Logic

[Subhotosh Khan]

Agree!

We've asked Ryan$ to follow the guidelines, so tutors can know what he's talking about. I intend to start enforcing those requests.

→ Not a basic face - neither an acidic face - just a neutral face (pH = 7) - as clear as water

 

[Ryan$]

Hello every one, ! thanks alot for your answers !!!

for whom you think I troll, I said I'm not trolling in advance, I said I'm not that much smarter and everytime I post here I apologize before about thats questions !
I admit that I have a suck mind, but whatever, I want to learn, not to still dull.

 

[Ryan$]

Hi guys, I'm sorry to post like this post here but I find it hard to accept what's I'm going to say !
if A=B , is it the same to say B=A? if it's, then why? what's confusing me A=B isn't in writings the same B=A so how we determined that A=B is the same as B=A!!!!!
I know it's equal, but who said we just care on equal? maybe also on the order of writing the elements of two sides of equation .. who said not?!

 

[topsquark]

In any Math system I have studied we have to have that A = B implies B = A.

For example: If we know that x = 2 then we also know that 2 = x. The order is unimportant.

-Dan

 

[pka]

I find it hard to accept what's I'm going to say !
if A=B , is it the same to say B=A? if it's, then why? what's confusing me A=B isn't in writings the same B=A so how we determined that A=B is the same as B=A!!!!!

You know that Samuel Clemens is Mark Twain.
Would you be bothered by someone saying Mark Twain is Samuel Clemens?

 

[Dr.Peterson]

if A=B , is it the same to say B=A? if it's, then why? what's confusing me A=B isn't in writings the same B=A so how we determined that A=B is the same as B=A!!!!!
I know it's equal, but who said we just care on equal? maybe also on the order of writing the elements of two sides of equation .. who said not?!

Since another question hints that you know something about computer programming, it may be worth pointing out that "=" can be used in different ways. Sometimes order doesn't matter, sometimes it does.

"Who says?" This is just an implicit agreement among users of the notation, in a particular context -- like all language! If you choose to be part of a community, and communicate with them, you use the words and symbols they use, in the way they use them.

In math, "=" by itself always means merely "is equal to". When we say that A = B, it means that quantities A and B are equal -- both quantities play the same role; it doesn't matter which order you put them. You can think of it as bidirectional, or symmetrical. (It can also be used in a definition, as in "let n = 3", which is not symmetrical; but even there, "=" just means these two quantities are equal; it is the sentence in which it is found that changes it from a mere statement of equality to a definition.)

This is what all mathematicians accept as the meaning. And this is what the symbol has meant since it was first invented, as described here.

But some programming languages use "=" in a different way, meaning "is assigned", similar to the usage in definitions. There, writing A = B means "put the current value of variable B into variable A". This is one-directional; it actually tells the computer to do something to variable A, and not to B. For example, in programming you can say "x = x + 1", which in math would be nonsense, but in a program changes the value of x by adding 1 to it.

Similarly, in English, the word "is" can be used in different ways. Taking pka's example, "Samuel Clemens is Mark Twain" means that they are two names for the same person; but "Samuel Clemens is an author" describes one aspect of Clemens, and identifies only one of many people who are authors -- it doesn't equate all that Samuel Clemens is with all authors. Or, returning to math, we can say "a square is a rectangle" but "a rectangle is a square" means something very different; this usage is asymmetrical.

But many students think of "=" in math as if it meant "has the answer", as in "2 + 3 = 5" meaning "if you add 2 + 3, the answer you get is 5". That is a misunderstanding; it really means merely that 2 + 3 and 5 are two "names" for the same quantity. When you move from arithmetic into algebra, you have to leave that earlier way of thinking behind.

 

[Ryan$]

Hi guys ! I hope this thread wouldn't be closed because I really struggle that and I need to understand and LEARN!

if I have any equations doesn't matter what it's , for instance f(x) = x^2+6 , I've a problem that I deeply know if I want to move forward in the solution then I must assign into the equation , I mean, I have a case in my problem/question which I must assign into the equation x0 (*specific case* but it's satisfy the equation) , then my question can I assign f(x0)=(x0)^2 + 6 and continue with my solution? what I mean by that I have the equation in general "f(x)=x^2+6 " , in whatever way in my question I arrived to conclusion that x0 in my question must satisfy the equation, then can I assign it into the equation? what's confused me x0 is a specific case that must satisfy the equation .. but the equation is general case .. so how do we assign speicfic x0 into the equation which it's general?





another question, the teacher in the video said if we have X(Y+Z) and we already know that Y is regardless to Z then X(Y+Z) is approximated to X*Z , why it's right? I mean why it's right to disregard Y which it's inside the parentheses .. if it was Y+Z explicitly without any parentheses then I accept the approximation .... what's confusing me why it's allowed to disregard what's inside the parentheses without worrying what's going out of the parentheses !

 

[Harry_the_cat]

Firstly you need to now the difference between:
1. an expression like x²+6 ….. possibly can be factorised, expanded or simplified (not in this case)
2. an equation like x²+6 =10 …… can be solved (ie what value of x makes it true)
3. a function f(x) = x² + 6 OR y = x²+6 …. shows a relationship and can be graphed

 

[topsquark]

In general we have the definition \(\displaystyle f(x) = x^2 + 6\) so if we have any specific value of x in mind, such as \(\displaystyle x_0\), then \(\displaystyle f(x_0 ) = (x_0 )^2 + 6\), as you say. Any value of \(\displaystyle x_0\) will do.

\(\displaystyle f(1) = 1^2 + 6 = 1 + 6 = 7\)
\(\displaystyle f(-3) = (-3)^2 + 6 = 15\)
\(\displaystyle f(85786) = (85786)^2 + 6 = 85786^2 + 6 = 7359237802\)
\(\displaystyle f(0.34699) = (0.34699)^2 + 6 = 6.1204020601\)

etc.

-Dan

 

[Harry_the_cat]

Your second question:
Instead of saying that "Y is regardless to Z", I think what you mean is that "Y is relatively small compared to Z".

For example if Y = 0.0001 and Z = 100000 and X is whatever you like
then
X(Y+Z) = X(0.0001 + 100000) is approximately X*100000 ie X*Z, not exactly of course but approximately.

 

[Ryan$]

Your second question:
Instead of saying that "Y is regardless to Z", I think what you mean is that "Y is relatively small compared to Z".

For example if Y = 0.0001 and Z = 100000 and X is whatever you like
then
X(Y+Z) = X(0.0001 + 100000) is approximately X*100000 ie X*Z, not exactly of course but approximately.

I understand that, but I mean why we are approximating Y and Z , what about X .. isn't he affect the approximation of Y relative to Z? I mean if it was given Y+Z then I accept the approximation if Y smaller than Z then Y
but what about if X(Y+Z) ? who said it's correct to do first approximation to the first term "(Y+Z)" and then multiply it by X and we would get the approximated value of the whole term?!

 

[HallsofIvy]

No, if Y and Z are specific numbers then "approximation of Y relative to Z" is not "affected" by any other number. We can asses the relative sizes of Y and Z by "Y/Z" or "Z/Y". That has nothing to do with any other number, "X". And the fact that it is "correct to do first approximation to the first term "(Y+Z)" and then multiply it by X" comes from the meaning of parentheses in mathematics: X(Y+ Z) means "first do Y+ Z, then multiply the result by X.

 

[Ryan$]

No, if Y and Z are specific numbers then "approximation of Y relative to Z" is not "affected" by any other number. We can asses the relative sizes of Y and Z by "Y/Z" or "Z/Y". That has nothing to do with any other number, "X". And the fact that it is "correct to do first approximation to the first term "(Y+Z)" and then multiply it by X" comes from the meaning of parentheses in mathematics: X(Y+ Z) means "first do Y+ Z, then multiply the result by X.

I almost got your point ! but still confused really about how it's not affected .. I'm not totally convinced !
if in math we do first the term of (Y+Z), and then multiply the result by X, doesn't that mean they are related? I mean the result is related to X because we want to multiply the result by X .... so the total after multiplying is related to "result" .. isn't it? so there's a relation ..
I mean the approximation that I make in first term is affecting on the total and the X also affecting the total, so X is related to total, and Y+Z is related to total, so we conclude that X is related to Y+Z ..


still not totally convinced on how really the approximation of the fird st terms isn't affecting the other terms..

 

[topsquark]

If you want to do it this way as well, then...
X(Y + Z) = XY + XZ. Since Y is small compared to Z then XY will be small compared to XZ. If you don't see this directly, put some numbers in like Harry_the_cat suggested in post 4.

-Dan

 

[JeffM]

You need to understand the difference between an approximation with a low absolute error and an approximation with a low relative error.

\(\displaystyle \text {Let } \epsilon \text { be an arbitrary real number} > 0.\)

\(\displaystyle \left |f(x_1,\ y_1,\ z_1) - f(x_1,\ y_1,\ 0) \right | < \epsilon \iff f(x_1,\ y_1,\ 0) \text { approximates } f(x_1,\ y_1,\ z_1) \text { with a low absolute error.}\)

\(\displaystyle \therefore x_1(y_1 + 0) = x_1y_1 \text { approximates } x_1(y_1 + z_1) = x_1y_1 + x_1z_1 \text { with a low absolute error} \implies \)

\(\displaystyle \left |(x_1y_1 + x_1z_1) - x_1y_1 \right | < \epsilon \implies |x_1z_1| < \epsilon.\)

With this expression, an approximation with a low absolute error depends only on the relation between x_1 and z_1 once epsilon is specified. However, what is more usually of interest is an approximation with a low relative error.

\(\displaystyle \left | \dfrac{f(x_1,\ y_1,\ z_1)}{f(x_1,\ y_1, 0} - 1 \right | < \epsilon \implies f(x_1,\ y_1,\ 0) \text { approximates } f(x_1,\ y_1,\ z_1) \text { with a low relative error.}\)

\(\displaystyle \therefore x_1(y_1 + 0) = x_1y_1 \text { approximates } x_1(y_1 + z_1) = x_1y_1 + x_1z_1 \text { with a low relative error} \implies \)

\(\displaystyle \left | \dfrac{x_1y_1 + x_1z_1}{x_1z_1} - 1 \right | < \epsilon \implies\)

\(\displaystyle \left | \dfrac{y_1 + z_1}{y_1} - 1 \right | < \epsilon \implies \left | 1 + \dfrac{z_1}{y_1} - 1 \right | < \epsilon \implies\)

\(\displaystyle \left | \dfrac{z_1}{y_1} \right | < \epsilon.\)

With this expression, an approximation with a low relative error depends only on the relation between y_1 and z_1 once epsilon is specified.

 

[Ryan$]

If you want to do it this way as well, then...
X(Y + Z) = XY + XZ. Since Y is small compared to Z then XY will be small compared to XZ. If you don't see this directly, put some numbers in like Harry_the_cat suggested in post 4.

-Dan

So my question is how to do that directly ....? it takes me alot of time to understand and do the approximation , in that case what can I do?

 

[Otis]

So my question is how to do that directly …

It requires mental effort, Ryan.

Are you willing to do some work?

\(\;\)

 

[Dr.Peterson]

So my question is how to do that directly ....? it takes me a lot of time to understand and do the approximation, in that case what can I do?

Actually, the approximation takes no time at all; it's just ignoring the Y, right?

And the decision whether to do that is not hard; but it depends on context. If a 1% error in your answer is acceptable, and Y is less than 1% of Z (actually, of their sum), then it's acceptable to ignore it. Unfortunately, you haven't told us the context of the video, so we have no idea what conditions led to the statement that Y can be ignored.

As for X, the idea is that multiplication does not change the relative error. The relative (percent) error in Z relative to Z+Y is the same as the relative error in XZ relative to X(Z+Y), because relative error is proportional; the X can be canceled. So, again, the decision whether to ignore a small number is easy: you only need to compare to the other addend, and don't have to consider any other parts of the calculation.

But by not clearly showing the specific problem you are asking about, you have complicated things. What was actually said in that video? What were they doing? How did they know that Y was negligible?

 

[Ryan$]

It requires mental effort, Ryan.

Are you willing to do some work?

\(\;\)

but lets assume you have EXP(LOG(EXP(LOG(EXP(X-Y))))) and X is neglected relative to Y, so my problem how do I do mental Effort on EXP(LOG(EXP(LOG(EXP(x-y))))))) i'M NOT robot ..what you're talking about ...

I need to understand and then apply what I understand ..

 

[topsquark]

So my question is how to do that directly ....? it takes me alot of time to understand and do the approximation , in that case what can I do?

Both methods have been given to you.
1) X(Y + Z). Given Y << Z (that's Mathspeak for Y is much smaller than Z), means Y + Z is very close to Z. Thus X(Y + Z) is close to X(0 + Z) = XZ.

2)X(Y + Z) = XY + XZ, Y<<Z means that XY<< XZ, so X(Y + Z) = XY + XZ is very close to XZ.

There are no other ways to look at this. Try some numbers: Let X = 1650, Y = 10, Z = 144 and see what happens in both cases.

-Dan

Addendum: Why the heck would you be trying to think of something like EXP(LOG(EXP(LOG(EXP(X-Y))))) ?? That would be like asking you to solve \(\displaystyle x^3 + 3x^2 + x + 2 = 0\) after you were shown how to use the quadratic formula. Potentially possible but completely useless as a teaching aid for that level. You need to get the basics down before you hit something like this. You are over-reaching.... Learn the small stuff first.