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## general methods for linear temperament notation

cmloegmcluin
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### Re: general methods for linear temperament notation

Dave Keenan wrote: Tue Mar 09, 2021 3:10 pm
cmloegcmluin wrote: Sat Feb 20, 2021 7:50 am Also, I suspect @Dave Keenan may eventually ask me to write a program which allows a regular (or is linear the preferred term?) temperament notation to be generated from some inputs, as a feature of a web-based Sagittal notation calculator...
Actually. It should be a simple extension to your existing code base, to at least do the grunt work of determining how many periods and generators each Olympian symbol's default comma maps to, for a given temperament.
It can already do this for an EDO, so yes, I agree it should be relatively simple to make it work for temperaments, too.
= [0 0⟩

That's not a monzo. It's a generator-count vector. So we see that the 5-comma is zero periods and zero generators in meantone and so and are of no use in notating meantone.

I note that Herman and I both use plus signs "+" for positive generator counts, which serves as a clue that they are not monzos. Herman further distinguished his generator count vectors above (but not everywhere) by using parentheses instead of square and angle brackets.
I don't see any plus signs in that one, though...? I prefer the version with parentheses. If math hacks like me are a reasonable representation of the microtonal community, I don't think there's room for two different structures enclosed in [ and ⟩.
Graham Breed has a tutorial here:
http://x31eq.com/matrices.htm
and if that isn't basic enough, he has a matrix math tutorial here:
http://x31eq.com/matritut.htm
Those are great, thanks.

I also stumbled across this by Mike Battaglia just the other day, which starts off much the same as your explanation above: https://en.xen.wiki/w/Diatonic,_Chromat ... bchromatic

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### Re: general methods for linear temperament notation

cmloegcmluin wrote: Thu Mar 18, 2021 3:44 am It can already do this for an EDO, so yes, I agree it should be relatively simple to make it work for temperaments, too.
Of course EDOs are temperaments ("rank 1" or "equal" temperaments). You mean it should be relatively simple to make it work for "rank 2" or "linear" temperaments (or perhaps "higher rank" temperaments).
I don't see any plus signs in that one, though...?
You mean in my [0 0⟩? That was because none of the counts are positive. But you're right. I could have written [+0 +0⟩. But there are still going to be gen count vectors whose terms are all negative, and so can't be distinguished as such by plus signs.
I prefer the version with parentheses. If math hacks like me are a reasonable representation of the microtonal community, I don't think there's room for two different structures enclosed in [ and ⟩.
But in the linear algebra formulation they are both column vectors. It's like with any measurements — you need to give the "units", either after each vector, e.g. "[0 0> gens", or in the column heading of a table. I note that George uses "G" on his graph.
I also stumbled across this by Mike Battaglia just the other day, which starts off much the same as your explanation above: https://en.xen.wiki/w/Diatonic,_Chromat ... bchromatic
Right. I notice that he not only uses the same notation |a b c ...> for prime exponents and generator counts, he doesn't use plus signs and he even calls the latter "tempered monzos" — yikes — helping making my case that Gene's monzo/val terminology is unhelpful and just raises a barrier to entry.

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### Re: general methods for linear temperament notation

Dave Keenan wrote: Thu Mar 18, 2021 6:40 am
cmloegcmluin wrote: Thu Mar 18, 2021 3:44 am It can already do this for an EDO, so yes, I agree it should be relatively simple to make it work for temperaments, too.
Of course EDOs are temperaments ("rank 1" or "equal" temperaments). You mean it should be relatively simple to make it work for "rank 2" or "linear" temperaments (or perhaps "higher rank" temperaments).
Yes.
I don't see any plus signs in that one, though...?
You mean in my [0 0⟩? That was because none of the counts are positive. But you're right. I could have written [+0 +0⟩. But there are still going to be gen count vectors whose terms are all negative, and so can't be distinguished as such by plus signs.
Oh, good point. Then I wouldn't bother with presence or absence of plus signs at all. I would then strongly wish for the parentheses.
I prefer the version with parentheses. If math hacks like me are a reasonable representation of the microtonal community, I don't think there's room for two different structures enclosed in [ and ⟩.
But in the linear algebra formulation they are both column vectors. It's like with any measurements — you need to give the "units", either after each vector, e.g. "[0 0> gens", or in the column heading of a table.
Okay, I see the problem, but I'm still pretty disturbed.

I never learned this [ ⟩ angle bracket notation when I was studying linear algebra in college. We always just wrote things with plain brackets [ ]. Why not just use those for generator-count vectors?
I note that George uses "G" on his graph.
Which graph? Oh in his derivation of the secor graph that I shared earlier. Got it.

Hm. Those look to me like just parenthetical notes re: the harmonics and how many generators it took to reach them. Besides, in a generator-count vector, doesn't the period come first, then the generator(s)?

Even if George was somehow using G in this way, would you recommend it? I think it would be really cumbersome to put P G₀ G₁ ... after each term in a generator-count vector as a means only to distinguish it from a monzo.
I also stumbled across this by Mike Battaglia just the other day, which starts off much the same as your explanation above: https://en.xen.wiki/w/Diatonic,_Chromat ... bchromatic
Right. I notice that he not only uses the same notation |a b c ...> for prime exponents and generator counts, he doesn't use plus signs and he even calls the latter "tempered monzos" — yikes — helping making my case that Gene's monzo/val terminology is unhelpful and just raises a barrier to entry.
Yeah, I admit I don't get the idea of "tempered monzo".

I didn't realize until you said that that you're suggesting monzos/vals are bad. That surprises me because in my ~15 years of experience, microtonal music is impossible without them. They're all I've ever known. So what you're proposing is an alternative to them? Not merely an alternative to "wedgies" or whatever higher-level Gene stuff I never bothered trying to learn that I assumed was what you were opposed to?

Or are you just proposing we stop calling them monzos and vals/mappings and instead call them column vectors and matrices? And an equal temperament's (rank-1) matrix would always just have a single row. In that case perhaps you're not changing anything about my understanding about any of this works, just renaming everything.

I'll admit that there's still something deeply mysterious and unsettling to me about how those six numbers arranged that way represent "meantone", first of all because meantone is just one of those things that I've read about hundreds of times and had explained to me dozens of times but I just can't ever really seem to understand it well enough to retain it or explain it to anyone. Maybe what weirds me out the most is that with a rank-1 temperament I can figure out how big a step is by dividing 1200¢ by the first term, the one for prime 2. But with a rank-2 or more, how are you supposed to know how big each step is? I dunno. I'm probably a hopeless case on this one. I just don't think I've liked or understood a piece of music using a linear temperament well enough to get a grip on it. This probably isn't the forum to try for the upteenth time to explain it to me. Hopefully other people shouldn't have this sort of reaction to this material.

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### Re: general methods for linear temperament notation

Okay, I see the problem, but I'm still pretty disturbed.

I never learned this [ ⟩ angle bracket notation when I was studying linear algebra in college. We always just wrote things with plain brackets [ ]. Why not just use those for generator-count vectors?
I never learned it in math either, but I learned the Dirac notation (or bra-ket notation) in physics where we needed to distinguish vectors in 3-dimensional physical space (or 4D-space-time) from those in more abstract mathematical spaces of quantum mechanics. And so when Gene introduced them from Exterior algebra, I actually thought this was a good notation. Now, I'm not so sure.

When you use square brackets in linear algebra, you are supposed to use the correct 2D arrangement of the numbers. If you don't list the numbers vertically, then it is not a vector but a (single-row) matrix. You can however use the superscript "T" for transpose. So
⎡a⎤ = [a b⟩ could be written [a b]T.
⎣b⎦

But sure. You can use whatever brackets you want, so long as you spell out what they mean in your particular context.
Hm. Those look to me like just parenthetical notes re: the harmonics and how many generators it took to reach them. Besides, in a generator-count vector, doesn't the period come first, then the generator(s)?

Even if George was somehow using G in this way, would you recommend it? I think it would be really cumbersome to put P G₀ G₁ ... after each term in a generator-count vector as a means only to distinguish it from a monzo.
OK. Bad example. Forget George's "G"s.

Yes the period comes first. But the period is also a generator. It's only special because it's chosen so you can make up the (possibly approximate) interval-of-equivalence by using it alone.

No I wasn't suggesting putting anything after each individual number. As per my example in quotes, I suggest putting the word "generators", or the abbreviation "gens", or even just a big "G", after the whole vector.

You'd do better to call them P G1 G2 ... so P can also be considered as G0.
I didn't realize until you said that that you're suggesting monzos/vals are bad. That surprises me because in my ~15 years of experience, microtonal music is impossible without them. They're all I've ever known. So what you're proposing is an alternative to them?
Good grief no. As I've said before, it's the terminology, the actual words "monzo" and "val" that are completely unnecessary and obscure the relationship to the math that at least some people (like yourself) are likely to have already learned.
Or are you just proposing we stop calling them monzos and vals...
Yes.
.../mappings and instead call them column vectors and matrices?
Not that either. But at least that's closer to what I'm proposing. In the sense that it's in the right direction, but it goes too far toward pure math instead of stopping with the math as applied to our specific domain. I don't understand how you could have missed what I've actually been calling them in this thread. "prime exponent vectors", "generator count vectors" and "mappings" (or "mapping matrices").

Saying "column vector" in linear algebra is redundant. All vectors are columns. If it's a row, its a matrix, or a "linear form". But because we've gotten it all mixed up with exterior algebra terminology and notation, thanks to Gene, I felt I had to add "column" to be clear here.
And an equal temperament's (rank-1) matrix would always just have a single row. In that case perhaps you're not changing anything about my understanding about any of this works, just renaming everything.
I wouldn't be renaming everything. I'd be un-renaming them. Those were the terms we were using before Gene turned up and renamed them in an obscure manner. And now we even have people who should know better, like Mike, calling something a monzo just because it's a vector. We should ask Joe Monzo what a monzo is.

And the word "val" wasn't some pre-existing math term that Gene applied. It's just a word Gene made up — to replace the already perfectly serviceable word "mapping", which can be shortened to "map" if you really need it to be one syllable and 3 letters. And "patent val"? "Patent" here just means "obvious". The "patent val" was previously called the "standard mapping".
I'll admit that there's still something deeply mysterious and unsettling to me about how those six numbers arranged that way represent "meantone", first of all because meantone is just one of those things that I've read about hundreds of times and had explained to me dozens of times but I just can't ever really seem to understand it well enough to retain it or explain it to anyone.
That's because there are way too many numbers for a first pass at what meantone is. If you assume octave equivalence and an octave period, you only really need two numbers in the mapping [1, 4]. That says that 1 generator is octave-equivalent to prime 3, or 3/2, and 4 generators are octave-equivalent to prime 5, or 5/4. That tells you that the generator (which can always be made less than an octave) must be somewhere between a pure fifth and whatever octave-extension of 5 you get close to when you stack 4 pure fifths. 3^4/2^4 = 81/16 = 5.065. So the generator is between 3/2 = 1.5 = 702.0¢ and ⁴√5 = 1.495 = 696.6¢.

OK. Not quite true. You have to also look at the ratio between primes 5 and 3. The mapping of 5/3 is 4-1 = 3 generators. If you stack 3 fifths you get something near 10/3. So we could have a generator that was as small as ³√(10/3) = 1.494 = 694.8¢.
Maybe what weirds me out the most is that with a rank-1 temperament I can figure out how big a step is by dividing 1200¢ by the first term, the one for prime 2. But with a rank-2 or more, how are you supposed to know how big each step is? I dunno. I'm probably a hopeless case on this one. I just don't think I've liked or understood a piece of music using a linear temperament well enough to get a grip on it. This probably isn't the forum to try for the upteenth time to explain it to me. Hopefully other people shouldn't have this sort of reaction to this material.
To get the step sizes. You have to choose a specific size of generator (say a middling one like 697¢) and decide how many notes you're going to have. This is not a given like it is with an EDO. Then you have to multiply the size of your generator by every integer from 0 up to one less than your number of notes, reduce all values to the first octave, and sort them into ascending order. Then you can add 1200¢ to the end of the list and subtract each one from the next to see what the step sizes are.

If you want to know what numbers of notes give you a distributionally-even scale, you can ask Wolfram Alpha to give you the convergents of the generator divided by the period (both in cents),
https://www.wolframalpha.com/input/?i=c ... 697%2F1200
and look at the denominators.

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### Re: general methods for linear temperament notation

Dave Keenan wrote: Thu Mar 18, 2021 12:27 pm You can use whatever brackets you want, so long as you spell out what they mean in your particular context.
What I'm hearing is: vectors and matrices are sufficiently fundamental mathematical tools that it would be silly to attempt — even within the microtonal music domain — to standardize certain formats for certain applications and other formats for other applications. It would be of comparable silliness to asking people to, let's say, flare the edges of their vinculums when they are used for interval ratios as opposed to pitch ratios.

I think I get it then. Alright.

One of my takeaways from this is that I never want to see angle brackets again. And I may need to rig these ⎡ ⎤ ⎣ ⎦ characters up in WinCompose.
But the period is also a generator. It's only special because it's chosen so you can make up the (possibly approximate) interval-of-equivalence by using it alone.

...

You'd do better to call them P G1 G2 ... so P can also be considered as G0.
Understood. I see that my poor decision to assign the first generator after P to G0 may have given you the impression I didn't.
No I wasn't suggesting putting anything after each individual number. As per my example in quotes, I suggest putting the word "generators", or the abbreviation "gens", or even just a big "G", after the whole vector.
That's not bad.
"prime exponent vectors", "generator count vectors" and "mappings" (or "mapping matrices").
We use prime exponent vector together a lot. I'm comfortable enough with that I see through it as a synonym for monzo.

I think I'll do a renaming in the Sagittal code away from monzo and val then. It's the kind of thing that's so fundamental to the code that I cannot accept a three-word, nineteen-character long name. But I suppose "pev" will do, an acronym treated as if it's a word. That even buys us back a couple characters
Saying "column vector" in linear algebra is redundant. All vectors are columns. If it's a row, its a matrix, or a "linear form". But because we've gotten it all mixed up with exterior algebra terminology and notation, thanks to Gene, I felt I had to add "column" to be clear here.
Got it.
And an equal temperament's (rank-1) matrix would always just have a single row. In that case perhaps you're not changing anything about my understanding about any of this works, just renaming everything.
I wouldn't be renaming everything. I'd be un-renaming them. Those were the terms we were using before Gene turned up and renamed them in an obscure manner. And now we even have people who should know better, like Mike, calling something a monzo just because it's a vector.
Okay, now I see why you were unhappy to see "tempered monzo" in Mike's writing.
We should ask Joe Monzo what a monzo is.
What do you mean by this?
And the word "val" wasn't some pre-existing math term that Gene applied. It's just a word Gene made up — to replace the already perfectly serviceable word "mapping", which can be shortened to "map" if you really need it to be one syllable and 3 letters.
I'll take "mapping" for the code to avoid confusion with lower-level map operations in the code.
And "patent val"? "Patent" here just means "obvious". The "patent val" was previously called the "standard mapping".
I was previously aware you weren't a fan of that term, yes. Standard mapping is better.
That's because ...

OK. Not quite true. You have to also...
Thanks for trying, but my head's still spinning.
To get the step sizes...
That sounds like it's the answer assuming octave-equivalence. I can handle that case already without much trouble. Where I'm lost is when you've got two different generators. My gut says that 2×3 matrix describes a couple ranges of cent sizes for the two generators which you get to apportion between them.
If you want to know what numbers of notes give you a distributionally-even scale...
Thanks, yeah, I do know that trick.

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### Re: general methods for linear temperament notation

cmloegcmluin wrote: Thu Mar 18, 2021 2:08 pm We use prime exponent vector together a lot. I'm comfortable enough with that I see through it as a synonym for monzo.
That's hilarious, in a sad kind of way — that you have to see "through" a term that is self-explanatory to one that is arbitrary. But presumably, someone who had never had to learn the meaning of the arbitrary word, wouldn't have that problem.

BTW, Lots of people used prime exponent vectors to represent frequency ratios before (or independent of) Joe Monzo, including myself before I learned of the tuning list. It's kind of an obvious move if you've learned linear algebra, or if you draw JI lattices, as they are just the coordinates of the points on a JI lattice.
I think I'll do a renaming in the Sagittal code away from monzo and val then. It's the kind of thing that's so fundamental to the code that I cannot accept a three-word, nineteen-character long name. But I suppose "pev" will do, an acronym treated as if it's a word. That even buys us back a couple characters
Don't do that on my account. It sounds like a lot of work for little gain. But I think "pev" is a fine acronym.
We should ask Joe Monzo what a monzo is.
What do you mean by this?
I only meant "ask Joe" figuratively. I just meant that we should look at how he's defined it. Surely Joe's definition of a monzo should be authoritative, and if you click the link, you'll see that both Joe and Gene defined it as a prime exponent vector, and so I don't think it's valid for Mike to call a generator count vector any kind of monzo.

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### Re: general methods for linear temperament notation

Admittedly, monzo is pretty catchy name. I may never fully wean myself off it. And Joe's the man; he sang on a piece of mine the first time I ever had one recorded. But you've convinced me that using the term "monzo" is bad.

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### Re: general methods for linear temperament notation

cmloegcmluin wrote: Thu Mar 18, 2021 4:05 pm Admittedly, monzo is pretty catchy name. I may never fully wean myself off it. And Joe's the man; he sang on a piece of mine the first time I ever had one recorded.
I love Joe too. He's such a kind and generous man. He deserves to have the prime exponent vector named after him. He did the most to popularise them, particularly in easily-understood graphical form.

I'm kind of just wishing, for pedagogical reasons, that Gene had never coined the term, in which case Joe wouldn't miss what he never had. You're talking to the guy who refused, for the same pedagogical reasons, to have a temperament named after him (me), asking that it instead be given a descriptive name. (It wasn't, but that's beside the point).
But you've convinced me that using the term "monzo" is bad.
I'm not sure I meant to go that far. And there's no way we'd eliminate it now if we wanted to. But I certainly wanted to make a point about pedagogy, and pine for a lost past. And I definitely think it's bad to use "monzo" for generator count vectors, or anything other than prime exponent vectors.

But I'd be happy if I convinced you that the term "val" was evil and should be expunged from the internet and replaced with "mapping" or "edo mapping". But that's never going to happen either.

I note, with pleasure, that the terms monzo and val do not occur anywhere in Paul Erlich's Middle Path paper, but he teaches all about prime exponent vectors and mappings.

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### Re: general methods for linear temperament notation

cmloegcmluin wrote: Thu Mar 18, 2021 2:08 pm One of my takeaways from this is that I never want to see angle brackets again. And I may need to rig these ⎡ ⎤ ⎣ ⎦ characters up in WinCompose.
Perhaps a little extreme. But if so, you'll also need the extensions ⎢ ⎥

I copied them from this table: https://en.wikipedia.org/wiki/Bracket#E ... ital_media

Suitable WinCompose sequences might be:
⎡ = ⎄▲[
⎢ = ⎄◀[
⎣ = ⎄▼[
⎤ = ⎄▲]
⎥ = ⎄▶]
⎦ = ⎄▼]

In this forum, you can also use latex as follows:

$\begin{bmatrix} -4 \\ 4 \\ -1 \end{bmatrix}$

$$\begin{bmatrix} -4 \\ 4 \\ -1 \end{bmatrix}$$

$\begin{bmatrix} 1 & 2 & 4 \\ 0 & -1 & -4 \end{bmatrix}$

$$\begin{bmatrix} 1 & 2 & 4 \\ 0 & -1 & -4 \end{bmatrix}$$

Unfortunately there doesn't seem to be any simple way to right-align the elements.
That sounds like it's the answer assuming octave-equivalence. I can handle that case already without much trouble. Where I'm lost is when you've got two different generators. My gut says that 2×3 matrix describes a couple ranges of cent sizes for the two generators which you get to apportion between them.
Right. The topic of optimising those two generators, and how to decide what you should optimise for, is a complex one. But assuming you just take standard values in cents from Graham's finder, finding the step sizes, differs from my description above only in that, instead of reducing your generator multiples modulo 1200, you'll use say mod 1203.5. And you'll tack 1203.5 on the end of the list (to get the size of the last step).

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### Re: general methods for linear temperament notation

Oh wait. You can right-align the elements if you use "array" instead of "matrix" as follows.

$\left[ \begin{array} {rrr} -1 & -10 & 1 \\ 100 & 5 & 16 \\ 13 & 7 & 7 \end{array} \right]$

$$\left[ \begin{array} {rrr} -1 & -10 & 1 \\ 100 & 5 & 16 \\ 13 & 7 & 7 \end{array} \right]$$


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## Swipe to navigate through the articles of this issue

09-04-2020 | Original Paper | Issue 6/2020

# Impact of point source and mass loading sensitivity on the propagation of an SH wave in an imperfectly bonded FGPPM layered structure

Journal:
Acta Mechanica > Issue 6/2020
Authors:
A. K. Singh, Sonam Singh, Richa Kumari, Anusree Ray
Important notes

## Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Abstract

The aim of the present study is to analyze the propagation behavior of an SH wave in a layered structure constituted of a functionally graded piezo-poroelastic material (FGPPM) layer imperfectly bonded to an FGPPM half-space influenced by a stress point source of disturbance situated at the interface. Mass loading sensitivity of the considered structure is also analyzed by assuming a deposition of an infinitesimally thin layer at the free surface. A mathematical model for mechanical and electrical dynamics of the said material is developed for both considered structures in distinct cases. Appropriate Green’s functions for both layer and half-space are derived by using admissible boundary conditions. The dispersion relation of an SH wave is obtained for both cases, and the obtained results are also validated with classical results of Love wave. The impact of various influencing parameters like functional gradient parameters, imperfectness parameter, piezoelectric coupling parameter, and piezo-porous coupling parameter on the phase velocity of an SH wave is analyzed graphically for a BaTiO$$_3$$-crystal layer and PZT-5H half-space. Moreover, for the case of the mass loading sensitivity, the influencing parameters are also studied for the deposition of a sensitive ZnO layer. Variation of group velocity with wave number is also sketched out graphically.

format: plain


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# [texworks] How do I add figures in Texworks?

Joseph Wright joseph.wright at morningstar2.co.uk
Thu May 8 23:06:51 CEST 2014

On 08/05/2014 15:42, McKinstrie, Colin wrote:
> For several years, I have been using Miktex and Winedt. I added figures (in eps) using the following command.
>
> \begin{figure}[h!]
> \centerline{\includegraphics[width=3.0in]{fig4.eps}}
> \caption{\small Fidelity plotted as a function of link loss.} \label{f1}
> \end{figure}
>
> When I tried the same command using Texworks, space was left for the figures, but the figures did not display. Using pdf figures did not help. (I got a no bounding box error.)
>
> What should I do?

(Not really a TeXworks issue)

TeXworks uses a PDF-based workflow: in any case, you should never use an
extension with \includegraphics:

\includegraphics[width=3.0in]{fig4}

Assuming a recent MiKTeX or TeX Live, auto-magic conversion will then
take place.

Note also that in LaTeX you really shouldn't use \centerline, which is a
plain TeX command: use

\begin{figure}
\centering
\includgraphics

--
Joseph Wright


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# Which One Doesn’t Belong: Fractions

Fractions are a fundamental part of mathematics and a key concept for understanding more complicated mathematical operations. But when it comes to fractions, which one doesn’t belong? From the most basic fractions, to more complex operations, there are certain fractions that simply don’t fit. Let’s take a closer look.

## The Basics of Fractions

At its core, a fraction is a number or ratio that is represented by two different numbers that are divided. The top number is the numerator, and the bottom number is the denominator. For example, one-half is written as 1/2, three-fourths is written as 3/4, and so on. Fractions are used to express parts of a whole, or to express a ratio between two values. Fractions can be written as a whole number, a decimal, or a percentage.

## Which Fractions Don’t Belong?

When it comes to fractions, some fractions don’t belong because they are not mathematically valid. For example, a fraction like 5/1 does not belong because it is not a valid fraction. A fraction like 5/1 is equal to 5, and thus it is not a fraction. Other fractions that don’t belong are fractions with a denominator of 0, such as 4/0. This is not a valid fraction, as it is not possible to divide by 0. Similarly, fractions with a denominator of 1 cannot be considered a fraction, as it is equal to the numerator.

## Improper Fractions

Improper fractions are fractions where the numerator is larger than the denominator. For example, 4/3 is an improper fraction. Improper fractions can be converted to a mixed number, which is a combination of a whole number and a fraction. For example, 4/3 can be written as 1 1/3, which is a whole number plus a fraction. Improper fractions are valid fractions, but they are not always the fractions that are looking for when asking which one doesn’t belong.

## Proper Fractions

Proper fractions are fractions where the numerator is smaller than the denominator. For example, 2/3 is a proper fraction. Proper fractions are valid fractions, and are the fraction that are usually sought when asking which one doesn’t belong. However, it is important to note that all fractions are valid fractions, and it is up to the individual to decide which fraction does not belong in a given context.

##

When it comes to fractions, some fractions don’t belong because they are not mathematically valid. These fractions include those with a denominator of 0, those with a denominator of 1, and those with a numerator larger than the denominator (improper fractions). Proper fractions are those with a numerator smaller than the denominator, and are usually the fractions looked for when asking which one doesn’t belong. Ultimately, all fractions are valid fractions and it is up to the individual to decide which fraction does not belong in a given context.


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# Improper Subset (Definition, Symbol, Example, Venn Diagram)

## Definition of Improper Subset

A set $A$ is called Improper Subset of set $B$ only when all the elements of set $A$ and $B$ are equal to each other and there is no extra element in any of the sets. Improper Subsets can simply be called Subsets. If a set is an Improper Subset of another set, then both sets are equal and have the same Cardinality.

## Improper Subset Symbol

The Improper Subset Symbol is similar to the symbol of the subset. If set $A$ is an Improper Subset of $B$ then mathematically it can be written as,

$\displaystyle A\subseteq B$

## Example of Improper Subset

To understand the fundamentals of Improper Subset, let us consider two sets $A$ and $B$ as given below.

$A$ = {1, 2, 3}

$B$ = {1, 2, 3}

The above figure shows the Improper Subset Venn Diagram. From the figure, it is clear that all the elements of set $A$ are present in the set $B$ and both the sets are equal. There is no extra element in any of the given sets. Therefore, $A$ is called the Improper Subset of $B$ and vice versa. Also, $A$ and $B$ are called equal sets.

## Difference between Improper Subset and Proper Subset | Proper subset vs Improper subset

The Differences between Improper Subset and Proper Subset are as follows:

[sp_easyaccordion id=”3780″]


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Schur complement

(Juju) #1

Hi guys,
I am having an issue writing the Schur complement in CVX that keeps telling me “Disciplined convex programming error”. Here is my problem, I have the following matrix [X x; x’ 1] which is positive semidefinite. Note that X is a K \times K matrix and x column vector of size K. If I write
[[X x];[x’ 1]] == semidefinite(K+1), it works fine. For some reasons, I need to write it as
two constraints, i.e. X positive semidefinite and X-xx’ positive semidefinite. So when I write
X== semidefinite(K) and X-x
x’ == semidefinite(K)
CVX says that “Disciplined convex programming error”. I am almost certain that the problem is with the product "xx’ ". Does someone have a solution to write "xx’ " in CVX?

(Michael C. Grant) #2

You simply cannot. You will have to use the first approach. Why do you think you “need” to write it as two equations.

(Juju) #3

@mcg I am solving an optimization problem using SDR approach. I have provided an optimality condition (which permits to have a rank one solution) where the theoretical analysis was mainly based on the rank of the Lagrangian multipliers associated with the constraint X-x*x’. So I need to see if empirically, it is really true.


text:
# Editor

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text:
• # question_answer 78)   Structures of molecules of two compounds are given below :     (a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding? (b) The melting point of a compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point? (c) Solubility of compounds in water depends on power toform hydrogen bonds with water. Which of the abovecompounds will form hydrogen bond with water easilyand be more soluble in it?

(a) Intramolecular hydrogen bonding is observed in compound $(I)$ because $N{{O}_{2}}$ and OH groups are nearer to each other. In compound $(II)$ intermolecular hydrogen bonding is observed as different molecules are involved. (b) Compound $(II)$ will have higher melting point as it is an associated species. (c) Compound $(I)$ is less soluble than compound $(II)$. Compound $(II)$ easily forms hydrogen bonds with water.


text:
# StackOverflow - Solved

looking for some solutions? You are welcome.

## SOLVED: Confusion about Complex Continued Fraction – math.stackexchange.com

Larry: I faced some problems when I tried to evaluate the following continued fraction \$\$\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cfrac{1}{i+\cdots}}}}}}\ \$\$ The common trick as usual is ...

Posted in S.E.F
via StackOverflow & StackExchange Atomic Web Robots


text:
# Buck–boost converter

The basic schematic of an inverting buck–boost converter.

The buck–boost converter is a type of DC-to-DC converter that has an output voltage magnitude that is either greater than or less than the input voltage magnitude. It is equivalent to a flyback converter using a single inductor instead of a transformer.[1]

Two different topologies are called buck–boost converter. Both of them can produce a range of output voltages, from an output voltage much larger (in absolute magnitude) than the input voltage, down to almost zero.

The inverting topology
The output voltage is of the opposite polarity than the input. This is a switched-mode power supply with a similar circuit topology to the boost converter and the buck converter. The output voltage is adjustable based on the duty cycle of the switching transistor. One possible drawback of this converter is that the switch does not have a terminal at ground; this complicates the driving circuitry. Neither drawback is of any consequence if the power supply is isolated from the load circuit (if, for example, the supply is a battery) because the supply and diode polarity can simply be reversed. The switch can be on either the ground side or the supply side.
A buck (step-down) converter combined with a boost (step-up) converter
The output voltage is typically of the same polarity of the input, and can be lower or higher than the input. Such a non-inverting buck-boost converter may use a single inductor which is used for both the buck inductor and the boost inductor,[2][3][4] it may use multiple inductors but only a single switch as in the SEPIC and Ćuk topologies.

## Principle of operation

Fig. 1: Schematic of a buck–boost converter.
Fig. 2: The two operating states of a buck–boost converter: When the switch is turned-on, the input voltage source supplies current to the inductor, and the capacitor supplies current to the resistor (output load). When the switch is opened, the inductor supplies current to the load via the diode D.

The basic principle of the buck–boost converter is fairly simple (see figure 2):

• while in the On-state, the input voltage source is directly connected to the inductor (L). This results in accumulating energy in L. In this stage, the capacitor supplies energy to the output load.
• while in the Off-state, the inductor is connected to the output load and capacitor, so energy is transferred from L to C and R.

Compared to the buck and boost converters, the characteristics of the buck–boost converter are mainly:

• polarity of the output voltage is opposite to that of the input;
• the output voltage can vary continuously from 0 to $\scriptstyle -\infty$ (for an ideal converter). The output voltage ranges for a buck and a boost converter are respectively 0 to $\scriptstyle V_i$ and $\scriptstyle V_i$ to $\scriptstyle \infty$.

### Conceptual overview

Like the buck and boost converters, the operation of the buck-boost is best understood in terms of the inductor's "reluctance" to allow rapid change in current. From the initial state in which nothing is charged and the switch is open, the current through the inductor is zero. When the switch is first closed, the blocking diode prevents current from flowing into the right hand side of the circuit, so it must all flow through the inductor. However, since the inductor doesn't like rapid current change, it will initially keep the current low by dropping most of the voltage provided by the source. Over time, the inductor will allow the current to slowly increase by decreasing its voltage drop. Also during this time, the inductor will store energy in the form of a magnetic field.

### Continuous mode

Fig 3: Waveforms of current and voltage in a buck–boost converter operating in continuous mode.

If the current through the inductor L never falls to zero during a commutation cycle, the converter is said to operate in continuous mode. The current and voltage waveforms in an ideal converter can be seen in Figure 3.

From $\scriptstyle t=0$ to $\scriptstyle t=D\,T$, the converter is in On-State, so the switch S is closed. The rate of change in the inductor current (IL) is therefore given by

$\frac{\operatorname{d}I_{\text{L}}}{\operatorname{d}t}=\frac{V_i}{L}$

At the end of the On-state, the increase of IL is therefore:

$\Delta I_{\text{L}_{\text{On}}}=\int_0^{D\,T}\operatorname{d}I_{\text{L}}=\int_0^{D\, T}\frac{V_i}{L} \, \operatorname{d}t=\frac{V_i\,D\,T}{L}$

D is the duty cycle. It represents the fraction of the commutation period T during which the switch is On. Therefore D ranges between 0 (S is never on) and 1 (S is always on).

During the Off-state, the switch S is open, so the inductor current flows through the load. If we assume zero voltage drop in the diode, and a capacitor large enough for its voltage to remain constant, the evolution of IL is:

$\frac{\operatorname{d}I_{\text{L}}}{\operatorname{d}t}=\frac{V_o}{L}$

Therefore, the variation of IL during the Off-period is:

$\Delta I_{\text{L}_{\text{Off}}}=\int_0^{\left(1-D\right) T}\operatorname{d}I_{\text{L}}=\int_0^{\left(1-D\right) T}\frac{V_o\, \operatorname{d}t}{L}=\frac{V_o \left(1-D\right) T}{L}$

As we consider that the converter operates in steady-state conditions, the amount of energy stored in each of its components has to be the same at the beginning and at the end of a commutation cycle. As the energy in an inductor is given by:

$E=\frac{1}{2}L\, I_{\text{L}}^2$

it is obvious that the value of IL at the end of the Off state must be the same with the value of IL at the beginning of the On-state, i.e. the sum of the variations of IL during the on and the off states must be zero:

$\Delta I_{\text{L}_{\text{On}}} + \Delta I_{\text{L}_{\text{Off}}}=0$

Substituting $\Delta I_{\text{L}_{\text{On}}}$ and $\Delta I_{\text{L}_{\text{Off}}}$ by their expressions yields:

$\Delta I_{\text{L}_{\text{On}}} + \Delta I_{\text{L}_{\text{Off}}}=\frac{V_i \, D\, T}{L}+\frac{V_o\left(1-D\right)T}{L}=0$

This can be written as:

$\frac{V_o}{V_i}=\left(\frac{D}{D-1}\right)$

This in return yields that:

$D=\frac{V_o}{V_o-V_i}$

From the above expression it can be seen that the polarity of the output voltage is always negative (because the duty cycle goes from 0 to 1), and that its absolute value increases with D, theoretically up to minus infinity when D approaches 1. Apart from the polarity, this converter is either step-up (a boost converter) or step-down (a buck converter). Thus it is named a buck–boost converter.

### Discontinuous mode

Fig 4: Waveforms of current and voltage in a buck–boost converter operating in discontinuous mode.

In some cases, the amount of energy required by the load is small enough to be transferred in a time smaller than the whole commutation period. In this case, the current through the inductor falls to zero during part of the period. The only difference in the principle described above is that the inductor is completely discharged at the end of the commutation cycle (see waveforms in figure 4). Although slight, the difference has a strong effect on the output voltage equation. It can be calculated like follows:

Because the inductor current at the beginning of the cycle is zero, its maximum value $I_{L_{\text{max}}}$ (at $t=D\,T$) is

$I_{L_{\text{max}}}=\frac{V_i\, D\, T}{L}$

During the off-period, IL falls to zero after δ.T:

$I_{L_{\text{max}}}+\frac{V_o\, \delta\, T}{L}=0$

Using the two previous equations, δ is:

$\delta=-\frac{V_i\, D}{V_o}$

The load current $I_o$ is equal to the average diode current ($I_D$). As can be seen on figure 4, the diode current is equal to the inductor current during the off-state. Therefore, the output current can be written as:

$I_o=\bar{I_D}=\frac{I_{L_{\text{max}}}}{2}\delta$

Replacing $I_{L_{\text{max}}}$ and δ by their respective expressions yields:

$I_o=-\frac{V_i\, D\, T}{2L}\frac{V_i\, D}{V_o}=-\frac{V_i^2\, D^2\, T}{2L\, V_o}$

Therefore, the output voltage gain can be written as:

$\frac{V_o}{V_i}=-\frac{V_i\, D^2 \, T}{2L\, I_o}$

Compared to the expression of the output voltage gain for the continuous mode, this expression is much more complicated. Furthermore, in discontinuous operation, the output voltage not only depends on the duty cycle, but also on the inductor value, the input voltage and the output current.

### Limit between continuous and discontinuous modes

Fig 5: Evolution of the normalized output voltage with the normalized output current in a buck–boost converter.

As told at the beginning of this section, the converter operates in discontinuous mode when low current is drawn by the load, and in continuous mode at higher load current levels. The limit between discontinuous and continuous modes is reached when the inductor current falls to zero exactly at the end of the commutation cycle. with the notations of figure 4, this corresponds to :

$D\, T + \delta \, T=T$
$D + \delta = 1\,$

In this case, the output current $\scriptstyle I_{o_{\text{lim}}}$ (output current at the limit between continuous and discontinuous modes) is given by:

$I_{o_{\text{lim}}}=\bar{I_D}=\frac{I_{L_{\text{max}}}}{2}\left(1-D\right)$

Replacing $\scriptstyle I_{L_{\text{max}}}$ by the expression given in the discontinuous mode section yields:

$I_{o_{\text{lim}}}=\frac{V_i\, D\, T}{2L}\left(1-D\right)$

As $\scriptstyle I_{o_{\text{lim}}}$ is the current at the limit between continuous and discontinuous modes of operations, it satisfies the expressions of both modes. Therefore, using the expression of the output voltage in continuous mode, the previous expression can be written as:

$I_{o_{\text{lim}}}=\frac{V_i\, D\, T}{2L}\frac{V_i}{V_o}\left(-D\right)$

Let's now introduce two more notations:

• the normalized voltage, defined by $\scriptstyle \left|V_o\right|=\frac{V_o}{V_i}$. It corresponds to the gain in voltage of the converter;
• the normalized current, defined by $\scriptstyle \left|I_o\right|=\frac{L}{T\, V_i}I_o$. The term $\scriptstyle \frac{T\, V_i}{L}$ is equal to the maximum increase of the inductor current during a cycle; i.e., the increase of the inductor current with a duty cycle D=1. So, in steady state operation of the converter, this means that $\scriptstyle \left|I_o\right|$ equals 0 for no output current, and 1 for the maximum current the converter can deliver.

Using these notations, we have:

• in continuous mode, $\scriptstyle \left|V_o\right|=-\frac{D}{1-D}$;
• in discontinuous mode, $\scriptstyle \left|V_o\right|=-\frac{D^2}{2\left|I_o\right|}$;
• the current at the limit between continuous and discontinuous mode is $\scriptstyle I_{o_{\text{lim}}}=\frac{V_i\, T}{2L}D\left(1-D\right)=\frac{I_{o_{\text{lim}}}}{2\left|I_o\right|}D\left(1-D\right)$. Therefore the locus of the limit between continuous and discontinuous modes is given by $\scriptstyle \frac{1}{2\left|I_o\right|}D\left(1-D\right)=1$.

These expressions have been plotted in figure 5. The difference in behaviour between the continuous and discontinuous modes can be seen clearly.

## Non-ideal circuit

### Effect of parasitic resistances

Fig 6: Evolution of the output voltage of a buck–boost converter with the duty cycle when the parasitic resistance of the inductor increases.

In the analysis above, no dissipative elements (resistors) have been considered. That means that the power is transmitted without losses from the input voltage source to the load. However, parasitic resistances exist in all circuits, due to the resistivity of the materials they are made from. Therefore, a fraction of the power managed by the converter is dissipated by these parasitic resistances.

For the sake of simplicity, we consider here that the inductor is the only non-ideal component, and that it is equivalent to an inductor and a resistor in series. This assumption is acceptable because an inductor is made of one long wound piece of wire, so it is likely to exhibit a non-negligible parasitic resistance (RL). Furthermore, current flows through the inductor both in the on and the off states.

Using the state-space averaging method, we can write:

$V_i=\bar V_{\text{L}} + \bar V_S$

where $\scriptstyle \bar V_{\text{L}}$ and $\scriptstyle \bar V_S$ are respectively the average voltage across the inductor and the switch over the commutation cycle. If we consider that the converter operates in steady-state, the average current through the inductor is constant. The average voltage across the inductor is:

$\bar V_{\text{L}}=L\frac{\bar{dI_{\text{L}}}}{dt}+R_{\text{L}}\bar I_{\text{L}}=R_{\text{L}}\bar I_{\text{L}}$

When the switch is in the on-state, $\scriptstyle V_S=0$. When it is off, the diode is forward biased (we consider the continuous mode operation), therefore $\scriptstyle V_S = V_i - V_o$. Therefore, the average voltage across the switch is:

$\bar V_S=D\, 0 + (1-D)(V_i-V_o)=(1-D)(V_i-V_o)$

The output current is the opposite of the inductor current during the off-state. the average inductor current is therefore:

$\bar I_{\text{L}}=\frac{-I_o}{1-D}$

Assuming the output current and voltage have negligible ripple, the load of the converter can be considered purely resistive. If R is the resistance of the load, the above expression becomes:

$\bar I_{\text{L}}=\frac{-V_o}{(1-D)R}$

Using the previous equations, the input voltage becomes:

$V_i= R_{\text{L}}\frac{-V_o}{(1-D)R}+ (1-D)(V_i-V_o)$

This can be written as:

$\frac{V_o}{V_i}=\frac{-D}{\frac{R_{\text{L}}}{R(1-D)}+1-D}$

If the inductor resistance is zero, the equation above becomes equal to the one of the ideal case. But when RL increases, the voltage gain of the converter decreases compared to the ideal case. Furthermore, the influence of RL increases with the duty cycle. This is summarized in figure 6.


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# M09-18

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15 Sep 2014, 23:39
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How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution of acid?

(1) There are 50 grams of the 35%-solution.

(2) In the 35%-solution of acid the ratio of acid to water is 7:13.

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15 Sep 2014, 23:40
Official Solution:

Say there are $$x$$ grams of 35%-solution of acid and $$y$$ grams of water should be added to obtain a 10%-solution of acid. Equate amount of acid: $$0.35*x=0.1*(x+y)$$.

Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)).

(1) There are 50 grams of the 35%-solution of acid. Given $$x=50$$, so $$0.35*50=0.1*(50+y)$$. We have a linear equation with only one unknown, hence we can get the single numerical value of $$y$$. Sufficient.

(2) In the 35%-solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%-solution of acid means that the ratio of acid to water is 35:65=7:13. Not sufficient.

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### Show Tags

27 Sep 2014, 06:27
Bunuel wrote:
Official Solution:

Say there are $$x$$ grams of 35%-solution of acid and $$y$$ grams of water should be added to obtain a 10%-solution of acid. Equate amount of acid: $$0.35*x=0.1*(x+y)$$.

Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)).

(1) There are 50 grams of the 35%-solution of acid. Given $$x=50$$, so $$0.35*50=0.1*(50+y)$$. We have a linear equation with only one unknown, hence we can get the single numerical value of $$y$$. Sufficient.

(2) In the 35%-solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%-solution of acid means that the ratio of acid to water is 35:65=5:13. Not sufficient.

Are there anymore Questions on similar lines which are Not part of Gmat Club Tests but are of the forum?
I tend to get lost in x%solution of y kind of questions.

Math Expert
Joined: 02 Sep 2009
Posts: 50623

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27 Sep 2014, 12:04
1
earnit wrote:
Bunuel wrote:
Official Solution:

Say there are $$x$$ grams of 35%-solution of acid and $$y$$ grams of water should be added to obtain a 10%-solution of acid. Equate amount of acid: $$0.35*x=0.1*(x+y)$$.

Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)).

(1) There are 50 grams of the 35%-solution of acid. Given $$x=50$$, so $$0.35*50=0.1*(50+y)$$. We have a linear equation with only one unknown, hence we can get the single numerical value of $$y$$. Sufficient.

(2) In the 35%-solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%-solution of acid means that the ratio of acid to water is 35:65=5:13. Not sufficient.

Are there anymore Questions on similar lines which are Not part of Gmat Club Tests but are of the forum?
I tend to get lost in x%solution of y kind of questions.

DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114

Hope it helps.
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27 Sep 2014, 12:12
(1) Given x=50

0.35*50=0.1*(50+y).

linear equation, one unknown,Sufficient.

(2) acid to water ratio, 7:13. same as information given that it is a 35%-solution. Not sufficient.
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13 Dec 2014, 11:45
1
Bunuel wrote:
Official Solution:

Say there are $$x$$ grams of 35%-solution of acid and $$y$$ grams of water should be added to obtain a 10%-solution of acid. Equate amount of acid: $$0.35*x=0.1*(x+y)$$.

Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)).

(1) There are 50 grams of the 35%-solution of acid. Given $$x=50$$, so $$0.35*50=0.1*(50+y)$$. We have a linear equation with only one unknown, hence we can get the single numerical value of $$y$$. Sufficient.

(2) In the 35%-solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%-solution of acid means that the ratio of acid to water is 35:65=5:13. Not sufficient.

It seems that there is a small typing error in your explanation for (2) the ratio should be 35:65=7:13.
Math Expert
Joined: 02 Sep 2009
Posts: 50623

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15 Dec 2014, 07:36
haihai89 wrote:
Bunuel wrote:
Official Solution:

Say there are $$x$$ grams of 35%-solution of acid and $$y$$ grams of water should be added to obtain a 10%-solution of acid. Equate amount of acid: $$0.35*x=0.1*(x+y)$$.

Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)).

(1) There are 50 grams of the 35%-solution of acid. Given $$x=50$$, so $$0.35*50=0.1*(50+y)$$. We have a linear equation with only one unknown, hence we can get the single numerical value of $$y$$. Sufficient.

(2) In the 35%-solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%-solution of acid means that the ratio of acid to water is 35:65=5:13. Not sufficient.

It seems that there is a small typing error in your explanation for (2) the ratio should be 35:65=7:13.

Thank you. Edited.
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19 Feb 2016, 13:42
2
With an allegation chart it might look like this:

35%-----------0%
25-(10)-10

So the ratio of Solution to Pure Water is 2:5

A) 2X=50 x= 25 5*25= 125 grams of water need to be added
B) 2x=7x/13x InSuf

Ans. A
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30 Mar 2016, 07:42
I'm super confused here. Look there are 35% of acid and 65% of water and as per the statement 1, there is 50gms of acid.

Now 35% gives 50gms of acid and 65% gives to 92gms of water. Approx 142.86 is the total amt of mixture of acid and water.

35/100( total mixture of acid and water)=50gm of acid

Hence total mixture= 142.86

Now I add 125L of water (as per the soln) to 142.86 which gives 267. Now if i calculate 50/267= 18% of acid and it is not giving me 10% of acid. Where am I going wrong?
Math Expert
Joined: 02 Aug 2009
Posts: 7036

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30 Mar 2016, 08:29
1
1
harish1986 wrote:
I'm super confused here. Look there are 35% of acid and 65% of water and as per the statement 1, there is 50gms of acid.

Now 35% gives 50gms of acid and 65% gives to 92gms of water. Approx 142.86 is the total amt of mixture of acid and water.

35/100( total mixture of acid and water)=50gm of acid

Hence total mixture= 142.86

Now I add 125L of water (as per the soln) to 142.86 which gives 267. Now if i calculate 50/267= 18% of acid and it is not giving me 10% of acid. Where am I going wrong?

Hi,

few points..

1)when we say 50 gms of 35%, it means the entire solution is 50gm
so acid= 35*0.5=17.5gm and water=65*0.5=32.5
2) when we add x quantity of water, acid becomes 10%
so $$\frac{17.5}{(x+50)} = \frac{10}{100}$$..
so 175=x+50..
x=125..
this means 125 gm of water is to be added

so A is correct
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28 Nov 2016, 14:41
I think this is a high-quality question and I agree with explanation. Interesting example to understand the diference between ratio and percentage
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28 Aug 2017, 02:43
Bunuel wrote:
Official Solution:

Say there are $$x$$ grams of 35%-solution of acid and $$y$$ grams of water should be added to obtain a 10%-solution of acid. Equate amount of acid: $$0.35*x=0.1*(x+y)$$.

Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)).

(1) There are 50 grams of the 35%-solution of acid. Given $$x=50$$, so $$0.35*50=0.1*(50+y)$$. We have a linear equation with only one unknown, hence we can get the single numerical value of $$y$$. Sufficient.

(2) In the 35%-solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%-solution of acid means that the ratio of acid to water is 35:65=7:13. Not sufficient.

Hi Bunuel

Acid+x/Total+x=Desired Qty/Total
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28 Aug 2017, 02:54
himanshukamra2711 wrote:
Bunuel wrote:
Official Solution:

Say there are $$x$$ grams of 35%-solution of acid and $$y$$ grams of water should be added to obtain a 10%-solution of acid. Equate amount of acid: $$0.35*x=0.1*(x+y)$$.

Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)).

(1) There are 50 grams of the 35%-solution of acid. Given $$x=50$$, so $$0.35*50=0.1*(50+y)$$. We have a linear equation with only one unknown, hence we can get the single numerical value of $$y$$. Sufficient.

(2) In the 35%-solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%-solution of acid means that the ratio of acid to water is 35:65=7:13. Not sufficient.

Hi Bunuel

Acid+x/Total+x=Desired Qty/Total

Check here: https://gmatclub.com/forum/how-much-wat ... 88479.html

18. Mixture Problems

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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14 Nov 2017, 22:29
High quality question!
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01 Jan 2018, 20:30
WRONG SENTENCE STRUCTURE! If you add the water to the strong acid, you will get a BIG explosion! That is a CHEM 101 lab test question! Moron test writers...
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Re: M09-18 &nbs [#permalink] 01 Jan 2018, 20:30
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# M09-18

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text:
Conditional probability given two variables from conditional probabilities of one variable

I have a question which comes from an example of a textbook, but all I am concerned with is how we go from having probabilities $P(X|A)$ and $P(X|B)$ to having $P(X|A,B)$. In the example events $A$ and $B$ are independent, each influencing $X$. The example comes from Pattern Classification by Duda et. al, edition 2.

The example gives us a table defining $P(x_i|a_j)$ and $P(x_i|b_j)$. Shortly thereafter the authors have a summation over $P(x_1|a_i,b_j)$:

$$\sum\limits_{i,j}P(x_1|a_i, b_j)$$

Later, they expand this sum into the particular values, which appear to me to be the values of $P(x_i|a_j)$. The numbers end up adding up to 4 which makes sense since there are four rows in $P(x_i|a_j)$ (each which must sum to 1).

The probability $P(x_1|a_i, b_j)$ makes a second appearance in the example, but unfortunately the authors have left a magic step and I am unsure of the values they used.

My question is why is this OK to do? What is the relationship between $P(X|A)$, $P(X|B)$, and $P(X|A,B)$? Do we need more information to get $P(X|A,B)$ and are they doing some sort of approximation?

The section is dealing with Bayesian belief networks. If you happen to have the book, it is Example 4 in Section 2.11.

-
I may have figured something out something unrelated to this question that may help me, though this question is still on the table and I'd really be interested to know the answer. – OEP Mar 22 '13 at 4:49
In case anyone is struggling with the same problem, I'm pretty certain now it is an error with the book. There is a revised section 2.11 with an example that's been almost completely rewritten. – OEP Mar 22 '13 at 13:55
I can see this Example 4 with Google Books but I don't see this summation. – Stéphane Laurent Mar 22 '13 at 14:18
You are right. I think I have an old version of the book which has errors in it. I'm currently separated from my newer version so that's probably the problem. – OEP Mar 22 '13 at 16:48


text:
# Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{c}}\ge2\sum\limits_{cyc}\sqrt{\frac{c}{a+b}}$

Let $a,b,c$ be positive numbers. Then we need to prove

$\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\ge2\left(\sqrt{\frac{c}{a+b}}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}\right).$

I have an idea to set $x=\frac a{b+c}$, $y=\frac b{c+a},z=\frac c{a+b}$ then $\frac1{1+x}+\frac1{1+y}+\frac1{1+z}=2$ and we need to prove $\frac1{\sqrt x}+\frac1{\sqrt y}+\frac1{\sqrt z}\ge2\left(\sqrt x+\sqrt y+\sqrt z\right)$

But I could not go further.

It is a consequence of Chebychev's inequality: $$\sum_{cyc}\sqrt{\frac{a+b}{c}}≥2\sum_{cyc}\sqrt{\frac{c}{a+b}}\iff\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}≥0$$ Since the $a+b-2c$ and $\frac{1}{\sqrt{c(a+b)}}$ are ordered in the same way, we can apply Chebychev's inequality to obtain: $$\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}≥\frac{1}{3}\left(\sum_{cyc}a+b-2c\right)\left(\sum_{cyc}\frac{1}{\sqrt{c(a+b)}}\right)=0$$
If we consider two real sequences $a_1,a_2,…,a_n$ and $b_1,b_2,…,b_n$ for which $a_1≤a_2≤…≤a_n$ and $b_1≤b_2≤…≤b_n$, then Chebyvhev tells us, that: $$\frac{a_1b_1+a_2b_2+…+a_nb_n}{n}≥\frac{a_1+a_2+…+a_n}{n}\cdot\frac{b_1+b_2+…+b_n}{n}$$ We are allowed to use it in this case, because by symmetry, we can assume $a≥b≥c>0$. This implies: $$a+b-2c≥a+c-2b≥b+c-2a$$ And: $$ab≥ac≥bc\iff a(b+c)≥b(a+c)≥c(a+b) \iff\\ \frac{1}{\sqrt{c(a+b)}}≥\frac{1}{\sqrt{b(a+c)}}≥\frac{1}{\sqrt{a(b+c)}}$$ So the two sequences we have in the above inequality are indeed ordered in the same way.
$$\sum_{cyc}\sqrt{\frac{a+b}{c}}-2\sum_{cyc}\sqrt{\frac{c}{a+b}}=\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}=$$ $$=\sum_{cyc}\frac{b-c-(c-a)}{\sqrt{c(a+b)}}=\sum_{cyc}(a-b)\left(\frac{1}{\sqrt{b(a+c)}}-\frac{1}{\sqrt{a(b+c)}}\right)=$$ $$=\sum_{cyc}\frac{(a-b)^2c}{\sqrt{ab(a+c)(b+c)}\left(\sqrt{a(b+c)}+\sqrt{b(a+c)}\right)}\geq0.$$ Done!


text:
On the linear independence of $p$-adic $L$-functions modulo $p$  [ Sur l’indépendance linéaire des fonctions $L$ $p$-adiques modulo $p$ ]
Annales de l'Institut Fourier, Tome 60 (2010) no. 5, pp. 1831-1855.

Soit $p\ge 3$ un nombre premier. Soit $n\in ℕ$ tel que $n\ge 1$, soient ${\chi }_{1},...,{\chi }_{n}$ des caractères de conducteur $d$ premier à $p$ ; notons $\omega$ le caractère de Teichmüller. Pour tout $i$ entre $1$ et $n$ et pour tout $j$ entre $0$ et $\left(p-3\right)/2$, on pose

 ${\theta }_{i,j}=\left\{\begin{array}{cc}{\chi }_{i}{\omega }^{2j+1}\hfill & \phantom{\rule{0.277778em}{0ex}}\text{si}\phantom{\rule{4pt}{0ex}}{\chi }_{i}\phantom{\rule{0.277778em}{0ex}}\mathrm{est}\phantom{\rule{4pt}{0ex}}\mathrm{impair}\phantom{\rule{4pt}{0ex}};\hfill \\ {\chi }_{i}{\omega }^{2j}\hfill & \phantom{\rule{0.277778em}{0ex}}\text{si}\phantom{\rule{4pt}{0ex}}{\chi }_{i}\phantom{\rule{0.277778em}{0ex}}\mathrm{est}\phantom{\rule{4pt}{0ex}}\mathrm{pair}.\hfill \end{array}\right\$

Soit $K={ℚ}_{p}\left({\chi }_{1},...,{\chi }_{n}\right)$ et soit $\pi$ un premier de l’anneau de valuation ${𝒪}_{K}$ de $K$. Pour tout $i,j$ notons $f\left(T,{\theta }_{i,j}\right)$ la série d’Iwasawa associée à ${\theta }_{i,j}$ et $\overline{f\left(T,{\theta }_{i,j}\right)}$ sa réduction modulo $\left(\pi \right)$. Finalement soit $\overline{{𝔽}_{p}}$ une clôture algébrique de ${𝔽}_{p}$. Nous montrons que si les caractères ${\chi }_{i}$ sont distincts modulo $\left(\pi \right)$, alors $1$ et les séries $\overline{f\left(T,{\theta }_{i,j}\right)}$, sont linéairement indépendantes sur un certain corps $\Omega$ qui contient $\overline{{𝔽}_{p}}\left(T\right)$.

Let $p\ge 3$ be a prime. Let $n\in ℕ$ such that $n\ge 1$, let ${\chi }_{1},...,{\chi }_{n}$ be characters of conductor $d$ not divided by $p$ and let $\omega$ be the Teichmüller character. For all $i$ between $1$ and $n$, for all $j$ between $0$ and $\left(p-3\right)/2$, set

 ${\theta }_{i,j}=\left\{\begin{array}{cc}{\chi }_{i}{\omega }^{2j+1}\hfill & \phantom{\rule{0.277778em}{0ex}}\text{if}\phantom{\rule{4pt}{0ex}}{\chi }_{i}\phantom{\rule{0.277778em}{0ex}}\mathrm{is}\phantom{\rule{4pt}{0ex}}\mathrm{odd};\hfill \\ {\chi }_{i}{\omega }^{2j}\hfill & \phantom{\rule{0.277778em}{0ex}}\text{if}\phantom{\rule{4pt}{0ex}}{\chi }_{i}\phantom{\rule{0.277778em}{0ex}}\mathrm{is}\phantom{\rule{4pt}{0ex}}\mathrm{even}.\hfill \end{array}\right\$

Let $K={ℚ}_{p}\left({\chi }_{1},...,{\chi }_{n}\right)$ and let $\pi$ be a prime of the valuation ring ${𝒪}_{K}$ of $K$. For all $i,j$ let $f\left(T,{\theta }_{i,j}\right)$ be the Iwasawa series associated to ${\theta }_{i,j}$ and $\overline{f\left(T,{\theta }_{i,j}\right)}$ its reduction modulo $\left(\pi \right)$. Finally let $\overline{{𝔽}_{p}}$ be an algebraic closure of ${𝔽}_{p}$. Our main result is that if the characters ${\chi }_{i}$ are all distinct modulo $\left(\pi \right)$, then $1$ and the series $\overline{f\left(T,{\theta }_{i,j}\right)}$ are linearly independent over a certain field $\Omega$ that contains $\overline{{𝔽}_{p}}\left(T\right)$.

DOI : https://doi.org/10.5802/aif.2573
Classification : 11R23,  11R18,  11S80,  11J72
Mots clés : fonctions $L$ $p$-adiques, transformée de Leopoldt $p$-adique, théorie d’Iwasawa, irrationalité
@article{AIF_2010__60_5_1831_0,
author = {Angl\es, Bruno and Ranieri, Gabriele},
title = {On the linear independence of $p$-adic $L$-functions modulo $p$},
journal = {Annales de l'Institut Fourier},
pages = {1831--1855},
publisher = {Association des Annales de l'institut Fourier},
volume = {60},
number = {5},
year = {2010},
doi = {10.5802/aif.2573},
mrnumber = {2766231},
zbl = {1219.11162},
language = {en},
url = {www.numdam.org/item/AIF_2010__60_5_1831_0/}
}
Anglès, Bruno; Ranieri, Gabriele. On the linear independence of $p$-adic $L$-functions modulo $p$. Annales de l'Institut Fourier, Tome 60 (2010) no. 5, pp. 1831-1855. doi : 10.5802/aif.2573. http://www.numdam.org/item/AIF_2010__60_5_1831_0/`

[1] Anglès, Bruno On the $p$-adic Leopoldt transform of a power series, Acta Arith., Volume 134 (2008) no. 4, pp. 349-367 | Article | MR 2449158 | Zbl pre05354502

[2] Lang, Serge Cyclotomic fields I and II, Graduate Texts in Mathematics, Volume 121, Springer-Verlag, New York, 1990 (With an appendix by Karl Rubin) | MR 1029028

[3] Sinnott, W. On the power series attached to $p$-adic $L$-functions, J. Reine Angew. Math., Volume 382 (1987), pp. 22-34 | Article | MR 921164 | Zbl 0621.12015

[4] Washington, Lawrence C. Introduction to cyclotomic fields, Graduate Texts in Mathematics, Volume 83, Springer-Verlag, New York, 1997 | MR 1421575 | Zbl 0966.11047


text:
Sharp PC-G850VS
09-21-2018, 05:18 PM (This post was last modified: 11-06-2018 09:01 PM by toml_12953.)
Post: #101
 toml_12953 Senior Member Posts: 1,795 Joined: Dec 2013
RE: Sharp PC-G850VS
(09-14-2018 04:22 PM)jwhsu Wrote:  I wasn't aware there was a STAT mode in the PC-G850VS. There was no mention of it in the German Manual. I have a pdf of the Japanese manual, however, a quick look didn't mention anything about a STAT mode.

Could you tell us where to get the Japanese PDF? It would be good to have a copy of the original manual as I learn Japanese.

Tom L
Cui bono?
09-21-2018, 05:34 PM (This post was last modified: 09-21-2018 05:35 PM by Zaphod.)
Post: #102
 Zaphod Member Posts: 270 Joined: Apr 2018
RE: Sharp PC-G850VS
Whilst looking for that manual (above - couldn’t find it right now) I found these pages which should be of interest to ‘850 owners:

09-22-2018, 02:00 AM
Post: #103
 jwhsu Member Posts: 76 Joined: Sep 2015
RE: Sharp PC-G850VS
(09-21-2018 05:18 PM)toml_12953 Wrote:
(09-14-2018 04:22 PM)jwhsu Wrote:  I wasn't aware there was a STAT mode in the PC-G850VS. There was no mention of it in the German Manual. I have a pdf of the Japanese manual, however, a quick look didn't mention anything about a STAT mode.

Could you tell us where to get the Japanese PDF? It would be good to have a copy of the original manual as I learn Japanese.

Tom L

I found it on foroplus’ website under manuals The link is here.
09-22-2018, 02:03 AM (This post was last modified: 11-06-2018 09:02 PM by toml_12953.)
Post: #104
 toml_12953 Senior Member Posts: 1,795 Joined: Dec 2013
RE: Sharp PC-G850VS
(09-22-2018 02:00 AM)jwhsu Wrote:
(09-21-2018 05:18 PM)toml_12953 Wrote:  Could you tell us where to get the Japanese PDF? It would be good to have a copy of the original manual as I learn Japanese.

Tom L

I found it on foroplus’ website under manuals The link is here.

Great! Thanks.

Tom L
Cui bono?
09-22-2018, 06:48 PM (This post was last modified: 09-22-2018 06:49 PM by Zaphod.)
Post: #105
 Zaphod Member Posts: 270 Joined: Apr 2018
RE: Sharp PC-G850VS
Known bug in Basic: LCOPY

One to add to the book perhaps to page 205 where LCOPY is explained.

It's not a statement/something you'd normally have in a program, but you can ... and it does crash the system completely, you'll need a hard reset to get out of this one.

Moral of the bug is: don't add it to a program line

Source:
10-01-2018, 02:25 AM
Post: #106
 megarat Junior Member Posts: 14 Joined: Jan 2014
RE: Sharp PC-G850VS
I'm interested in learning more about the Turbo mode that has been mentioned here. I've done some searching to see how to accomplish this, but so far I haven't found anything, perhaps because my searches have been in English.

I've been playing around with my PC-G850VS, thanks to Jack's excellent manual, and it's been a lot of fun. In a way, it feels like this device was designed for the purposes of education, with the simplified way that only one compiled C program can be held in memory, the implementation of the CASL/COMET system, etc. Does anyone know more about the target demographic for this device?
10-01-2018, 12:19 PM (This post was last modified: 10-01-2018 12:19 PM by Zaphod.)
Post: #107
 Zaphod Member Posts: 270 Joined: Apr 2018
RE: Sharp PC-G850VS
(10-01-2018 02:25 AM)megarat Wrote:  Does anyone know more about the target demographic for this device?

Unsurprisingly: japan (computing) education market
10-01-2018, 12:52 PM
Post: #108
 xerxes Member Posts: 116 Joined: Jun 2014
RE: Sharp PC-G850VS
(10-01-2018 02:25 AM)megarat Wrote:  I'm interested in learning more about the Turbo mode that has been mentioned here. I've done some searching to see how to accomplish this, but so far I haven't found anything, perhaps because my searches have been in English.

You need to replace the 8 MHz resonator to a higher value. To have no complications with data transfer, it would be practical to use 16 or 32 MHz and setting the baud rate to the half or quarter.

Calculator Benchmark
10-24-2018, 07:40 PM (This post was last modified: 10-24-2018 07:41 PM by Jlouis.)
Post: #109
 Jlouis Senior Member Posts: 671 Joined: Nov 2014
RE: Sharp PC-G850VS
Because of this thread and the englsh manual, i just grab an almost new VS specimen from Japan.

I'm playing with it and so far, a really pleasure to use.

One thing I notice is that you can choose to use postfix or prefix for sqr, cur, etc.

Thanks folks!

Cheers

JL
08-21-2019, 11:38 PM
Post: #110
 koikoi Junior Member Posts: 4 Joined: Aug 2019
RE: Sharp PC-G850VS
(09-18-2018 10:58 PM)jwhsu Wrote:  New version of the manual has been posted.
Thanks for all your work on this English manual! It's actually how I came to find this forum. I figured I'd register to say thanks.

(10-24-2018 07:40 PM)Jlouis Wrote:  Because of this thread and the englsh manual, i just grab an almost new VS specimen from Japan.
Do you mind me asking what was the price of your VS?

(04-16-2018 06:39 PM)mfleming Wrote:  There is a link to the source for an emulator called g800 below. It emulates the G850, G815 and E200. Who knows what a perusal of the source might reveal?

http://ver0.sakura.ne.jp/
This site also includes a lot of sample programs each of which have their own source code. Of particular interest to me is an implementation of Koi-Koi.

cheers!
09-01-2019, 11:16 AM
Post: #111
 Jlouis Senior Member Posts: 671 Joined: Nov 2014
RE: Sharp PC-G850VS
(08-21-2019 11:38 PM)koikoi Wrote:
(10-24-2018 07:40 PM)Jlouis Wrote:  Because of this thread and the englsh manual, i just grab an almost new VS specimen from Japan.
Do you mind me asking what was the price of your VS?

US$100 09-06-2019, 12:58 PM Post: #112  koikoi Junior Member Posts: 4 Joined: Aug 2019 RE: Sharp PC-G850VS Why thanks! 12-23-2019, 12:52 PM Post: #113  david16965 Junior Member Posts: 2 Joined: Dec 2019 RE: Sharp PC-G850VS I've recently acquired a working example of the Sharp PC G850 from Japan. It answers my desire for a portable C / ASM device with a calculator-like interface and battery life. I'm very grateful for the English language translation of the original Japanese language manual. One thing that is not clear to me is how to delete lines of C code in the text editor, and how to renumber the remainder. I can see that insertion of new lines between existing lines is easy because the editor sorts by line number; it's simply a matter of picking an intermediate value for the line number of the inserted line. Likewise, replacement of an existing line is easy; I just use the existing line number for the new line. I would be very grateful for any guidance on deletion / renumbering of C code. 12-23-2019, 04:21 PM Post: #114  david16965 Junior Member Posts: 2 Joined: Dec 2019 RE: Sharp PC-G850VS (12-23-2019 12:52 PM)david16965 Wrote: I've recently acquired a working example of the Sharp PC G850 from Japan. It answers my desire for a portable C / ASM device with a calculator-like interface and battery life. I'm very grateful for the English language translation of the original Japanese language manual. One thing that is not clear to me is how to delete lines of C code in the text editor, and how to renumber the remainder. I can see that insertion of new lines between existing lines is easy because the editor sorts by line number; it's simply a matter of picking an intermediate value for the line number of the inserted line. Likewise, replacement of an existing line is easy; I just use the existing line number for the new line. I would be very grateful for any guidance on deletion / renumbering of C code. I figured it out. The text editor commands like "A", "R" and "D" are case sensitive and are executed as soon as they are entered, without a line number, at any point when editing C code. Very neat. I think I'm going to enjoy using this little machine! 03-07-2020, 03:38 AM Post: #115  Luca Puntin Junior Member Posts: 4 Joined: Mar 2020 RE: Sharp PC-G850VS Hi, I'm new on this forum, so I present myself. First of all... excuse me in advance for my poor English, I'm italian and I don't like translators... I'm a 56 yo teacher of Mechanics (practice teacher) and electonics hobbyst. I've recently acquired a Sharp PC-G850VS directly from Japan (around 110 Euro) with the main target of PIC programming. I know that a very poor PC with PICKIT2 or 3 is more flexible, but the Sharp ultra-portable solution is somewhat fascinating... I've downloaded the english manual (thanks to who did a so good work, translating it from Japanese and/or German), but I've got two questions about programming a PIC: 1) how can I enter the "_" character (for the __CONFIG directive to assembler) into text editor ? 2) where is the file "p16f84a.inc" located ? About the first question, I don't see the underscore on G850 keyboard, but I don't know if there is a function like the "alt+nnn" keystroke on a PC numeric pad: what is sure is that "_" does exist into the G850: going into BASIC run mode, typing PRINT CHR$ 95 + (Enter) results in "_" visualization. But there is no "Copy and Past" to the Text editor.
I'm tempted to run a BASIC program to save a couple of "_" into Text editor working ram area, but I suppose that area is protected when reserved to editor, so a machine code (Z80 assembler) routine may be the solution.... But if someone has just programmed a PIC there must be a simpler way.

About second question, I suppose that "p16f84a.inc" is located into G850 firmware, so a bit hard to modify for change PIC model. PIC16F627 (that is declared usable) is a pin-to-pin substitute of the 16F84A, but I suppose to be difficult use internal oscillator, analog inputs an comparators that are not present into 16F84a. But it is obviously still possible do all "EQU's" into program, without use "include" assembler directive.

Bye.

Luca
03-07-2020, 03:58 PM
Post: #116
 mfleming Senior Member Posts: 750 Joined: Jul 2015
RE: Sharp PC-G850VS
The underscore character "_" is SHIFT-K which is not too obvious! As to the .inc file location, it is likely in ROM and thus not editable. You could incluude the header anyway, then override or add to the defines to get what you need.

Congratulations to member jwhsu for the translation and others for their contributions. The 850 is an interesting handheld!

Remember kids, "In a democracy, you get the government you deserve."
03-08-2020, 09:13 AM
Post: #117
 Luca Puntin Junior Member Posts: 4 Joined: Mar 2020
RE: Sharp PC-G850VS
Thanks Mfleming, now I can go on buildind the ICSP programming interface for PIC's!

I've just to order the DC/DC converter, all other stuff needed is somewhere in my repository (read: somewhere in drawers, boxes, garage, bedroom...).

Looking at page 241 of the manual (Appendix A: 11 pin interface) I see that pin #1 isn't used by any interface configuration, so I'm thinking to place a small piece of plastic into it (like polarization key in IDC flat female cable connectors) to avoid reversing insertion of an home-made connector to the port.

I think that overriding "p16F627.inc" definitions consists of include that file at the top of the assembler listate and then re-define what is needed ( I usually edit the .inc itself, but on a real PC it is all SO easy...).
I don't know if load from a PC a modified "pxxxx.inc" (waste of memory on the G850 apart) can works, but this need, first of all, to interface the two devices each to the other.
Have anyone do this with the USB adaptor cable shown here : http://basic.hopto.org/basic/g850v.php
It works ?
An other question: when I program PIC's, I usually download the proper ".inc" file and then edit it before use, modifing the BADRAM declarations, in order to avoid the BADRAM warnings of MPASMWIN compiler. I usually don't use the BANKSEL directive, switching BANK0 / BANK1 by setting or resetting RP0 bit into STATUS file register (or even RP1 if that PIC has memory BANK2 and/or BANK3 too); the built-in G850 compiler, with a modified ".inc" filke, will works in this way ?
I own the G850 in VS version, so the p16fxxx.inc files are probably into a flashable device, but I thing it is not a good idea to try flashing the firmware with new PIC's .inc files without known exactly what to do (or, better, NOT to do).
In Appendix F: system bus of the manual, he connections of the system bus are illustrated; years ago (first '90s) I've made a new dashboard for my car, converting it from analog to digital, implementing a Z80 CPU, CTC, one ADC0804 and some strange VFD (vacum fluorescent displays) with strange drivers (National MM58342N, serial input / parallel output cascadable register drivers with voltage output shiftable levels to drive VFD displays) and I programmed Z80 in Assembler.
As I remember, referring to 850VS system bus pinout, pins #6 (IORESET), #11(BNK1), #12(BNK0), #13 (CEROM2) and # 14 (CERAM2) are not part of the Z80 bus, but are probably used by 850VS firmware to switch external memory (see Appendix I: memory map), probably ROM banks #2 and #3), but signals names seems not corresponding exactly to this function.
I'm obviously thinking to a memory expansion cartridge full of ".inc" files.....

One last thing: SHIFT-U is the " ' " charachter, that is not to obvious too.....

I'll keep this forum updated about PIC programming with the PC 850VS.

Bye.

Luca
03-09-2020, 02:07 AM
Post: #118
 mfleming Senior Member Posts: 750 Joined: Jul 2015
RE: Sharp PC-G850VS
Good luck with your PIC programming. Regarding a usb connection to the Sharp's serial port - I have done it with an FTDI-based cable. The Sharp serial signal logic is inverted and FTDI provides a configuration utility to alter its own signal logic to match. Works great for moving files to/from the PC.

Remember kids, "In a democracy, you get the government you deserve."
03-09-2020, 01:59 PM
Post: #119
 Luca Puntin Junior Member Posts: 4 Joined: Mar 2020
PC G850 PIC programming
Well...
First of all (you've just confirmed it) the spanish way to interface the G850 to a PC via FTDI cable works, so this is the first accessory to build.
Then I must order a bunch (10 pcs.) of PIC 16F627A from China (waiting around one month to receive them, but price is SO MUCH competitive...).
Last thing: I've found a Canon S70 electronic typewriter and it has a serial port to use it like a printer: I'm very tempted to buy it and use like a printer for the G850...

Thanks for the tips !

Bye.

Luca
03-19-2020, 12:24 AM
Post: #120
 thenozone Junior Member Posts: 18 Joined: Mar 2017
RE: Sharp PC-G850VS
Android calculator, pocket computer emulator "PockEmul" has an emulation of the sharp
"PC-G850V" and downloadable english manual. the batch of newer sharp computers cost £1.79 on their store via android play store; and lots of other devices to play with too.
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