Today I Learned

[1oldman2]

TIL, That I have wanted to be a luthier for a very long time, I just didn't know the title of the job/craft. Over the years building things I have found "rough framing" ( tolerances of plus/minus 3/8 inch ) to be very unsatisfying, building log houses (tight fit) to be very interesting, finish work (Particularly the higher grades) is awesome, and that leads to contemplating building something like a stringed instrument. (beyond awesome)
I'm looking forward to hearing how your rebuild goes, keep us posted.

 

[Breanna Geise]

Today I learned how to cooked Seafood paella. My family loves it!

 

[Stephanus]

Good Breanna. And welcome to PF Forum

 

[Stephanus]

Today I learned, that you can get "likes" just from "Hm..."

 

[demelzareveley]

Today I learned to cook a simple recepie.

 

[Jonathan Scott]

TIL a violin string is thirteen inches long.
Well, some folks say 12$\frac{7}{8}$ inches .

I have several violins, and for the three full-size ones the part of the string which vibrates is 12 7/8 inches to the nearest 1/8 inch (between the bridge and the nut, which is between the finger-board and the pegbox). The string itself is quite a bit longer than that because of the part which attaches to the tailpiece and the part which is wound around the peg.

 

[Stephanus]

I have several violins...

You'd better believe him @jim hardy

 

[jim hardy]

You'd better believe him @jim hardy

Perhaps he'd post a picture or two ?

I have only two, well actually one violin and one violin project...
and no clue how to operate any musical instrument
is it something one can learn late in life ?

 

[Jonathan Scott]

Today I learned that it seems to be a very long time since I last played my viola; I accidentally opened the case when looking for random violins and found that one string had spontaneously broken and the others had partially untightened themselves. I can't even remember the last time I used it in performance; it must be more than a decade ago.

We have several smaller violins which the kids used for learning, and four full sized ones, one of which is electric and one of which I have lent to my daughter. I also have a viola, and I sometimes play my wife's second best cello. Apart from those we also have a baby grand piano, a digital piano, a couple of classical guitars, flute, clarinet, some recorders, some ocarinas and odd percussion bits such as triangle and tambourine.

I don't know much about how difficult it is to learn musical instruments late in life; I learned how to read music at 5 (with my mother's help) then started the piano at 6 and the violin at 8, plus many other random instruments later along the way including for example the cello, clarinet and tenor viol (which I learned at school) and trombone (which I learned in order to play duets with my youngest brother when he learned it). I learned the flute many years later, and although I learned it very quickly, I found that my lips get tired far too rapidly, so I can play only for a very short time. Similar things apply on all my instruments; learning is one thing, but having the speed, strength and stamina to play stuff properly takes a lot of practice. If it's arms, wrists and fingers (as it is for piano and strings) then I've got those working well now, but for things involving blowing I would need far more practice to be able to play properly.

I'd like to do more conducting, especially of a full symphony orchestra. It doesn't require fantastic speed or strength but is reasonably energetic anyway, and makes good use of my experience on the receiving end as an amateur orchestral player for most of my life (since age 10 anyway), together with my score-reading skills (admittedly quite hard earned from practice long ago) and my ability to make sense of a complex orchestral sound. Unfortunately, now I've managed to appoint permanent regular conductors for both of the orchestras which I run, plus a student assistant conductor for the larger orchestra, I'm not likely to have any more opportunities in the near future (although the regional rail system has recently made several attempts to give me an opportunity, only to be thwarted by the conductor managing to find an alternative route).

 

[jim hardy]

I really envy people with musical ability.
Our parents took sister and me to a children's concert when i was maybe seven - ca 1953?
and i never forgot the sensation when those violins started up - i felt as if i were standing in a waterfall showered by a cool soothing fluid completely engulfing me
afterward i'd sit outside the school band room just to hear them practice
and to this day live music gives me a thrill,
i once walked a whole block in Quebec to find the source of a synthesizer-like rendition of Beethoven's Ninth finale,
An old man sitting on the sidewalk was playing it on a saw ...
I of course put some money in his hat
and observed he was using a Disston D-23, which is a very fine saw..

http://www.disstonianinstitute.com/d23page.html

forty years later i took my own kids to a children's concert
found to my delight they're still playing "The Worried Drummer" at kids concerts.

old jim

 

[jim hardy]

That concert may have affected son
he seems to value good sound quality

his Christmas present one year was sixteen inexpensive speakers and a sheet of plywood,
kept him occupied until school started again

and they sounded doggone good !

 

[Jonathan Scott]

An old man sitting on the sidewalk was playing it on a saw ...

Playing a long and flexible saw with an old cello bow can produce a quite magical sound - I've tried it reasonably successfully, but found that the teeth tend to be a hazard!
My guess is that most (but not all) people are capable of learning to play some musical instrument reasonably well if they have the patience and persistence. I was lucky in learning young, and having the interesting challenge of playing the piano for weekly school hymns from age 8, which taught me how to get by effectively when I was somewhat out of my depth.

 

[Stephanus]

That concert may have affected son
he seems to value good sound quality

his Christmas present one year was sixteen inexpensive speakers and a sheet of plywood,
kept him occupied until school started again

and they sounded doggone good !

Wow, such a handsome son you have. And should you envy a person who has musical ability. Of all people, not JS. And you might learn something about envy

 

[jim hardy]

And you might learn something about envy

Full Definition of envy

transitive verb[/B]

1 : to feel envy toward or on account of

noun

1 : painful or resentful awareness of an advantage enjoyed by another joined with a desire to possesses the same advantage

it's painful not resentful, i assure you...

Thanks for the kind words !

old jim

 

[Stephanus]

it's painful not resentful, i assure you...

Thanks for the kind words !

old jim

Oh, sorry. English is my second language. But JS is very good. I play piano and guitar, too. But not near as good.

 

[Pepper Mint]

Oh, sorry. English is my second language. But JS is very good. I play piano and guitar, too. But not near as good.

What is JS ? Javascript ?

 

[jim hardy]

Oh, sorry.

?? nothing to be sorry about, just i wasn't sure what you meant about envy...

The real "haves" are they who can acquire freedom, self-confidence, and even riches without depriving others of them. They acquire all of these by developing and applying their potentialities.

eric hoffer

Resent my betters? Heck no, i try to emulate them !

old jim

 

[256bits]

Today ( actually yesterday ) I learned coffe is just as good with molasses.
Ran out of sugar and I was too lazy to go out and buy some.

 

[jim hardy]

I play piano and guitar, too.

Piano and guitar is a delightful combo. I once saw "Jacques Brel is Alive and Well..." performed with only those two instruments to accompany the singers. Very effective, it's still an intense memory after forty five years.

"Can music save your mortal soul? " Don McLean
i answer a resounding "Yes!"

 

[Samia qureshi]

Today i learned that why sky look black from Moon

 

[Stephanus]

Today I learned,
that I can overcome my sleep problem by counting numbers.
Instead of counting 1,2,3,4,5...
I try to solve this puzzle.
Supposed, you have 5 cent and 7 cent stamps.
What is the minimum sequential value by putting 5 and 7 cent stamps?
It's 24.

24 = 5*2 + 7 *2
25 = 5 * 5 + 7 * 0
26 = 5 * 1 + 7 * 3
27 = 5 * 4 + 7 * 1
28 = 5 * 0 + 7 * 4
29 = 24 + 5 * 1, etc...

It's 24, because you can't find any stamp combination to make 23.

What about 7, 40, and 41 cents?
It's

115 = 7 *  5 + 40 * 2 + 41 * 0
116 = 7 *  5 + 40 * 1 + 41 * 1
117 = 7 *  5 + 40 * 0 + 41 * 2
118 = 7 * 11 + 40 * 0 + 41 * 1
119 = 7 * 17 + 40 * 0 + 41 * 0
120 = 7 *  0 + 40 * 3 + 41 * 0
121 = 7 *  0 + 40 * 2 + 41 * 1
122 = 115 + 7, etc...

Because you can't make any combination to produce 114
You can try with any number. As long as ALL of the number don't have the same prime factor. For example: 6, 9, 15.
Any two primes in the number will always work.

Here is a sample for: (101, 103, 105, 107, 109)

It's 2625. Because there's no combination for 2624. Have to use computer, though.
And my problem is when I tried to calculate, say, (11, 90, 95)
I always slept away before (or after?) I found the answer. And when I woke up in the morning, I already forgot what numbers that I calculated the night before?
Was it, say, (12, 13, 14) or (21, 31, 41)? Or have I found the answer yet last night? Just don't remember. But most important is, I can sleep, no more restless!

What about 7, 50 and 51?
Anybody want to try on their bed?

 

[Sophia]

What helps me is watching thoughts as a film/radio but not trying to ruminate, control or assess them in any way. Sometimes it turns into a very fast chain of various free associations and helps to fall asleep when I wake in the middle of the night.

 

[mfb]

What is the minimum sequential value by putting 5 and 7 cent stamps?
It's 24.

24=(5-1)*(7-1). I would expect this pattern to work for every set of two integers without a common factor.
With more numbers this product is just an upper limit.

 

[Stephanus]

24=(5-1)*(7-1). I would expect this pattern to work for every set of two integers without a common factor.
With more numbers this product is just an upper limit.

Oh, my! You're genius @mfb! I just don't know how on Earth you do it!

I've countered 100 numbers. All of them show your pattern!

 

[mfb]

TIL: This rule is true in general. Proof, with an interesting approach I think:

Let's call the numbers a and b.
They key number here is a*b, which can be expressed as "a times b" or "b times a". It is the first number that has this ambiguity, as it is the least common multiple.

Let's consider the set of n*a+m*b mod (ab) for n=0 to b-1 and m=0 to a-1. We get (ab) different results (due to (ab) being the least common multiple - getting the same result twice would lead to a smaller common multiple) - but there are just ab possible results, so we get every result exactly once.
In absolute terms, the values n*a+m*b go from 0 to 2ab-a-b.

Let's consider (b-1)*a+(a-1)*b = 2ab-a-b. We know it is in the set above. But that means ab-a-b cannot be in this set, otherwise we would have a contradiction to the unique results mod (ab). We also cannot reach it with larger n or m - the sum would be too large. There is no way to get to ab-a-b with positive n,m.
What about all the values between 2ab-a-b and 2ab? They are not in the set above, But that means the values between ab-a-b and ab have to be in the set, because we know those values appear as remainder mod (ab).

=> ab-a-b = (a-1)(b-1)-1 cannot be reached, while (a-1)(b-1) to (ab) can be reached. It is a well-known result that all numbers above (ab) can be reached as well. Therefore, (a-1)(b-1)-1 is the last number that cannot be reached.

 

[NascentOxygen]

Today ( actually yesterday ) I learned coffe is just as good with molasses.
Ran out of sugar and I was too lazy to go out and buy some.

Recipes calling for molasses require caution. You may not be aware that "molasses" means different things in different countries.

 

[1oldman2]

TIL, a

is 6.022 × 10²³ units of anything.

 

[Astronuc]

TIL Paul (Louis-Toussaint) Héroult (10 April 1863 – 9 May 1914), a French scientist and the inventor of the aluminium electrolysis process that bears his name, developed the first successful commercial electric arc furnace. EAF melting and refining became important for clean steels.

 

[Stephanus]

TIL, a View attachment 103334is 6.022 × 10²³ units of anything.

Hahahahahaha. English is not my first language. Guess what when I tried:
https://www.google.co.id/search?q=mole&biw=1366&bih=643&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjt2vmn5fnNAhVFNI8KHRu6CKIQ_AUIBigB
I would expect pictures of laborartory apparatus, but... Well try it yourself.

Of course this will give the chemisty link.
Below is Google search
https://www.google.co.id/search?q=mole&biw=1366&bih=643&source=lnms&sa=X&ved=0ahUKEwj9zuCr5fnNAhXIKo8KHfHRCYUQ_AUIBygA&dpr=1
[Add: ahhh "mole hunt" is the term in spy novel about searching for a mole inside a secret service agency]

 

[1oldman2]

Hahahahahaha. English is not my first language. Guess what when I tried:
https://www.google.co.id/search?q=mole&biw=1366&bih=643&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjt2vmn5fnNAhVFNI8KHRu6CKIQ_AUIBigB
I would expect pictures of laborartory apparatus, but... Well try it yourself.

Of course this will give the chemisty link.
Below is Google search
https://www.google.co.id/search?q=mole&biw=1366&bih=643&source=lnms&sa=X&ved=0ahUKEwj9zuCr5fnNAhXIKo8KHfHRCYUQ_AUIBygA&dpr=1
[Add: ahhh "mole hunt" is the term in spy novel about searching for a mole inside a secret service agency]

https://www.khanacademy.org/science...oichiome/stoichiometry-ideal/v/stoichiometry#!

 

[Stephanus]

https://www.khanacademy.org/science...oichiome/stoichiometry-ideal/v/stoichiometry#!

Yep, 1 mole = 6 * 10²³ is NOT (added) a Today I learned thing
But a mole =

is really a Today I learned thing. Yeah , today I learned that that is a mole.

 

[Stephanus]

TIL: This rule is true in general. Proof, with an interesting approach I think:

Well this is not as straight forward for me as for everyone else. I have to break it down.

Spoiler: I'm going to need a proof for n*a+m*b mod ab will produce different results.

Let's consider the set of n*a+m*b mod (ab) for n=0 to b-1 and m=0 to a-1. We get (ab) different results ..

Fact 1. for n=0 to b-1 and m=0 to a-1 we get ab diffrent results. Will it give us different results?
Fact 2: For every n mod b where n<b there will be unique/different results.
Fact 3: for every n*a mod b*a where n<b then there will be unique results and some gaps. This will also work for n = b-1, so
Fact 4: for every (b-1)*a mod b*a there will be unique results and some gaps.
Fact 5: (b-1)*a + (a-1) * b might be more than ab, seems contradict fact 1. Will (a-1)*b fil the gaps?
5.1 Let's add (b-1)*a with sequences of m*b where m=0..a-1
Before that. Sequence of (b-1)*a mod a is 0.
((b-1)*a + a*b) will give an overlap result.
because ((b-1)*a + a *b) mod a = 0.
and
((b-1)*a + a*b) mod ab = -a; ((b-1)*a) mod ab = -a

But (b-1)*a + (a-1)*b will not give an overlap result.
((b-1)*a + (a-1)*b) mod a = -b
((b-1)*a + (a-1)*b) mod ab = -a-b.
So, for ((b-1)*a + m*b) mod a where m = 1..a-1 there wouldn't be overlap result.
So the (a-1)*b gaps will be reduced by a-1. So there will be (a-1)*b-(a-1) = (a-1)*(b-1) gaps.

5.2 Do 5.1 again with (b-2) * a
Again ((b-2) * a + m*b) mod a wil produce the exact result as in 5.1
But there wouldn't be any conflict with ((b-2) * a + m*b) mod ab, because it's like shifting/decreasing all the filled number by a
So there will be (a-1)*(b-1)-(a-1) = (a-1)*(b-2) gaps.

5.3 Do 5.1 again until (b-b) * a
So there will be (a-1)*(b-b) gaps. = 0 gaps!

So here is the proof for

Let's consider the set of n*a+m*b mod (ab) for n=0 to b-1 and m=0 to a-1. We get (ab) different results ..

There are ab different results.

Let's call the numbers a and b.
They key number here is a*b, which can be expressed as "a times b" or "b times a". It is the first number that has this ambiguity, as it is the least common multiple.

Let's consider the set of n*a+m*b mod (ab) for n=0 to b-1 and m=0 to a-1. We get (ab) different results (due to (ab) being the least common multiple - getting the same result twice would lead to a smaller common multiple) - but there are just ab possible results, so we get every result exactly once.

Okay...

In absolute terms, the values n*a+m*b go from 0 to 2ab-a-b.

Okay...

Let's consider (b-1)*a+(a-1)*b = 2ab-a-b. We know it is in the set above.

Okay..

But that means ab-a-b cannot be in this set, otherwise we would have a contradiction to the unique results mod (ab).

Okay..

Spoiler: Here is the proof

Sequence of ((b-1)*a+(a-1)*b) mod ab will give ab different results.

void Fill(int a, int b)
{
  int Numbers1[a*b]; // I know this is error :smile:
  int Numbers2[2*a*b];
  int c=0;
  int n,m;
  for (n=0;n<b;n++) for (m=0;m<a;m++) Numbers1[(a*n+b*m) % (a*b)]++;
  for (n=0;n<b;n++) for (m=0;m<a;m++) Numbers2[(a*n+b*m)]++;
  for (n=0;n<b;n++) for (m=0;m<a;m++) c++; // capturing the last element of the counter
}

All elements in Numbers1 will be filed by 1. But there elements in Numbers2 that are zero and the others are of course 1.
So there's no overlap here. I'm sorry. I'm not a mathematician, I don't know math language. But that's what I mean.
Because all Numbers1 element already filled with 1 this instruction
for(n=c;n<2*a*b;n++) if(!Numbers2[n]) {Numbers1[n%(a*b)]++; break;} will set one of the Numbers1 element by 2.
I swear I just write this code in the browser not in the compiler. But here's is the proof that ab-a-b cannot be in the set. I really can't express myself in math language!

We also cannot reach ab-a-b (edited by me) with larger n or m - the sum would be too large. There is no way to get to ab-a-b with positive n,m.

Yes. (ab-a-b) mod ab = -a-b. And we already have Numbers1[(-a-b)% (a*b)]==1 from (b-1)a + (a-1)b. I don't have the proof that we can reach ab-a-b from na+mb, but this proof is sufficient if we combine the previous proofs.

What about all the values between 2ab-a-b and 2ab? They are not in the set above,

Yes, sure.

But that means the values between ab-a-b and ab have to be in the set, because we know those values appear as remainder mod (ab).

Yes, all of them.

=> ab-a-b = (a-1)(b-1)-1 cannot be reached, while (a-1)(b-1) to (ab) can be reached. It is a well-known result that all numbers above (ab) can be reached as well. Therefore, (a-1)(b-1)-1 is the last number that cannot be reached.

Oh my. I've never thought of that. Although, about the "gaps" my software has already shown that. They are all before my eyes. I just don't realize them.

Reds row are the gap. But my software doesn't continue the proof until 2ab-a-b. It just counter from ab-a-b+1 to the least of a or b. Supposed if a is the smaller
then if there's a sequence from ab-a-b+1 to ab-a-b+1 + (a-1) or sequence from ab-a-b+1 to ab-b then it just cuts. And continue the next sequence in another window as I upload yesterday.
And today I learned that I do believe that all the mentors/staffs/science advisors inf PF Forum are really geniuses! What's been shown before my eyes a long time can be solved just in 5 minutes by them. I do hope that this community either direct/indirectly can contribute many things to human wealth and continuity.

 

[Stephanus]

Spoiler: Today I leaned

That we can use spoiler to shorten the details in our post although it's not intended as a spoiler.

Spoiler: Details

Because the details tends to divert our attention to the main message.
They are best kept in a spoiler so it does not crowd our message.
Only for those who really need to read the proof/detail can expand the spoiler.
Perhaps in the future, the PF Forum allow the word SPOILER to be replaced by user?

Spoiler: And another one

We can use spoiler inside spoiler

We can also insert graph in a spoiler

 

[mfb]

A mole of moles?

2ab-a-b is the largest number that has a unique way to being expressed as n*a+m*b with non-negative n,m, by the way, all larger numbers have at least two options (because you can replace "a times b" by " b times a" if n>=b or m>=a).

 

[jtbell]

When I was in high school many years ago, my chemistry teacher once posed this question on a test:

If a mole could dig a mole of holes in a day, how many holes could a mole of moles dig in a mole of days?