Author Topic: The Poetry of Logarithms (or, Be Drunk on Math)  (Read 1658 times)

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The Poetry of Logarithms (or, Be Drunk on Math)
« on: June 18, 2018, 02:00:14 pm »
"You have no idea how much poetry there is in a table of logarithms."
      ~ Karl Friedrich Gauss to His Students


I felt myself slipping ever so slightly into a depressive state, but I trudged along, focusing in on an older textbook, hoping that some novel exercises might revive my passion for compuation and calculation.   I want to make it clear that I perceive myself more as a calculator and a computer than anything resembling an actual mathematician.   Maybe this had to do with my training as a youth.  We were taught to calculate and compute using algorithms, and only later in life have I been disturbed by my ignorance when it comes to reading and writing mathematical proofs.   The closest I come to the spirit of writing mathematical proofs is in writing mathematical-computational programs.

OK, so, I want to express a certain appreciation for the poetry in logarithms, not specifically tables of logarithms.

I was testing out three programs I had made from scratch that can handle irrational logarithms.  Of course, the digitization of irrationals means there will be some kind of error, so it is only an approximation, or estimate.

I wish I could write the mathematical notation here, but the reader will just have to be patient as I make do with what we have.

I will show first how a problem might be approached with the computer programs, and then how much more elegant the same problem becomes when done by hand.

Let log_[a](y) = x mean "x is the logarithm of y to the base a."

The problem is:   log_[the 4th root of 3](3*sqrt(3))

So, here "the 4th root of 3" represents sqrt(sqrt(3))

The way I proceeded with using my programs at the command line was as follows:

sqrt 3 ---->  1.73205
echo 3*1.73205 |bc -l  ----->  5.19615
sqrt 1.73205 ----> 1.31607

So, 3*sqrt(3) = 5.19615 and sqrt(sqrt(3)) = 1.31607

Finally, logn 1.31607 5.19615 ----> 6.0001

That is, log_[1.31607](5.19615)  = 6


Now, what jolted me out of my depression was when I took up my pencil and went through the exercise, and discovered that it was actually MORE BEAUTIFUL and elegant this way:


log_[the 4th root of 3](3*sqrt(3)) = x   <------>  [the 4th root of 3]^x = 3*sqrt(3)

<---->  3^(x/4) = 3*3^(1/2)  <---->  3^(x/4) = 3^(1 + 1/2)  <---->  3^(x/4) = 3^(3/2)

<----> x/4 = 3/2   <----->  x = 12/2  <----->  x = 6


I have gotten a glimpse of what Karl Gauss was trying to tell his students ....  and with just this tiny glimpse, I revive my interest to the point where I might once again be drunk on computational mathematics.

Be Drunk
Charles Baudelaire



You have to be always drunk. That's all there is to it—it's the only way. So as not to feel the horrible burden of time that breaks your back and bends you to the earth, you have to be continually drunk.

But on what? Wine, poetry or virtue, as you wish. But be drunk.

And if sometimes, on the steps of a palace or the green grass of a ditch, in the mournful solitude of your room, you wake again, drunkenness already diminishing or gone, ask the wind, the wave, the star, the bird, the clock, everything that is flying, everything that is groaning, everything that is rolling, everything that is singing, everything that is speaking. . .ask what time it is and wind, wave, star, bird, clock will answer you: "It is time to be drunk! So as not to be the martyred slaves of time, be drunk, be continually drunk! On wine, on poetry or on virtue as you wish."



Holden, do you think it is possible to be drunk on math?

I think this would entail becoming quite indifferent to academic metrics, and to insist on nurturing a very personal relationship with mathematics itself.
« Last Edit: June 20, 2018, 07:27:36 pm by Kaspar Hauser »
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The below video is only 5 minutes long, but revealed in this is a mysterious constant 2.42 which, after some reckoning, I think might be better if it were simply 2.3.



In the notation below, let lg(x) represent the logarithm to the base 10.
Let ln(x) represent the logarithm to the base e ~=~ 2.718.

If one ever wanted to calculate common logarithms (base 10) by hand without scientific calculator nor access to Logarithm Tables, it might suffice to memorize (or write on a small index card) the following approximations to the logarithms of a handful of the first several prime integers.

lg(2) ~=~ 0.30103
lg(3) ~=~ 0.47712
lg(5) ~=~ 0.69897
lg(7) ~=~ 0.84510
lg(11) ~=~ 1.04139

In the video above, there is the magical constant 2.42 in the formula:

lg(a + b) = lg(a) + b/(2.42*a)

Notice the following:

e ~=~ 2.718

lg(2.718) ~=~ 1/ln(10) ~=~ 0.43429

ln(10) = 2.30258509299405 ~=~ 2.3

What happens when we replace 2.42 by 2.3?

Then the formula becomes:

lg(a + b) = lg(a) + b/(ln(10)*a)

or, demystifying the constant term even further,

lg(a + b) = lg(a) + lg(2.718)*b/a = lg(a) + lg(e)*b/a

I can understand how inviting it is to latch on to magical constants, but it gives more insight when you explain why you use that constant and can explain how it was derived (and why it works).

lg(1211) = lg(1210) + lg(1 + 1/1210)

Here, a + b = 1211

a = 1210, b = 1

so we have lg(a + b) = lg(a) + lg(1 + b/a)

lg(1 + x) ~=~ 10^m * x * lg[1 + 10^(-m)] for 10^m < x < 10^(m+1)

The value of m is simply that which makes the term (10^m * x) lie between 1 and 10.

We need to know lg(1 + 10^[-m]) as well:

lg(1.01) = 0.00432...
lg(1.001) = 0.000434...
lg(1.0001) = 0.0000434...
etc

As m increases, the digits of lg(1 + 10^[-m]) approach those of lg(e), where, e is Euler's constant, about 2.718, and the logarithm of 2.718 to the base 10, that is, lg(2.718) ~=~ 0.43429...

So,

lg(1211) = lg(1200) + lg(1 + 11/1200) = 3.07918 + 11/1.2*lg(1.001)

lg(1211) = 3.07918 + 0.00398 = 3.08316


Let's see how this would look if applied to the example in the ZooTube video.

lg(737) = lg(7.37 * 10^2) = lg(7.37) + lg(10^2) = lg(7.37) + 2

lg(737) = lg(7) + lg(1 + .37/7) + 2

We would still need some kind of table. I am just showing where the constant 2.42 comes into play and wondering if we might be better off using 2.3 instead.

With Sage:

sage: log(7, 10) + log(1 + .37/7, 10) + 2   ----> log(7)/log(10) + 2.02236944784479

sage: _.n()  -----> 2.86746748785905

So, it's actually only ever so slightly higher than 2.866

What about if we use 2.3 as the constant?

lg(a + b) = lg(a) + b/(2.3*a)

lg(737) = lg(7.37 * 10^2) = lg(7.37) + 2
lg(7.37) + 2 = lg(7) + .37/(2.3*7) + 2

Using our estimation for lg(7) = 0.8451, and with 2.3*7 = 16.1, the only arithmetic we need to carry out by hand is .37/16.1, or 37/1610.

We can round 0.02298... to 0.023 (which has nothing to do with the constant 2.3; it's just a coincidence)

So, lg(737) = 0.8451 + 0.023 + 2 = 2.8681

----------------------------------------------------------

Which constant gives the more accurate result? 

Using constant 2.3 (about ln(10)), we are over by only 0.00063
Using 2.42, the result is off in the other direction by 0.00146

2.3 is closer by 0.00083.

So, let's use 2.3

-----------------------------------------------
Some notes taken from Dead Reckoning: Calculating Without Instruments, Ronald W. Doerfler, c.1993, pp.116 to 118

So, write the following on a small business card and stick it in your pocket or wallet:

lg(2) ~=~ 0.30103
lg(3) ~=~ 0.47712
lg(5) ~=~ 0.69897
lg(7) ~=~ 0.84510
lg(11) ~=~ 1.04139

lg(a + b) = lg(a) + b/(2.3*a)


also, ln(10) ~=~ 2.3
lg(2.718) = 1/ln(10) ~=~ 0.43429

For whatever reason, replacing the constant 2.3 with 2.33 works best.
This is quite interesting and will lead to deeper reflection on how one might use methods similar to the past generations when one is without access to calculators or tables.   This kind of knowledge (craft) would be valuable in a certain situations, for sure.
__________________________________________________________
An aside:  Holden, if I were ever held in some kind of mental hospital, this is the kind of information I would really appreciate.  So, we will have to come up with some kind of KEYWORD that you might search for in "Why Mathematics?" to find such "Jail Bird Math Hacks Essentials".

How about JAILBIRD MATH HACKS, or, to be safe, JAILBIRD MATHSHACKS

I'll hear "Math Shacks" whereas you may remember math(s) hacks - but one word.

Got it:  Key TAG to search for JBMH.

I'll test it out.
« Last Edit: June 21, 2018, 02:57:47 pm by Kaspar Hauser »
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Logarithms with pencil and paper and no tables!
« Reply #3 on: June 21, 2018, 02:53:54 pm »
I may try to contact the person who made the video, Anshul Kumar, to inquire about the significance of the constant 2.42.

2.3 gets a closer result than 2.42, but I found that tweaking the constant to 2.33 got it a couple more digits of accuracy when compared to calculator and pretty much exactly what one gets when using a 4-place table with rightmost Proportional Part as well.

So I will make a note above to this effect.
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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #4 on: June 21, 2018, 04:35:42 pm »
I left a comment at the ZooTube channel for Anshul Kumar, so I will report any insight he is willing to share with us.

In the meantime, I would suggest using the formula log10(a + b) = log10(a) + b/(2.33*a) only in rare cases when you have exhausted other methods.

First of all, for large numbers, such as 1458000, you can factor this into primes.

1458000 = 2^4 * 3^6 * 5^3

So log10(1458000) = log10(2^4) + log10(3^6) + log10(5^3)
 = 4*log10(2) + 6*log10(3) + 3*log10(5)

Note that I did some experimenting (playing, tinkering), and I found that the extra couple "memorized" digits DOES make a difference.  You get a couple extra digits accuracy in the result.

So, instead of Anshul Kumar's suggested:

log10(2) ~=~ 0.301
log10(3) ~=~ 0.477
log(5) ~=~ 0.699
log(7) ~=~ 0.845

I would opt for:

log10(2) ~=~ 0.30103
log10(3) ~=~ 0.47712
log10(5) ~=~ 0.69897
log10(7) ~=~ 0.84510
log10(11) ~=~ 1.04139

log10(1458000) = 4*log10(2) + 6*log10(3) + 3*log10(5)

= 4*(0.30103) + 6*(0.47712) + 3*(0.69897)

= 1.20412 + 2.86272 + 2.09691 = 6.16375 ~=~ 6.1638 (we only aim for 4 digits)

Great estimate!  and no tables or calculator!

With the tables, you would have looked up log10(1458000) as follows.

log10(1458000) = log10(1.458 * 10^6) = log10(1.458) + 6

To find log10(1.458), you would find row 14, column 5, and see: 1614

Then, in the far right columns (not on all tables), there may be Proportional Parts.  There, you look to column 8, the last digit in 1.458: the number is 24.

You add 1614 + 24 = 1638, so log10(1.458) = 0.1638

That is called the mantissa.  The characteristic is 6, so

log10(1458000) = 6.1638 using the tables.

Using a scientific calculator:  6.16376

Using my homemade C++ program: logn 10 1458000 ----> 6.1638

In my code, there is an option to display more digits.  I choose 14.
It returns: LOG_BASE_10(1458000) = 6.16375752398196

So, you see, if you use the above approximations for the logarithms of the first several primes, you can use common logarithms without tables or calculators (if you have to).

You may wonder what the use is.  Well, it reduces multiplication and division problems into addition and subtraction, and you can even use it to find weird roots and exponents of expressions.
« Last Edit: June 21, 2018, 04:58:58 pm by Kaspar Hauser »
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Gorticide @ Nothing that is so, is so DOT edu

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Just for the record:



Let's get down and dirty.  I, for one, find the method described in this video, while fascinating, not realistic for me.  I won't remember all the details.

Let's try this:



I'm not too crazy about this technique either as it gives little insight into what is going on.  I will once again consult Dead Reckoning: Calculating Without Instruments, Ronald W. Doerfler, c.1993; this time, pp 133-144 (Approximate Logarithmic Inverses).
« Last Edit: June 21, 2018, 08:30:34 pm by Kaspar Hauser »
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Gorticide @ Nothing that is so, is so DOT edu

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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #6 on: June 22, 2018, 10:37:09 pm »
I plan to eventually transcribe some of my notes a systematic approach from the research I have been involved with over the last few days.

I have opted to use the method shown in Doerfler's Dead Reckoning: Calculating Without Instruments, a method based upon Bemer's method which was used in 1958 in early "decimal" computers.

The idea here for finding logarithms (without tables) is to first extract powers of 10 until the number N is less than 1.  Then N is multiplied by a series of simple numbers until N is just greater (or less) than 1, at which point the formula for ln(1 + x) is invoked.

ln(1 + x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...

Working with base 10, this becomes log10(1 + x) ~=~ 0.4343*[x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...]

where the constant 0.4343  is from log10(e) = 1/ln(10) ~=~ 1/(2.3025) = 0.4343

Doerfler alters Bemer's method only in making it more precise by directly using the multipliers 1.1/.2 = 11/2, 1.1/.3 = 11/3, etc

The multipliers are simply obtained as the highest two-digit numbers that, when multiplied by the maximum value in the range, will result in a new value of N <= 1.1

For example, the range from 0.1 to 0.2 would have a maximum value of .2, so the multiplier is 1.1/.2 = 11/2

A value of N that will be between 1.0 and 1.1 will be reached in at most three multiplications.  The series for ln(1 + x) is then taken to the desired accuracy, with, as mentioned, log10(1 + x) ~=~ 0.4343*[x - (x^2)/2 + ...].

The following table is needed, and, over time, might be memorized.

I will abbreviate log10(x) as lg(x), log of x to the base 10, or "common logarithm"
___________________________________________________

N           Multiplier    Log (Multiplier)
to One
Digit
_____________________________________________________
.1          11/2           lg(11) - lg(2) = 0.740363 or just 0.7404
.2          11/3           lg(11) - lg(3) = 0.564271 or just 0.5643
.3          11/4           lg(11) - lg(4) = 0.439333 or just 0.4393
.4          11/5           lg(11) - lg(5) = 0.342423 or just 0.3424
.5          11/6           lg(11) - lg(6) = 0.263241 or just 0.2632
.6          11/7           lg(11) - lg(7) = 0.196295 or just 0.1963
.7          11/8           lg(11) - lg(8 ) = 0.138303 or just 0.1383
.8          11/9           lg(11) - lg(9) = 0.087150 or just 0.0872
.9          11/10         lg(11) - lg(10) = 0.041393 or just 0.0414
___________________________________________________________     

Note that lg(11/2) = lg(11) - lg(2)

Another important formula to remember is 10^d = 1 + 2.3*d

I personally use 10^d = 1 + 2.303*d

For cases where the exponent of 10 ends up relatively large, say, a magnitude greater than 0.005,  we can introduce correction values for 10^(a + b), where we add to the term "1 + 2.3*d".  I think this may be related to the source of Anshul Kumar's "magic trick".   I don't really care for tricks.  I like to see why it works.

These cases amount to 10^(a+b) where a = 0.02, 0.01, -0.01, or -0.02 and b <= 0.005

We will significantly improve our approximation of 10^d (by breaking d into a + b) =
 
10^(a + b) if we always add to the prescribed quantity 1 +/- 2.3*(a+b) the term (c*b + d), where c and d are given in the table (to be memorized) below:

CORRECTION VALUES FOR 10^(a + b)
__________________________
SIGN    a          c          d
------------------------------------
+        0.02     0.1     0.00113
+        0.01    0.1/2   0.00029
-         0.01    0.1/2   0.00024
-         0.02    0.1      0.00099
__________________________


After using this method on a substantial amount of exercises, I will see if I might be able to run through a couple complete examples, but it would be very frustrating to do with typing.

As you know, I prefer pencil and paper for ease of notation and drawing arrows with side-notes.

This is not the best medium for such communications.
I just wanted to be sure to get some of the numbers posted.

tag: JBMH

I will admit that there is a definite satisfaction in breaking the numbers into prime factors and using a short list of values for the logarithms of the primes, and to be able to estimate 10^x, the antilogarithm, is also quite liberating.

I still use a calculator for the arithmetic and a computer to check logarithms and 10^x and for all sorts of things.   I even check log tables to compare results.  What I found interesting is that, in a crunch, when it comes to a discrepancy in a result in a text, I have found that doing these by hand and using the correction values and longer 6 decimal place representations, that my results, when they differ from the log table and a solution given in a textbook, agree with the calculator and computer programs (which I defer to as King Authorities).

In other words, sometimes the textbook's accuracy is limited to the log tables used, where by hand I am free to reach for extra digits of accuracy that I am able to verify with computer code and CAS-driven calculators. This was a revelation in that often, when it comes to the last significant digit, my wrong answers were more correct than the textbook "answers" AND the Log Tables.  How empowering.

Poetry in Logarithms?  No, that's quite a stretch, Mr. Gauss; but maybe what he is referring to is the feeling of detached madness, that drunkenness Baudelaire was talking about?  To be so engrossed in such things in one's own mental life - kind of immune to the propaganda booming from the Idiot Box.
« Last Edit: June 22, 2018, 11:04:47 pm by Kaspar Hauser »
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

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I just found the sources on line for some of the material I was reading from a book, so the interested reader need not wait for this ape-man to type up from his notes what they can print out with a printer.

Main site: 

Area of Current Interest:  Dead Reckoning, Chapter 4: Logarithms and Their Inverses

Bemer's technique

It is not out of laziness that I point out this resource.  It is valuable, and the author even provides links to hard to track down sources.

Very fascinating stuff ...

A link at the bottom of the page on Bemer's technique for finding "exponential" - that is, antilogarithms, points to John McIntosh, aka urticator.

 I'm not bragging when I say that my hand-written notes showing the work in pencil could prove invaluable to someone, and especially invaluable to me.  What this implies is that the best way to learn these techniques is to find an old book with problems meant to be solved using tables, and then, instead of using the tables, solve from scratch using the formulas and techniques.  Then compare results and notice that your results may actually be more accurate than those listed as "solutions" in the text which thinks you are using 4-place tables.

This can be a very satisfying experience.

Of course, I realize that it is certainly not the instant gratification we "moderns" are used to.  It entails work and concentration, and, of course, undisturbed leisure where one is not plagued with anxiety and worry.
« Last Edit: June 26, 2018, 11:47:11 pm by Kaspar Hauser »
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

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« Reply #8 on: June 27, 2018, 03:51:10 am »
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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #9 on: June 27, 2018, 09:50:48 am »
I've seen it before, but I haven't found a need to use it.  It's interesting in that it does not use any trigonometric functions.  Even though the usual formula for area of a triangle does not explicitly include any trigonometric functions, as it is Area = (1/2)*base*height, in finding the height, the sine or cosine is used to help find the lengths of height or base.

I suppose if you were able to measure each side of a triangle, supposing you could actually find a triangle in Nature, then you might approximate the area with this trick.

I think I may start referring to mathematical formulas as "tricks".

As you know, I am currently under the spell of logarithmic trickery ... Trichery.
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #10 on: June 29, 2018, 11:30:29 pm »
Thanks for the response.I am sure you are aware of the convex and concave quadrilaterals.Very interesting.
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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #11 on: June 30, 2018, 06:18:32 am »
It is frustrating discussing mathematics, don't you think?

Well, it is certainly less frustrating than discussing unknowable unknowns such as whatever the word "God" or "the gods" means to anyone in particular, but consider discussing something like the trigonometric functions with someone who has just learned how to "use" them.

They may insist that they know all about the sine and the cosine, the tangent, and their inverses.   In reality, these are transcendental functions that require the notion of limits to be defined ... infinite series which converge; but learning about these is a prerequisite to studying Calculus.   The trigonometric functions are transcendental, depending on the idea of approaching a limit, depending on the idea of infinite, in short, depending on Calculus to be defined; and yet they are presented as a prerequisite to studying the very thing on which their identities depend.

It's rather circular, isn't it?  It's no wonder we are initially mystified by them.

Now, the problem with discussing something like "God" is that so many people think they know what they are talking about, and - as such, it's not such a great topic of conversation, except for blasphemy and heresy.

Now, when it comes to math, I don't mind so much discussing it as I do appreciate the challenge it presents in trying to communicate.  How does one communicate unknowing?  It is said that a little knowledge is dangerous.

I don't think so.  A little knowledge is ok as long as we have the confidence to stand up to those who wish to intimidate us.  After all, as long as we are willing to confess our general ignorance, or the possibility that not a great deal of knowledge and understanding can be retained in such matters, that it is quite possible that mathematics education may be somewhat of a sham, where so many people do a lot of posturing and pretending.

Do you get the gist of the direction in which I am leaning?

We are capable monkeys who can be trained to pass tests and to focus on the task at hand, and the to answer certain questions and solve certain problems based upon the material we may have recently been exposed to.

Have you seen all the texts available at Library Genesis?

Have you gathered more than you will be able to fathom in 100 lifetimes?

One of the reasons I like to focus on a handful of old high school math books is that it tends to keep me grounded.  Sure I have the books on Computational Physics as well as the verbose and systematic works of Kant; but when i want to feel authemtic, which is most of the time, I like to continually go over the fundamentals.

I hate pretensious snobs, and I would never wish to be one of them.

I am an honest man, and it is for this reason that employment is problematic.

One is required to lie to oneself in order to function is society.

The builders of the houses cannot confess any ignorance.   People are trained to hide their ignorance, and not to show weakness and vulnerability.

I like to be honest.  Unfortunately, honesty is not always polite and kind.  When put into practice, honesty may come off sounding hostile.  It may be offensive.

I like to say what I mean and mean what I say, and I am hoping one day to be able to write about mathematics in such a way that does not cause more confusion than already exusts in the world.

Most likely, I may only add to the confusion, unless I can muster enough honesty to force the reader to join me in this quest to uncover the truth that we will forever be in the dark and may only get glimpses here and their of a limited understanding.

Most importantly, we may suspect our entire civilization depends upon the lot of us deceiving ourselves and allowing ourselves to be deceived.

Most people do not really want to know the truths we are trying to uncover.

I vow to study math for the rest of my days even if the main thing I learn is that we are not machines that can program ourselves to understand.  We can learn how to compute and to calculate, but undertanding seems to be one of these hocus pocus mystical voodoo phenemenon based on delusion, self-deception, and pretension.

When an honest thinker comes along and confesses to knowing nothing, the world will take him at his word and hand him a bucket and a mop and some tools for cleaning toilets.

Meanwhile, the gorilla who thinks he knows everything pops open the hood off an automobile and proceeds to play mad scientist with the engine, investing money into tools and such.  He fancies himself a mechanic.

Another ape fancies himself an engineer who "designs Volvos".

And yet .... how much of this Colony of Industrial Civilizations runds as a kind of mysterious hive that not one individual ape-man can comprehend?

Each individual has enough trouble just getting through the day and eating meals, just like the creepy crawling giant mammalian insects that we actually are.

What do you find interesting about convex and concave quadrilaterals?

Is there something specific you want to mention?

I know writing in plain honest language can be a challenge when we are so used to these verbose philosophers who fill hundreds of pages explaining their "system of philosophy" as if it were necessary for others to validate the system they have come up with for describing our situation.

If only we could discover a way of communicating where we might be able to mean what we say and say what we mean.  If I have nothing to say, it might take 500 words to express this.  So be it!

One man says he knows how to build a house.  Can he do this without a shovel?

Who makes the shovel?  Another man designs an operating system to be loaded onto a hard-drive of a computer.  Who places the chips on the circuit board?  Who designs the circuit board?

When I install a web browser in Funtoo, a branch off of Gentoo created by the founder of Gentoo, I witness "it," the operating system that is, compiling all the code into a "program" which will be launched.  It takes several, if not many, hours.  It's nothing like watching a setup.exe program in Windows install a precompiled loader. 

Do you know how I feel when I issue the command to install such software?

It does not make me feel "smart".

It is all very humbling since I am only expected to know the command, "sudo emerge www-client/chromium" or whatever, and then I sit back and "pray" (to the gods of programming?) that all goes well, that these mother-scratchers know what they're doing.  I am amazed that there are so many super-intelligent monkeys who can pull off such feats as designing operating systems and motorcycles.

I am proud of myself if I can grow zucchini, squash, tomatoes, and cucumbers.

 :P

... and yet i can calculate and compute and follow some algebraic proofs.   I don't respect people who think having a Masters in Business Administration is a big deal.  I have more respect for, well, someone like myself, who has absolutely no "business-sense" and does not care to design bridges either.

I do like mathematics, a very limited amount that will keep the gears in my head spinning for the rest of my life, but I do not care for "NASA" and am totally not impressed at all with astronauts or people who study "rocket science".

In actuality, when the kids study trigonometry in high school, they are studying "rocket science" even if they won't even use that knowledge to earn a living constructing buildings.

One can know how to read and write and do a little math, but we have no choice but to be mystified by the complicated artifice all around us, at our fingertips even.  We are truly apes on wheels.

When I mention my interest in mathematics to a nosy passerby who inquires as to what it is I am "studying" there with my book on a stand and notebook on my lap, an immediate followup question is, "What are you studying for?  Why don't you attend a school or try one of those on-line courses?"

I become more than a little annoyed and irritated, and I don't feel like giving my life stroy, so I simply say that I prefer to study what I want, when I want, and how I want, and don't really think that there exists any "programmed" course of training.

Holden, like your namesake, I think there are many PHONIES in this world.

People that don't know what the hell they are talking about may be getting paid to tell others what to do with their lives.  They advise someone to attend community college or "upgrade their skills".

What if I want to learn how to calculate logarithms without a calculator?   Same with the arctangent.  I want to calculate that by hand as well?  Does the teacher know how?  Well, then why "upgrade my skills"?   What does that mean, anyway, be trained how to design "apps" for smartphones and smart televsions?  Build apps for the gorts?  Are they fuucking kidding me???

No, I want to build console programs in C/C++ and explore the computer algebra systtems Sage and SymPy created by the "scientific open source community".  They are the mysterious priests in my tribe, a tribe in which I have been playing the role of village idiot in.   :P

I'm "touched" in the head?   No.  I'm just an a-s-s-h-o-l-e who doesn't give a shiit anymore.
Would you think less of me if I went from "being drunk on math" to "doing math drunk"?   :D

I am as Raul describes those who watch tragedies unfold on the television news.  I am only concerned with the math book under my nose.  The thing is, I want to reach a point where I one can be upfront about being immune to the propaganda machine.  I want to think like a non-human creature.



What does it mean to say, "I don't want to cry for somebody else's need."   ?

In the context of:

"My Country I & II"


Something horrid....
I would not fight for my country
I would not work to no rule
Don't have no truck with no power game
Won't be some other jerk's tool
Don't have no part of no partisans
Won't have no part 'cos one party and another's all the same
(Workin' for some other man)
All the same
Gonna play it
(Workin' for some other man)
At my own game

I would not fight for my country
I would not work to no rule
Don't have no truck with no power game
Won't be some other jerk's tool
Don't have no part of no partisans
Won't have no part 'cos one party and another's all the same
All the same
Gonna play it at my own game

Don't wanna die for some old man's crusade
Don't wanna hear what they feed
Don't wanna kill for some cause of the age
Don't wanna cry for somebody else's need
Don't want no piece of no flag in the breeze
Don't want no part 'cos one party and another's all the same
All the same
Gonna play it at my own game

Ooooooh not me
You don't get me
Oi you

Ooooooh not me
Oi you
You don't get me
Oi you
« Last Edit: July 02, 2018, 07:53:03 pm by Kaspar Hauser »
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

~ Tabak und Kaffee Süchtigen ~

Nation of One

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Re: The Poetry of Logarithms (or, Be Drunk on Math)
« Reply #12 on: July 19, 2018, 09:14:09 pm »
Quote from: Holden
What do you think of Hero's formula?

Take Two

When going through exercises in old books, I usually limit myself to the more "novel" problems as I do not get anything out of plugging values into formulas.   Lo and behold a set of 3 exercises will constitute a proof of Hero's formula.  It is right in my path … Coincidence?

Maybe tomorrow, if I am pleased with what I come up with tonight, I will see if I can motivate myself to type it up - or, if I find that too frustrating, I can just scan the pages from my scratchpad and email the PDF files to you.

It is in exploring a way to prove the formula that I will be able to respond to this question in a more meaningful manner.
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

~ Tabak und Kaffee Süchtigen ~

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Constituting a Proof of Hero's Formula
« Reply #13 on: July 19, 2018, 10:40:08 pm »
[Note:  Also called Heron's formula.]

(1) In geometry we learn that the area of a triangle is given by the formula

K = (1/2)(base)(height)

I won't draw any diagrams, obviously, although I wish I could.  We don't have that kind of message board.

If we let the length of the base = b, and the height of the triangle = a*sin(C), we obtain:

K = (1/2)*a*b*sin(C)

To prove Hero's formula, K = sqrt[s(s-a)(s-b)(s-c)],
we first show that K^2 = (1/4)*a^2*b^2*sin^2(C)

That's kind of trivial:  K^2 = [(1/2)*a*b*sin(C)]^2 = (1/4)*a^2*b^2*sin^2(C)

(2)  In Hero's formula, s = (1/2)*(a + b + c) so that 2s = a + b + c


Hence   2*(s - a) = 2s - 2a = a + b + c - 2a = -a + b + c

2*(s - b) = 2s - 2b = a + b + c - 2b = a - b + c

2*(s - c) = 2s - 2c = a + b + c - 2c = a + b - c



(3) This part has 9 parts lettered a through i.


(a) a^2 + b^ - c^2 = 2*a*b*cos(C)


Why?  By the law of cosines, c^2 = a^2 + b^2 - 2*a*b*cos(C)

 So, a^2 + b^2 - (a^2 + b^2 - 2*a*b*cos(C)) =  2*a*b*cos(C)

(b) (a^2 + b^2 - c^2)^2 = 4*a^2*b^2*cos^2(C) = 4*a^2*b^2*[1 - sin^2(C)]


since sin^2(C) + cos^2(C) = 1, cos^2(C) = 1 - sin^2(C)


(c)  (a^2 + b^2 - c^2)^2 = 4*a^2*b^2*[1 - sin^2(C)]
= 4*a^2*b^2* - 4*a^2*b^2*sin^2(C)] = 4*a^2*b^2 - 16*K^2

In part (1) we defined K = (1/2)*a*b*sin(C) and K^2 = (1/4)*a^2*b^2*sin^2(C)

16*K^2 = 16*[(1/4)*a^2*b^2*sin^2(C)] = 4*a^2*b^2*sin^2(C

(d) 16*K^2 = 4*a^2*b^2 - (a^2 + b^2 - c^2)^2


Since (a^2 + b^2 - c^2)^2 = 4*a^2*b^2 - 16*K^2,

by the Additive Property of Equality, 16*K^2 = 4*a^2*b^2 - (a^2 + b^2 - c^2)^2


(e) 16*K^2 = [2ab + (a^2 + b^2 - c^2)]*[2ab - (a^2 + b^2 - c^2)]


reason:  x^2 - y^2 (x + y)*(x - y)


Therefore 4*a^2*b^2 - (a^2 + b^2 - c^2)^2 = (2ab)^2 - (a^2 + b^2 - c^2)^2
where x = 2ab and y = a^2 + b^2 - c^2



(f) 16*K^2 = [(a + b)^2 - c^2]*[c^2 - (a - b)^2]


Additive Property for Addition:

x^2 + 2xy + y^2 = (x + y)^2  and  x^2 - 2xy + y^2 = (x - y)^2


(g)  16*K^2 = [(a+b) + c]*[(a+b) - c]*[c + (a-b)]*[c - (a-b)]

Since x^2 - y^2 = (x + y)*(x - y)


(h) 16*K^2 = [2s]*[2*(s - c)]*[2*(s - b)]*[2*(s - a)]


By part (2), recall: 

2s = a + b + c

2*(s - c) =  a + b - c

2*(s - b) = a - b + c

2*(s - a) = -a + b + c

(i)  K = sqrt[s*(s - a)*(s - b)*(s - c)]


since 2*2*2*2 = 16 and K^2 = s*(s - a)*(s - b)*(s - c)

__________________________________________________

 :-\ 

Heron did not pull that one out of his hat.

The other 6 formulas for finding the area of a triangle are much easier to show.






« Last Edit: July 20, 2018, 11:07:27 pm by Kaspar Hauser »
Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

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C Program for finding Area of a Triangle
« Reply #14 on: August 18, 2018, 01:18:00 am »
C Program I created today for finding area of a triangle (includes Heron's formula for the case when 3 sides are given):

/****************************************************************************/
#include <stdio.h>
#include <math.h>



  double radians = M_PI/180.0;
  double degrees = 180/M_PI;
  int second_solution = 0;
  int no_solution = 0;
  double a, b, c, A, B, C;

  struct Area_bh
  {
   double base;
   double altitude;
  };

  struct Area_3s
  {
     double a, b, c;
  };

  struct Area_bh K;
  struct Area_3s H;
  struct Area_bh K2;

int display_menu ()  {
  int choice;

  printf("What is your option?\n");
  printf("<1> 2 sides and included angle\n");
  printf("<2> 2 angles and included side\n");
  printf("<3> 3 sides\n");
  printf("<4> 2 sides and angle opposite one of them\n");
  printf("ENTER: ");
  scanf("%i", &choice);
  printf("\n");
  return choice;

}

int get_data (int o)  {

  switch (o)  {

    case 1:   printf("Enter the length of a side: ");
              scanf("%le", &a);
              printf("Enter the length of the other side: ");
              scanf("%le", &c);
              printf("Enter the measure of the included angle: ");
              scanf("%le", &B);
              K.base = a;
              K.altitude = c*sin(B*radians);
              break;

    case 2: printf("Enter the measure of an angle: ");
            scanf("%le", &A);
            printf("Enter the measure of the other angle: ");
            scanf("%le", &B);
            C = 180 - (A + B);
            printf("Enter the length of the included side: ");
            scanf("%le", &c);
            a = (c*sin(A*radians))/sin(C*radians);
            K.base = a;
            K.altitude = c*sin(B*radians);
            break;

    case 3: printf("Enter the length of a side: ");
            scanf("%le", &a);
            printf("Enter the length of another side: ");
            scanf("%le", &b);
            printf("Enter the length of the third side: ");
            scanf("%le", &c);
            H.a = a;
            H.b = b;
            H.c = c;
            break;

    case 4: printf("Enter the measure of the angle: ");
            scanf("%le", &B);
            printf("Enter the length of the side opposite this angle: ");
            scanf("%le", &b);
            printf("Enter the length of the other side: ");
            scanf("%le", &c);
         

            /* check to see if data is valid  */
            if ( ((c*sin(B*radians)/b) < -1) || ( (c*sin(B*radians)/b) > 1) )
            {
                printf("\nThe number %g is outside the range of the sine function.\n", (c*sin(B*radians)/b));
                printf("Therefore, there is no triangle to find the area of!\n");
                no_solution = 1;
                return 0;
            }
            else {
              C = asin((c*sin(B*radians)/b))*degrees;
              A = 180 - (B + C);
              a = (b*sin(A*radians))/sin(B*radians);
              K.base = a;
              K.altitude = c*sin(B*radians);

            }
           
           if (B < 90)  {
                 if ( b < (c * sin(B*radians)) ) {
                    printf("NO SOLUTION");
                    no_solution = 1;
                    printf("angle < 90 AND length %g < %g*sin(%g)", b, c, B);
                  }
                  else if (b == (c * sin(B*radians))) {
                      K.altitude = c;
                      printf("There's just one possible area since this is a right triangle.\n");
                  }
                  else if ( (c * sin(B*radians) < b) && (b < c) ) {
                     second_solution = 1;
                     C = 180 - C;
                     A = 180 - (B + C);
                     a = (b*sin(A*radians))/sin(B*radians);
                     K2.base = a;
                     K2.altitude = c*sin(B*radians);
                  }
                  else if (b >= c)  {
                      printf("There's just one possible area for this triangle.\n");
                  }

            }               
            else if ( (90 <= B) && (B < 180) )  {
              if (b <= c) {
                  printf("NO SOLUTION\n");
                  printf("90 <= angle < 180 AND %g <= %g", b, c);
                  no_solution = 1;
              }
              else printf("There's just one possible area for this triangle.\n");
            }

            break;

    default:
            printf("Not a valid option.\n");
            break;

  }
  printf("\n");

}


int display_results(int o) {
    double area;

  if (no_solution) printf("No results to display.\n");
  else
     {
      if (o == 3)  {
       
        if (H.a + H.b < H.c) {
            printf("NO SOLUTION\n");
            no_solution = 1;
            printf("Since %g + %g = %g < %g,\n", a, b, a + b, c);
            printf("there is no triangle with the given sides.\n");
            return 0;
        }
 
        if (!no_solution)  {
           /* compute semiperimeter s */
           double s = (H.a + H.b + H.c) / 2.0;
           area = sqrt(s * (s - H.a) * (s - H.b) * (s - H.c));
           printf("The area of the triangle is %g\n", area);
        }
       }
      else  {
          area = 0.5 * K.base * K.altitude;
          printf("The area of the triangle is %g\n", area);
      }
      printf("\n");
     }

  if (second_solution)  {
           printf("OR\n");
           area = 0.5 * K2.base * K2.altitude;
           printf("\nThe area of the triangle is %g\n\n", area);
  }
  return 0;
}

int main()
{
 
  int option;
 
  option = display_menu();
  get_data(option); 
  display_results(option);
}







Things They Will Never Tell YouArthur Schopenhauer has been the most radical and defiant of all troublemakers.

Gorticide @ Nothing that is so, is so DOT edu

~ Tabak und Kaffee Süchtigen ~