logic of math's equations

Ryan$

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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 !
 
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Harry_the_cat

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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

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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

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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$

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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

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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$

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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

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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

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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$

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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

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So my question is how to do that directly …
It requires mental effort, Ryan.

Are you willing to do some work?

\(\;\)
 

Dr.Peterson

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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$

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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

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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.
 
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Ryan$

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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?
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?
It was given that Y was negligible ..

my problem is that .. who said that relative error wouldn't change if we continue with other operands/elements?(in my case multiplication .. ) ?! ..may please gimme an example for that? thanks alot! exactly that's my problem, why wouldn't relative error change if we continue to other elements/operands ?!
 

JeffM

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It was given that Y was negligible ..

my problem is that .. who said that relative error wouldn't change if we continue with other operands/elements?(in my case multiplication .. ) ?! ..may please gimme an example for that? thanks alot! exactly that's my problem, why wouldn't relative error change if we continue to other elements/operands ?!
You are trying to generalize from one specific case. I gave you the definitions for both the relative and absolute cases. Those definitions are what is general. Then you must apply them individually to each specific case.

We keep telling you to give specific cases. I have a feeling that you have tried to deduce a general rule that y is always relevant and x is always irrelevant to an approximation. There is no such general rule.
 

Dr.Peterson

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my problem is that .. who said that relative error wouldn't change if we continue with other operands/elements?(in my case multiplication .. ) ?! ..may please gimme an example for that? thanks alot! exactly that's my problem, why wouldn't relative error change if we continue to other elements/operands ?!
In math, the issue is not "who said?". We don't go by authority. If you question what you're told, you can check it out (that is, make up an example for yourself!), or examine the reasons that were given. I showed you a reason:
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.
Did you try working through what I said here, seeing why the X has no effect? Did you try working through an example (which has been mentioned in this thread long ago)? You're not going to understand anything fully if you don't do the thinking for yourself.

Now, in general, if there is something other than multiplication involved, it might turn out that relative error is changed. No one here (and presumably not the person in the video) has claimed any broad generalization. In order to discuss how relative error is affected by various functions, we'd have to look at error propagation and calculus. But that's not part of your question.

Now, your questions might be very much worth discussing in the context of the video (which you have told us nothing about). It may be that there are things that weren't said that should have been, or even things that were done that are not fully justified. But if you can't show us the video, then you should just be writing to its author, rather than to us, asking what his justifications were. Don't be looking for extreme generalizations that don't exist.
 

Otis

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… X is neglected relative to Y , so my problem how do I do mental Effort on EXP(LOG(EXP(LOG(EXP(x-y))))))) …
The same way you were shown in posts #3 and #4: Experiment, by substituting numbers for X and Y and evaluating to compare results. Choose values for X that are relatively small compared to Y.

Your expression simplifies to e^(X-Y), so you may experiment using that simpler version.

Are you willing to try?

\(\;\)
 

Ryan$

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In math, the issue is not "who said?". We don't go by authority. If you question what you're told, you can check it out (that is, make up an example for yourself!), or examine the reasons that were given. I showed you a reason:

Did you try working through what I said here, seeing why the X has no effect? Did you try working through an example (which has been mentioned in this thread long ago)? You're not going to understand anything fully if you don't do the thinking for yourself.

Now, in general, if there is something other than multiplication involved, it might turn out that relative error is changed. No one here (and presumably not the person in the video) has claimed any broad generalization. In order to discuss how relative error is affected by various functions, we'd have to look at error propagation and calculus. But that's not part of your question.

Now, your questions might be very much worth discussing in the context of the video (which you have told us nothing about). It may be that there are things that weren't said that should have been, or even things that were done that are not fully justified. But if you can't show us the video, then you should just be writing to its author, rather than to us, asking what his justifications were. Don't be looking for extreme generalizations that don't exist.

what do you mean by "relative error" here I didn't understand that term
 

JeffM

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what do you mean by "relative error" here I didn't understand that term
For goodness' sake ryan, do you bother to try understanding what is written in response to your questions?
The term is defined in post 9. Dr. P used it in post 11. Suddenly, in post 19, you notice that you have not been aware of the topic being discussed. It is possible that the video was unclear, but, as usual, you do not give a link or anything that gives specific context.
 
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