Why does math have to suck so badly?

SkyHog

SkyHog

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Everything Offends Me
Ok guys - I need some serious help here, and I don't really have the time for a tutor...I hope someone can help explain something to me.

Today's class pretty much went over my head. I'll start with what I understood, and follow with what I don't get:

I understand that f(x) + g(x) = (f+g)(x) and I understand that f(x) - g(x) = (f-g)(x). I also understand that f(x) * g(x) = (f*g)(x) and I understand that f(x) / g(x) = (f/g)(x).

That makes perfect sense to me.

I also understand that on the coordinate plane, (x,y) is the same as (x,f(x)).

What I don't get is why to get x to equal x+h, you also have to add h to f(x), as in (x+h, f(x)+h). And why this point somehow defines the secant of a parabola.

And then.....the fun of the "difference quotient." f(x+h)-f(x)/h

My notes are essentially what was written on the board, so I need some help with that also:

f(x) = x2 - 3x +2
f(x+h) = (x+h)2 - 3(x+h) + 2 - (x2 - 3x + 2) / h
f(x+h) = x2 + 2hx + h2 - 3x - 3h +2 - x2 +3x - 2 / h
f(x+h) = 2hx + h2 - 3h / h
f(x+h) = h(2x + h - 3) /h
f(x+h) = 2x + h - 3

I get the simplifing part. What I don't really get is why you need to sub f(x) with f(x+h) to begin with, and why the difference quotient matters.

Whatever happened to the days when I could easily solve the entire homework assignment (like 50 probs) on one page. Now it seems to take one page to solve one problem. Sigh.
 
Huh? ...

(The ... was to satisfy the minimum post length) :D
 
SkyHog said:
I get the simplifing part. What I don't really get is why you need to sub f(x) with f(x+h) to begin with, and why the difference quotient matters.
Click to expand...

They are prepping you for calculus, integrals and differentials where it will make more of a different. Kinda like flying, do something a certain way in private, and you don't have to unbreak a habit getting your IR.
 
N2212R said:
They are prepping you for calculus, integrals and differentials where it will make more of a different. Kinda like flying, do something a certain way in private, and you don't have to unbreak a habit getting your IR.
Click to expand...

Right - but how am I supposed to know that I'm supposed to do that? Here's a sample problem that requires it:

Find the difference quotient of f for each function. Be sure to simplify
73. f(x) = 4x + 3

My guess at the solution is:
f(x+h) = 4(x + h) + 3 - (4x + 3) / h
f(x+h) = 4x + 4h + 3 - (4x + 3) / h
f(x+h) = 4h / h
f(x+h) = 4

Am I even close to being right?
 
SkyHog said:
Ok guys - I need some serious help here, and I don't really have the time for a tutor...I hope someone can help explain something to me.

Today's class pretty much went over my head. I'll start with what I understood, and follow with what I don't get:

I understand that f(x) + g(x) = (f+g)(x) and I understand that f(x) - g(x) = (f-g)(x). I also understand that f(x) * g(x) = (f*g)(x) and I understand that f(x) / g(x) = (f/g)(x).

That makes perfect sense to me.

I also understand that on the coordinate plane, (x,y) is the same as (x,f(x)).

What I don't get is why to get x to equal x+h, you also have to add h to f(x), as in (x+h, f(x)+h). And why this point somehow defines the secant of a parabola.

And then.....the fun of the "difference quotient." f(x+h)-f(x)/h

My notes are essentially what was written on the board, so I need some help with that also:

f(x) = x2 - 3x +2
f(x+h) = (x+h)2 - 3(x+h) + 2 - (x2 - 3x + 2) / h
f(x+h) = x2 + 2hx + h2 - 3x - 3h +2 - x2 +3x - 2 / h
f(x+h) = 2hx + h2 - 3h / h
f(x+h) = h(2x + h - 3) /h
f(x+h) = 2x + h - 3

I get the simplifing part. What I don't really get is why you need to sub f(x) with f(x+h) to begin with, and why the difference quotient matters.

Whatever happened to the days when I could easily solve the entire homework assignment (like 50 probs) on one page. Now it seems to take one page to solve one problem. Sigh.
Click to expand...

OK, I'm just a lowly EE and I took my college math classes over 30 years ago (and haven't really used them since college - how's that for encouragement to learn this stuff?), but I think part of the problem is "What I don't get is why to get x to equal x+h, you also have to add h to f(x), as in (x+h, f(x)+h)." I would expect the last term to be f(x+h), not f(x)+h.

Difference quotient? I have no idea what that is, but you are approaching taking the first derivitive of the equation (which when evaluated at a point gives you the slope of the line at that point).

The first derivitive of the equation you present, f(x) = x2 - 3x +2 , is 2x-3. X can be any value we want, so if you feed in x+h you get 2(x+h) -2 which equals 2x + 2h -3 for the slope of the line at that offset point.

Does this make sense?
 
Over My head Nick, been out of school to many years 34 to be exact.
Dave G.
 
SkyHog said:
Right - but how am I supposed to know that I'm supposed to do that? Here's a sample problem that requires it:

Find the difference quotient of f for each function. Be sure to simplify
73. f(x) = 4x + 3

My guess at the solution is:
f(x+h) = 4(x + h) + 3 - (4x + 3) / h
f(x+h) = 4x + 4h + 3 - (4x + 3) / h
f(x+h) = 4h / h
f(x+h) = 4

Am I even close to being right?
Click to expand...

Nick, I can't remember what the difference quotient is, but based on your example above and my mental algebra, this is correct.

Tony,
Wannabe Aerospace Engineer
Survivor of Lots of Math, up to Differential Equations
 
First Nick let me applaud you for learning a new language. Math is a universal language that can explain a great many things. You are almost to the point where it all starts to make sense. Once you get into calculus, differential equations, and vector calc, you will be amazed.

The difference quotient you are quoting has a small mistake in it.

It is NOT f(x+h)-f(x)/h at all,

It is [f(x+h)-f(x)]/h and that makes a big difference and is what you actually did in your operation and simplification. But what the heck does it mean?

It is a way of calculating the slope along a line at any point. In linear equation the slope is constant but in non linear equations it is constantly changing. What this is doing is calculating that rate of change at any given point. In calculate you will learn a more elegant method of doing this.


The h term is ment to be slightly away from point x, so small in fact that it is almost right next to the x value. If you move just slightly away from x you can see the affect on f(x) and then measure that rate of change or slope.

In you example the slope of x^2-3x+2 is simply 2x-3 plus a little fudge factor that is so small we can ignore it and still have a correct answer. In calc that remainder will be called d/dx instead of h and it means the change due to a change in x.

This is also called the first derivative.

In the linear example you have f(x)=mx+b where m is the slope. If you applied the difference quotient to that formula you would have

[m(x+h)+b-mx-b]h
=(mx-mx+b-b+mh)/h
=mh/h
=m

Think back to algebra and the first discussions of slope. It was something like (y2-y1)/(x2-x1)

f(x) is a fancy name for y
So if f(x) =y1
and f(x+h)=y2

x=x
x+h=x2

Then we subsitute and get
[f(x+h)-f(x)]/[(x+h)-x] simplify and we get our difference quotient

[f(x+h)-f(x)]/h

Does that help?
 
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SkyHog said:
Right - but how am I supposed to know that I'm supposed to do that? Here's a sample problem that requires it:

Find the difference quotient of f for each function. Be sure to simplify
73. f(x) = 4x + 3

My guess at the solution is:
f(x+h) = 4(x + h) + 3 - (4x + 3) / h
f(x+h) = 4x + 4h + 3 - (4x + 3) / h
f(x+h) = 4h / h
f(x+h) = 4

Am I even close to being right?
Click to expand...
That is right, see the proof in my previous note. but do get your brackets in teh right spot. what you are writing would only devide f(x) by h not the sum of f(x+h)-f(x) by h

The solution should look like this
= [4(x + h) + 3 - (4x + 3)] / h
= [4x + 4h + 3 - (4x + 3) ]/ h
= 4h / h
= 4

If you leave off the brackets your math is wrong and you should be ending up with 4x+4h+3 -(4x+3)/h which is the wrong answer
 
Last edited:
smigaldi said:
First Nick let me applaud you for learning a new language. Math is a universal language that can explain a great many things. You are almost to the point where it all starts to make sense. Once you get into calculus, differential equations, and vector calc, you will be amazed.

The difference quotient you are quoting has a small mistake in it.

It is NOT f(x+h)-f(x)/h at all,

It is [f(x+h)-f(x)]/h and that makes a big difference and is what you actually did in your operation and simplification. But what the heck does it mean?

It is a way of calculating the slope along a line at any point. In linear equation the slope is constant but in non linear equations it is constantly changing. What this is doing is calculating that rate of change at any given point. In calculate you will learn a more elegant method of doing this.


The h term is ment to be slightly away from point x, so small in fact that it is almost right next to the x value. If you move just slightly away from x you can see the affect on f(x) and then measure that rate of change or slope.

In you example the slope of x^2-3x+2 is simply 2x-3 plus a little fudge factor that is so small we can ignore it and still have a correct answer. In calc that remainder will be called d/dx instead of h and it means the change due to a change in x.

This is also called the first derivative.

In the linear example you have f(x)=mx+b where m is the slope. If you applied the difference quotient to that formula you would have

[m(x+h)+b-mx-b]h
=(mx-mx+b-b+mh)/h
=mh/h
=m

Think back to algebra and the first discussions of slope. It was something like (y2-y1)/(x2-x1)

f(x) is a fancy name for y
So if f(x) =y1
and f(x+h)=y2

x=x
x+h=x2

Then we subsitute and get
[f(x+h)-f(x)]/[(x+h)-x] simplify and we get our difference quotient

[f(x+h)-f(x)]/h

Does that help?
Click to expand...

That helps immensely. That's getting printed and going into my notes :)

So - for the time being, the only way I'll know to use the difference quotient is if the problem is written as "Find the difference quotient....."?

I think I can handle that.

As for the missing brackets - oops - I was copying it directly from my written work, in which I put the first term over the other. Didn't even think about missing the bracket when changing to "/"

Thanks man!
 
Your welcome and positive feedback is apreciated.

Here is some extra credit.

The only way that the difference quotient really works for non-linear equations is if h is very close in value to x but not quit equal to x.

If we were to take h and keep moving it closer and closer to x without making them equal we would have the exact slope at that point. This is called taking the limit as h approaches x. in calculus this is the derivative of the equations.

A simple way to find the derivative is as follows.

For an value of Ax^n you would take (A*n)x^(n-1) for each term for any constant without an x term just set it to zero. With this you can check you answers.

So for Ax^2+Bx+C the solution is 2Ax+B
[NOTE:if h is not sufficiently small enough you would have a remainder of C/h]

Notice for the 2nd order equation that the slope on any point is a straight line with either a positive or negative slope. Think of what a 2nd order equation represents, a parabola and then envision the slope at any given point.
 
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If calculus is in your future, then it is something to really look forward to. Especially if you get a good calculus teacher!

I was once trying to explain to someone (a non-pilot) what it was like to train for the instrument rating. The fact that it's intense and hard, but opens your mind and lets you look at the world a little differently and with greater clarity... The best I could come up with was "It's like learning calculus!" :)

Anyway, Nick keep those cool problems coming. I dig 'em.

--Kath
 
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