# Logic

Staff member
Agree!

We've asked Ryan$to follow the guidelines, so tutors can know what he's talking about. I intend to start enforcing those requests. Not a basic face - neither an acidic face - just a neutral face (pH = 7) - as clear as water #### Ryan$

##### Full Member

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.

##### Full Member
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

##### Senior Member
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

##### Full Member
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

##### Senior Member
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.

##### Full Member
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

##### Full Member
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

##### Elite Member
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.

##### Full Member
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

##### Full Member
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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