Lloyd Blog

[Lloyd]

Derivative of sqrt(x)

Airman said:
The derivative of
y = x (1/2)
is
y’ = (1/2)x(-1/2)

Airman and Cr6, can you show in brief detail how to get that answer (y' = (x/2)^-1/2) from Mathis' method?
Also, would either of you like to quote briefly from Mathis' papers (such as where links are given above by Airman) to show a little about how he solves log(x), sin(x) etc? Does he solve them by finding constant differentials as above?

David said Mathis' method is:
x: 1, 2, 3, 4, 5
sqrt(x): 1, 1.414, 1.732, 2, 2.236
∆sqrt(x): 0.414, 0.316, 0.268, 0.236
∆∆sqrt(x): -0.098, -0.048, -0.032
∆∆∆sqrt(x): 0.05, 0.016
∆∆∆∆sqrt(x): -0.034

The triangles mean "delta" which I think means "rate of change of". Right?
And "rate of change of" means the rate at which the differences between integer values of x increase or decrease. Right?
Does David's table indicate that the sqrt(x) never comes to a constant difference between values of x?
Or does it show that the constant difference is -.034 at the fourth level?
Can each of the lines of text in David's table be shown on a graph, if y = sqrt(x)?

Is the Derivative the Slope of the Tangent to a Curve?
Here's a graph of y = x^2 {y = x squared}, y = x {the diagonal dashed line}, and y = x^1/2 {y = square root of x, shown as x = y^2, which is the same}.

Does the derivative mean the slope of the tangent to a curve?
The graph of y = x doesn't curve, so the tangent is the same as the entire line, but the graphs of the other two equations do curve and it looks like they curve identically, but just in different directions on the graph. So is the solution for the derivative of sqrt(x) (which is also written x^1/2) similar to that for x^2?
(The slope is 90 degrees at x = 0, then it quickly diminishes toward 0 degrees, nearly parallel to the x-axis. Are derivatives expressible as degrees of slope of the tangent to a curve?)

P.S., I plan to discuss other interesting things here eventually, besides Mathis' stuff.

[LongtimeAirman]

David said: Hold on, I never said the power rule (y'=nx^n-1) doesn't work. Of course it works. It was discovered by Cavalieri in 1630, before the advent of calculus. But it only appies to polynomials, and cannot be used for sin(x), ln(x), exp(x), etc.To solve the problem using the Mathis method you must first construct a table for the sqrt(x); that is, showing the table of differences. The table would contain sqrt(1), sqrt(2), sqrt(3), and so on. You must build this table for sqrt(x).All you have done is use Cavalieri's power rule, which was discovered before Newton and Leibniz were even born. You must provide the Mathis table of differences for sqrt(x) as a solution. Otherwise, you are just using Calarieri's solution; not Mathis'.

The power rule is the most important derivative formula in calculus. It fell directly out of Miles' table, Miles didn't just borrow it. Cavalieri, Newton and Leibniz used it but they couldn't explain it because they were thinking of infinitesimals. Miles has done something new here and you refuse to accept it.

This table is a great simplification. I used it and found what you said wasn't there. Miles is using it as a basis to find the derivatives of all the functions you say it cannot be used for. He is putting a solid new base on the calculus. Is it your complaint that Miles hasn't reproduced 2000 years of work in the spare time he has devoted to it in the last 10 years.

David said: If his mathematics is garbage (and it is my contention that it is), then his theories are sitting on a rotten foundation.Mathis has redefined calculus, and has developed new methods for finding the derivative of a function. So let's closely examine this new calculus that Mathis has created and see if it can hold up to scrutiny. Isn't that the way science is suppose to proceed; by actually testing the theory?

You put forward a sqrt(x) table and call it a Mathis table of differences and say "good luck". Is that the spirit of science?

sin(x) is addressed in http://milesmathis.com/trig.html
log(x) in: http://milesmathis.com/ln.html
exp(x) in: http://milesmathis.com/expon.html

REMCB

[David]

LongtimeAirman,

Mathis has mistakenly assumed that he can use the power rule for all functions. But you can't. It only works polynomials. And that is the reason why his method fails; it cannot be used for non-polynomials.

For example: find the derivative of (sin(x))^2

Using the power rule, the answer would be 2sin(x). However, that's the wrong answer. You cannot use the power rule for a non-polynomial. The correct answer is 2sin(x)cos(x).

Incidentally, the above example is the very mistake that Mathis makes in his trig functions article.

[Chromium6]

LongtimeAirman,

Mathis has mistakenly assumed that he can use the power rule for all functions. But you can't. It only works polynomials. And that is the reason why his method fails; it cannot be used for non-polynomials.

For example: find the derivative of (sin(x))^2

Using the power rule, the answer would be 2sin(x). However, that's the wrong answer. You cannot use the power rule for a non-polynomial. The correct answer is 2sin(x)cos(x).

Incidentally, the above example is the very mistake that Mathis makes in his trig functions article.

David, you aren't MM' mom are you? Even approximately approaching her? Like going to zero or infinity you can get close (to his Mom) but not actually become her even via a good tailor with pant-suits in series using the same fitting? Where is your zero in the real world? On your belly button?

Even my mother, who is a professional mathematician, has failed to see how I can incorporate my table into an analysis of all functions. She never got angry, but she has used my silence as proof that my method has limited use.

http://milesmathis.com/trig.html

[Lloyd]

If you guys would give specific examples with somewhat detailed explanations, it would be easier to get an idea if you're making sense.

David said: For example: find the derivative of (sin(x))^2
Using the power rule, the answer would be 2sin(x). However, that's the wrong answer. You cannot use the power rule for a non-polynomial. The correct answer is 2sin(x)cos(x).

Here's a graph of (sin(x))^2:

From the graph the derivative of (sin(x))^2 (assuming the derivative is the tangent to the curve at a point) appears to be 0 degrees when x is 0. Am I right? Can the derivative be given in degrees? Or is it given as a fraction, or what? When x is 0, pi/2, pi, etc the derivative or tangent or the slope appears to be 0 or horizontal. Am I doing this right? So can someone explain what 2sin(x)cos(x) means and show that it conforms to some of the tangents to the curves on this graph? Can you say what 2sin(x)cos(x) equals when x = 1, 2, 3 etc? Or does x need to be stated as fractions of pi?

[David]

Lloyd,

Yes, I agree with everything you just said. When x is 0 or pi/2, the tangent to the curve is horizontal and would have a slope of zero.

You can use any value for x; it doesn't have to be a multiple of pi.

The derivative is the slope at any point on the curve. To find the slope at x=0, just plug in the value: 2sin(0)cos(0) = 0.

Similarly, when x=pi/2:

2sin(pi/2)cos(pi/2) = 0.

[LongtimeAirman]

Derivative of sqrt(x)7

    Airman said:
    The derivative of
    y = x (1/2)
    is
    y’ = (1/2)x(-1/2)

Airman and Cr6, can you show in brief detail how to get that answer (y' = (x/2)^-1/2) from Mathis' method?

First, Miles said (probably in calcsimp, forgive me if I paraphrase), "Everyone fails calc-1 (and two),  even nerds. Someone who can memorize formulas without trying to understand them may do well, but everyone else fails it." It boils down to the fact that calculus is described in infinitesimals, and infinitesimals really don't exist. So calculus is a sort of a gateway math where you have to give up some of your physical certainty and just calculate.  This table, however, is something else.  It's easy to interpret, while rich in all the possibilities of higher power functions. Spend some time reading his papers and studying his table and you will understand calculus. Well, a lot more than most mathematicians currently do.

Miles' original calculus paper: A Re-definition of the Derivative (why the calculus works—and why it doesn’t. http://milesmathis.com/are.html
Miles' simplified calculus paper: http://milesmathis.com/calcsimp.html

I'll start where the table has been trimmed down to to show the relationship between the exponential curve and the rate of change of that exponential curve
2x = Δx2 
3x2 = Δx3 
4x3 = Δx4 
5x4 = Δx5 
6x5 = Δx6

The last part just needs to be reversed to

Δx2 = 2x
Δx3 = 3x2
Δx4 = 4x3
Δx5 = 5x4
Δx6 = 6x5

The rate of change, Δ, of x to the nth power, equals n times x to the (n-1) power,   Δy^n = y’ = nx^(n-1).

The assignment is to find the derivative of sqrt(x).
Find y' when: y = x^(1/2).

We are given, n = (1/2), so,

y’ = nx^(n-1)
y' = (1/2)x^((1/2)-1)
y' = (1/2)x^(-1/2)

Also, would either of you like to quote briefly from Mathis' papers (such as where links are given above by Airman) to show a little about how he solves log(x), sin(x) etc? Does he solve them by finding constant differentials as above?

David, Take note. Miles has created new tables, and, every function doesn't necessarily require a new table. New tables (I think) include: ln (x), 1/x, and e.

I've cited the MM papers, multiple times. You're making arguments without support. I'll look at your assertion concerning the derivative of (sin(x))^2, even though Miles makes the substitutions I mention below.

In " Trig Derivatives foundwithout the old Calculus", http://milesmathis.com/trig.html, Miles starts with,
y = sin x = ±√(1 - cos2 x) , then substitutes
y = a = ± (1- b^2)^(1/2)
And then uses the table already described. He differentiates both a and b to get:
y' = b = cos x

In "My Calculus Applied to Exponential Functions", http://milesmathis.com/expon.html, Miles uses both the table already given as well as a new table listings for e. The notion that a new table is necessary for each function is false.

In, " The Derivatives of the Natural Log and 1/x are wrong", http://milesmathis.com/ln.html, Miles makes new ln (x) and 1/x tables.

    David said Mathis' method is:
    x: 1, 2, 3, 4, 5
    sqrt(x): 1, 1.414, 1.732, 2, 2.236
    ∆sqrt(x): 0.414, 0.316, 0.268, 0.236
    ∆∆sqrt(x): -0.098, -0.048, -0.032
    ∆∆∆sqrt(x): 0.05, 0.016
    ∆∆∆∆sqrt(x): -0.034

Nope. I disagree. This is David's table, far as I know.

The triangles mean "delta" which I think means "rate of change of". Right?
And "rate of change of" means the rate at which the differences between integer values of x increase or decrease. Right?

I would say, "... differences between EXPONENTIAL values of x increase or decrease.

Does David's table indicate that the sqrt(x) never comes to a constant difference between values of x?
Or does it show that the constant difference is -.034 at the fourth level?
Can each of the lines of text in David's table be shown on a graph, if y = sqrt(x)?

There are no constant differentials in that table. It's not Mile's table. It has only one "variable". Why would we need a new exponential table when we already have one?

Is the Derivative the Slope of the Tangent to a Curve?

Here's a graph of y = x^2 {y = x squared}, y = x {the diagonal dashed line}, and y = x^1/2 {y = square root of x, shown as x = y^2, which is the same}.
Image
Does the derivative mean the slope of the tangent to a curve?

That's what we've been taught, but there are several problems with comparing lines on graphs to reality, especially at "points" in reality or on the curve.

Distance, velocity, acceleration, and changing acceleration of any order represent the sort of varying behavior that calculus was intended to describe.

So, Yes! (I think). That's all for now.

REMCB

[David]

LongtimeAirman,

Alright, if you don’t trust the table that I created, then let’s take a look at a table that Mathis created. This is taken from his ln(x) article:

ln(x) 0, .6931, 1.099, 1.386, 1.609, 1.792, 1.946, 2.079, 2.197
Δln(x) .6931, .4055, .2877, .2231, .1823, .1542, .1335, .1178
ΔΔln(x) .2876, .1178, .0646, .0408, .0281, .0207, .0157
ΔΔΔln(x) .1698, .0532, .0238, .0127, .0074, .0050
ΔΔΔΔln(x) .1166, .0294, .0111, .0053, .0024
ΔΔΔΔΔln(x) .0872, .0183, .0058, .0029

Now using the above table, can you find the derivative of ln(x)?

Of course you can’t; and neither could Mathis. It’s just a bunch of meaningless numbers. If you read his article you will discover that Mathis admits that he can’t use this table to solve the problem; instead, he computes a rough approximation based on average values using this formula:

slope @ (x,y) = [y@(x + 1) - y@(x - 1)]/2

Which is the whole point that I have been making; his tables will not work for non-polynomials.

[David]

Chromium6,

Personally, I think it is inappropriate to drag Mathis' mother into this discussion. But if that is where you want to steer this conversation, then please discuss that topic with someone else.

[Lloyd]

Thank you all for trying to answer my questions. Feel free to carry on with the discussion, if yous like, but I personally want to start discussing something else now, though I'll continue to read whatever yous may say about Mathis' calculus etc.

Proof against Geological Dating
I read at least a couple years ago that coastal erosion disproves mainstream geological dating, but I had forgotten about it.

Continental drift very likely did occur, but it occurred very rapidly and very recently and gradually. See http://newgeology.us for details.

Coastal erosion and continental shelf sedimentation are proof against continental drift having occurred more than about 5,000 years ago.

This website, http://oceanworld.tamu.edu/resources/oc ... rosion.htm says: Average erosion rates are 6 feet per year along the Gulf and 2 to 3 feet per year along the Atlantic. [] Beatley, Brower, and Schwab (2002).

This video, http://youtube.com/watch?v=OCR6-nH3WMA, says most coasts worldwide are eroding faster than the Atlantic coast is, so that would be more than 2 feet per year. It also says the continental shelves are gaining a few feet of sediment per year.

Conventional science claims that most continental drift occurred over 200 million years ago. But if erosion has been going on for all that time, the coasts would have eroded 380 miles in 1 million years and 3,800 miles in 10 million years. So the continents would have all eroded into the oceans within 20 million years and there would be no land above sea level.

There is a lot of other evidence that the continents are young. Several major rivers are known to be less than 20,000 years old. That includes the Mississippi River.

I could include evidence from my old thread on the Grand Canyon etc too.

[Lloyd]

Grand Canyon Discussions
These are from my previous discussions of the Grand Canyon.
- Earth's Surface
http://www.thunderbolts.info/wp/forum/phpB ... 135#p67787
http://www.thunderbolts.info/wp/forum/phpB ... 165#p71350
- Northwest
http://www.thunderbolts.info/wp/forum/phpB ... 086#p71222
- Cook
http://www.thunderbolts.info/wp/forum/phpB ... =10&t=7294
- Saturn Flares
http://thunderbolts.info/wp/forum/phpBB3/v ... =30#p79239
- C14 Dating
http://www.thunderbolts.info/wp/forum/phpB ... =60#p72183
- Cardona Interview
http://www.thunderbolts.info/wp/forum/phpB ... 4&start=90

[Lloyd]

Continents Are Under 20,000 Years Old
The above info proves that the present continents are very young, but there was a supercontinent which broke up, leaving the present smaller continents etc. Determining the age of the supercontinent is harder to do. The breakup likely occurred about 4,500 years ago. Before that the supercontinent must have been rather flat, since all of the present mountain chains were likely formed during the breakup. The continents are formed from sedimentary rock, mostly shale, i.e. mud, which must have been deposited during the Flood. The supercontinent must have consisted mostly of clay soil.

Submarine Canyons
The submarine canyons on the east and west coasts of North America, near NYC and LA, must have formed while the oceans were down in those areas, because canyons don't form under water. When the ocean was down in the Atlantic, it was probably high in the Pacific and vice versa, as a huge tsunami moved around the globe and left a deep trough behind it, until the seas eventually settled into about the present level. The submarine canyons must have formed as the flood waters on the continents drained off into the nearly empty ocean basins. Tsunamis are said to travel at about 500 mph. Since the Earth is about 25,000 miles in circumference, it would take tsunamis about 50 hours, or about 2 full days, to go all the way around the globe.

Tsunami During Continental Drift
When North America broke away from Europe (due to an asteroid impact near east Africa), the Appalachian Mountains would have formed first on the east coast and a flood from the Pacific would have flowed across North America (that was just before the Rocky Mountains formed on the west coast, so there was nothing to stop the flood). This tsunami flood possibly formed the St. Lawrence seaway, the Mississippi River, the Hudson River and the Hudson canyon in the continental shelf near NYC, while the Atlantic Ocean basin was opening up and ocean waters were gradually starting to fill it from the south and the north. After about 12 hours of North America sliding over the Moho layer, the west coast would have begun to run over the east Pacific ridge, causing western North America to become elevated, and as the plate movement slowed, friction would have increased, causing the Rocky Mountains for form. When the plate stopped, some of the flood waters would have poured off of California as the Pacific Ocean waters now were moving away from shore, due to inertia, and the canyon near LA would have formed from the retreating flood waters.

Grand Canyon
The flood waters left two large lakes just north and east of the present Grand Canyon location, called Grand and Hopi Lakes. When the lakes breached the plateau, maybe 200 years after the supercontinent breakup event, the canyon formed from the ensuing water flow through the breach in the plateau, when the sediments were still somewhat soft.

[LongtimeAirman]

David said:
Now using the above table, can you find the derivative of ln(x)?Of course you can’t; and neither could Mathis. It’s just a bunch of meaningless numbers ... his tables will not work for non-polynomials.

David,

The Table is not a bunch of meaningless numbers. It is also not a make-or-break piece of gotcha. The Table is a list of differentials based upon the integer number line applied to the function. It is a tool for examining the functions behavior.

The power function diminishes nicely, and there are constant differentials, and y’ = nx^(n-1) falls out. For that alone, the methodology is justified.

With that in mind, I agree, but I don't know if it will not work for all non-polynomials. Maybe that could be a definition of well behaved. The tables indicate that the derivatives (and integrals) of 1/x and ln (x) diverge, (if I understand correctly). But more importantly, for those two functions, Miles shows that the historical derivation of those derivatives yielded close approximations, but were not the true numbers.

Miles' larger mission is to share his investigations and ideas. Like a kid, every rock he turns over exposes its own little world. There's a lot that needs exposure. Let's move on.

REMCB

[David]

“My method applies to all of calculus and all functions, not just differentials or polynomials. It applies to trig functions, logarithms, integrals, and so on." -- Miles Mathis

The following are Mathis created tables, and posted at his website:

log(x) 0. .301, .477, .602, .699, .778, .845, .903, .954
Δlog(x) .301, .176, .125, .097, .079, .067, .051
ΔΔlog(x) .125, .051, .028, .018, .012, .016
ΔΔΔlog(x) .074, .023, .01, .006, .004
ΔΔΔΔlog(x) .051, .013, .004, .002

ln(x) 0, .6931, 1.099, 1.386, 1.609, 1.792, 1.946, 2.079, 2.197
Δln(x) .6931, .4055, .2877, .2231, .1823, .1542, .1335, .1178
ΔΔln(x) .2876, .1178, .0646, .0408, .0281, .0207, .0157
ΔΔΔln(x) .1698, .0532, .0238, .0127, .0074, .0050
ΔΔΔΔln(x) .1166, .0294, .0111, .0053, .0024
ΔΔΔΔΔln(x) .0872, .0183, .0058, .0029

1/x 1, .5, .3333, .25, .2, .1667, .1429, .125, .1111, .1, .0909
Δ1/x .5, .1667, .0833, .05, .0333, .02381, .0179
ΔΔ1/x .3333, .0833, .0333, .01667, .00952, .00591
ΔΔΔ1/x .25, .05, .01667, .00714, .00361
ΔΔΔΔ1/x .2, .0333, .00953, .00353

The above tables are clear and obvious examples where his method doesn’t work. A far more difficult task is trying to find a table where his method does actually work for a non-polynomial. So far, I have yet to find even one working example. Which makes the following quote, laugh out loud hilarious:

“My method applies to all of calculus and all functions, not just differentials or polynomials. It applies to trig functions, logarithms, integrals, and so on." -- Miles Mathis

[LongtimeAirman]

“My method applies to all of calculus and all functions, not just differentials or polynomials. It applies to trig functions, logarithms, integrals, and so on." -- Miles Mathis http://milesmathis.com/trig.html .

I believe he is reffering to his paper: A REDEFINITION OF THE DERIVATIVE(Why the calculus works—and why it doesn't) http://milesmathis.com/are.html

Miles has redefined the derivative, based on intervals instead of points. The Table in that paper justified his approach. He is well along in his further developments, though "lacking" seems to be a main complaint.

REMCB