definition of the derivative as a limit,
and the basic properties of limits.

Let h(x) = f(x) g(x), and suppose that f and g are each differentiable at x0.
(Note that x0 will remain fixed throughout this proof ... but that, REALLY, x0 represents ANY point where both f and g are differentiable!)
We want to prove that h is differentiable at x0 and that its derivative h'(x0) is given by
f'(x0) g(x0) + f(x0) g'(x0).

(Note!  We again are using  D  to represent the  delta  operator.)
Let Dh = h(x0+Dx) - h(x0); note that although x0 is fixed, Dh depends on the value of Dx,
which is thought of as being "small."

The function h is differentiable at x0 if the limit

lim Dx->0 of Dh/Dx exits ...

When it does exist, h'(x0) is defined to be the value of the limit.


As with Dh ...

let
Df = f(x0+Dx) - f(x0)
and
Dg = g(x0+Dx) - g(x0)

which, like Dh, also depends on Dx.

Then
f(x0+Dx) = f(x0) + Df
and
g(x0+Dx) = g(x0) + Dg.

It follows that
h(x0+Dx)
= f(x0+Dx) * g(x0+Dx)

=  (f(x0) + Df) * (g(x0) + Dg) (and multipying out we see that ...)

=  f(x0)  * g(x0) + Df * g(x0) + f(x0) * Dg + Df * Dg
 
To get the value of Dh, subtract h(x0) = f(x0) * g(x0) .... from above:

Dh  =  Df * g(x0) + f(x0) * Dg + Df * Dg

So ... as Dx->x

Dh/Dx -> f' * g(x0) + f(x0) * g' + 0  (Note that df*dg is '2nd order' so to goes to zero much faster that dx)

Thus ...

dh/dx  =  df/dx * g  +  dg/dx * f   (  where h(x)  =  f(x) * g(x)  )

q.e.d.