Economist questioned for doing math on a flight

[JusKeepBreathin 19,311]

1 hour ago, LebaneseDude said:

Algebra is named after an Arabic mathematician called Al-Jaber.

You've already been indoctrinated. 

I knew there was a logical explanation for my hatred and fear of algebra. 

[LebaneseDude 6,146]

10 minutes ago, just4fun said:

I knew there was a logical explanation for my hatred and fear of algebra. 

Good thing I don't consider myself Arab otherwise I would have taken offense at your Arabophobia 

[ThisGuyTony 35,349]

I can't.  

[JusKeepBreathin 19,311]

21 minutes ago, LebaneseDude said:

Good thing I don't consider myself Arab otherwise I would have taken offense at your Arabophobia 

Actually my fear is called Arithmophobia. lol Arabs are hot. 

[Florian 20,017]

50 minutes ago, Licq said:

You can add a constant to the left side, then you move the constant on the left side to the right side, you still end up with a c on the right side, it doesn't matter what value c really is..

My point was more about adding the constant which is completely useless. In all my studies it has never been useful. 

[LebaneseDude 6,146]

1 hour ago, Florian said:

My point was more about adding the constant which is completely useless. In all my studies it has never been useful. 

What?

To put it simply, the derivative of a function gets you the rate of change with respect to the variable. Integrating that function returns you to the higher order equation.

For example, in the case of linear equations, the derivative of the function is the slope of the graph. Integrating the slope gets you the line again.

Here's a function: f(x) = 2x + 3

Get the derivative: f'(x) = 2

Now integrate to return to the function without the constant : f(x) = 2x

What's missing? Where did the 3 go?

The point of the constant is to point out that there could have been (and in this case that there was) a constant in the equation. f(x) = 2x is not the same as f(x) = 2x + 3

If you had put a constant, f'(x) = 2 integrates to f(x) = 2x + c 

If you know a point on the graph, you can replace the values to find c.

So if we had known that (0,3) was on the graph, replacing would yield:

3 = 2(0) + c  so c = 3

so f(x) = 2x + 3

Voila...back to the original equation.

[Lord Temptation 11,209]

2 hours ago, Florian said:

My point was more about adding the constant which is completely useless. In all my studies it has never been useful. 

What have you been studying? 

[LetsGetHigh 52,683]

6 hours ago, Florian said:

What scares me is that he's adding a constant on the right side but not on the left side when he integrates  This isn't logical 

That's how differential equations work.  

[Florian 20,017]

7 hours ago, LebaneseDude said:

What?

To put it simply, the derivative of a function gets you the rate of change with respect to the variable. Integrating that function returns you to the higher order equation.

For example, in the case of linear equations, the derivative of the function is the slope of the graph. Integrating the slope gets you the line again.

Here's a function: f(x) = 2x + 3

Get the derivative: f'(x) = 2

Now integrate to return to the function without the constant : f(x) = 2x

What's missing? Where did the 3 go?

The point of the constant is to point out that there could have been (and in this case that there was) a constant in the equation. f(x) = 2x is not the same as f(x) = 2x + 3

If you had put a constant, f'(x) = 2 integrates to f(x) = 2x + c 

If you know a point on the graph, you can replace the values to find c.

So if we had known that (0,3) was on the graph, replacing would yield:

3 = 2(0) + c  so c = 3

so f(x) = 2x + 3

Voila...back to the original equation.

Thanks for the lessons, but I know how to do that for way more years than you seem to think. The fact is that finding an original equation isn't useful (at least in the fields I've been studying) except in the first lesson when you learn what intégration is. 

6 hours ago, Lord Temptation said:

What have you been studying? 

Math, physics, mechanics and automation. Needless to say that dérivation and intégration was required most of the time for almost 5years now.

[LebaneseDude 6,146]

3 hours ago, Florian said:

Thanks for the lessons, but I know how to do that for way more years than you seem to think. The fact is that finding an original equation isn't useful (at least in the fields I've been studying) except in the first lesson when you learn what intégration is. 

Oh  

Right because unless constants are big enough, we engineers like to assume they don't exist.  

Still I didn't know your background and your statement isn't true when applied to high school math lol

[Florian 20,017]

Just now, LebaneseDude said:

Oh

Right because unless constants are big enough, we engineers like to assume they don't exist.  

Still I didn't know your background and your statement isn't true when applied to high school math lol

Yeah, I'm studying to become an engineer that's probably why. 

But even in my high scool math class we almost always set the constant to 0 cause it was useless to the problems studied.