Calculus explanation

thepikminman

Can somebody try to explain to concept of differentiation and integration? I know the first derivative is the slope of the graph, and integration is the sum of all the infinitesimal points, but I don't really know HOW it works. I get what the lim as x --> 0 means, but why not use x = 0? How does the function know what you're talking about? I really wish they explained this better in class, because everytime I do one of these question (correctly or incorrectly) I feel like I'm cheating because I don't really understand it, and I know that it is really amazing. You would be helping me more than you can imagine if you answer this, because I'm studying engineering, and if I can understand this, it will fuel my interest, and increase my thirst for knowledge for the entire .

Yakuza

I think of differentiation as "rates of change" - how a the dependent variable(y) of a function changes relative to changes in its independent value (x) (using the standard notation of y=f(x). It's not quite [latex] \lim_{x \to 0}[/latex] as you put it, but rather [latex] \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/latex]

The top line is basically the change in the value of the function when the dependent value (input variable) is x taken away from the value of the function when the input variable is (x+h).
The bottom line in the change (delta) in the independent variable (h).

You can't just let h = 0 as then you've no change in either variable so there is no way to measure the rate of change (dividing by zero is not defined).

Take the case of y = x²:
How does y change when x changes by h?
f(x+h) = (x+h)² = x² + 2xh + h².

So, f(x+h) - f(x) = x² + 2xh + h² - x² = 2xh + h².
Dividing this by h (we can do this as h will never equal zero (but will get infinitessimally close to it) yields : 2x +h

Taking the limit of (2x+h) as h-->0 yields 2x (as x is unaffected by h)
This, by definition, is the derivative of x².

When I first saw this in 4th year, I'll be honest and say I was a bit baffled by it initially, for a few weeks I just took it on faith and the concept just gradually sank in.

Have you looked at the Khan Academy? A brilliant online resource: https://www.khanacademy.org/math/calculus/differential-calculus
Wikipedia has plenty of articles too, but they are fairly technical/dry.

bnt

This is how I got my head around it: remember how you learned to find the Slope of a graph? You take Δy/Δx, so e.g. in an x range 1..2, Δx = 1. If, in that range f(x) (or y) varies from 2..4, then Δy = 2. So the Slope is Δy/Δx = 2/1 = 2.

That's easy when the graph is a straight line, but what if it's not e.g. if f(x) = x² ? Then, if you use that formula, you don't get the slope of the whole function, but the slope between two points. What if you want the slope at a single point? In the example above, Δx = 1, but at a single point Δx = 0. You'd have Δy = 0 too. So Δy/Δx = 0/0 = a nonsense value. The old method breaks down when you try to find the slope at a point.

If you visualise the function f(x) = x² , then visualise the slope, you can see that the slope is not constant: it's also a function of x. So you need a way to take a function, and from that extract another function for its slope. That's Differentiation.

The function that Yakuza gives, [latex] \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/latex], is the foundation of this. It's just another way of expressing the slope function Δy/Δx, but with h instead of Δx. You can't go straight to h = 0, because then you have 0 as the denominator and the function blows up to infinity. So, what you do instead is replace f(x) with the actual function you're differentiating. In my example:
f(x) = x²
f(x+h) = (x+h)²

Substitute those in the formula, and simplify algebraically until it's safe to substitute h = 0 without it going to infinity. The result is 2x, and it's usually written as f'(x) = 2x. So we've taken a function, differentiated it, and gotten another function for its slope. We can now find the slope at any point, so e.g.
x = 3
=> f(x) = x² = 9
=> f'(x) = 2x = 6

Apologies if that has you going "well, duh" ..! :cool:

Blackpanther95

the answers above are good, so I'll just address the question why can't x=0 instead of x-->0.
The question is really philosophical, but in simple terms, the whole point of calculus is to consider infinitesimal changes, i.e. changes which are small than every number but bigger than zero. These are sometimes called 1/infinity. You could think of them as 0.000...001 with a one somewhere down the line. Now these are great because if we want the slope at x we have two points where {y=f(x)} (x , f(x)), and a point infinitesimally close to it, which is good because we can use simple co-ordinate geometry, but unfornately the infinitesimals aren't numerical, you can write them down, but if they were all to cancel that would be great, because it wouldn't matter what they were, all we need to know is: they exist, and not what they are. Remember the infinitesimal is non-zero, otherwise we couldn't use simple 2nd year junior cert co-ordinate geometry to find the slope.

bnt

Blackpanther95 wrote: »

the answers above are good, so I'll just address the question why can't x=0 instead of x-->0.
...

I disagree with the rest of it: for calculus to work, the infinitesimal must be zero, not "almost zero", no matter how close it would be. That was the nut that Newton and Leibnitz were trying to crack, and it explains why it's expressed the way it is.

I read the h-->0 bit as "h tends to zero", meaning that it will reach zero at some point. You want the true slope of the function at a point, not an approximation over a small portion. No matter how close to zero the infinitesimal gets, it's not quite there.

In the fundamental equation of calculus, if you substitute h=0 (Δx = 0), the equation goes straight to 0/0, which is a useless result. So, by framing it as h-->0, instead of h=0, you're saying something like "it will go to zero, but only when it's safe to go there without blowing up the equation". :cool:

edit: maybe it will become clearer if I complete working out that differential from first principles:
[latex]f(x) = x^2 [/latex]
[latex]f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/latex]
[latex] = \lim_{h \to 0} \frac{(x+h)^2-x^2}{h}[/latex]
[latex] = \lim_{h \to 0} \frac{x^2+2xh+h^2-x^2}{h}[/latex]
[latex] = \lim_{h \to 0} \frac{2xh+h^2}{h}[/latex]
[latex] = \lim_{h \to 0} {2x+h}[/latex]
Now, since I've eliminated the denominator, it's safe to set h=0:
[latex]f'(x) = 2x [/latex]

Blackpanther95

If the infinitesimal is zero then you're just adding zero, so your considering two points, x,y and x+0,y+0, i.e the same point so you can't use co-ordinate geometry. But don't just take my word for it http://en.wikipedia.org/wiki/Infinitesimal_calculus its in there, but also 1/infinity > 0 , but also note an infinitesimal is not a real number (its smaller than any real number).

bnt

Blackpanther95 wrote: »

If the infinitesimal is zero then you're just adding zero, so your considering two points, x,y and x+0,y+0, i.e the same point so you can't use co-ordinate geometry. But don't just take my word for it http://en.wikipedia.org/wiki/Infinitesimal_calculus its in there, but also 1/infinity > 0 , but also note an infinitesimal is not a real number (its smaller than any real number).

Not sure what you're replying to there. Yes, if the infinitesimal is zero then your two points are the same, the slope is 0/0, and so that coordinate geometry method doesn't work. Yet the maths I added to my last post shows you how to work around that problem, using a Limit of h --> 0, instead of h=0 (Δx = 0), and that's how Differentiation was invented. Remember the OP - no need to overcomplicate matters. :P

Blackpanther95

bnt wrote: »

Not sure what you're replying to there. Yes, if the infinitesimal is zero then your two points are the same, the slope is 0/0, and so that coordinate geometry method doesn't work. Yet the maths I added to my last post shows you how to work around that problem, using a Limit of h --> 0, instead of h=0 (Δx = 0), and that's how Differentiation was invented. Remember the OP - no need to overcomplicate matters. :P

Yeah but none of your replies explain what the limit is,its might be a little overkill for engineers. I think its easier to think of it in terms of infinitesimalls rather than limits especially for physics. But I'm not saying your answer isn't valid. I'd say the question is answered.

FoxT

When calculus was invented, the question of x--> 0 was fraught. Is x = zero or is it not? How can a point have a slope? etc.

They way it is taught, these questions are glossed over. I'm an engineer myself, and mentally I have thought of it as :

"x is more or less equal to zero, for all intents & purposes."

This does two things: It makes sure your equation won't burp with a divide by zero error, but it also guarantees that when you differentiate from first principles you get the slope AT THE EXACT POINT ON THE CURVE, because, x is more or less equal to zero.

It's a have cake + eat it situation, works for me!

I dont mean to be facetious, and pure mathematicians will probably recoil in horror at what I have written, but I have found it a practical way to deal with it.

As an aside, the world is full of stuff like this. Light, for example , has a wave /particle duality, just like x-->0 is zero sometimes, and sometimes it isn't. Also there are multiple different infinities...it all works out grand in the end!

Blackpanther95

FoxT wrote: »

When calculus was invented, the question of x--> 0 was fraught. Is x = zero or is it not? How can a point have a slope? etc.

They way it is taught, these questions are glossed over. I'm an engineer myself, and mentally I have thought of it as :

"x is more or less equal to zero, for all intents & purposes."

This does two things: It makes sure your equation won't burp with a divide by zero error, but it also guarantees that when you differentiate from first principles you get the slope AT THE EXACT POINT ON THE CURVE, because, x is more or less equal to zero.

It's a have cake + eat it situation, works for me!

I dont mean to be facetious, and pure mathematicians will probably recoil in horror at what I have written, but I have found it a practical way to deal with it.

As an aside, the world is full of stuff like this. Light, for example , has a wave /particle duality, just like x-->0 is zero sometimes, and sometimes it isn't. Also there are multiple different infinities...it all works out grand in the end!

Well arguably there are only 2 infinities haha but that philosophical debate is unwinnable on both sides. Yes it does work out grand in the end, but it should be noted that h is not an infinitesimal, just a non-zero variable, the limit is the magical part, not the variable, also the limit notation assumes (which is reasonable for an engineer) that the curve is continuous.

FoxT

Ah, but sure, if it wasn't a continuous curve, engineers wouldn't want to differentiate it

Blackpanther95

FoxT wrote: »

Ah, but sure, if it wasn't a continuous curve, engineers wouldn't want to differentiate it

And pure mathematicians would

bnt

I'm not looking at the Limit as a "thing", so I don't know how to answer a question like "what is it?". It's a bit of mathematical notation that says "this variable tends to a value", but we can't just substitute the value straight away, since that leads to an undesirable result.

Wikipedia uses a non-Calculus example to illustrate why we need a Limit:

For example, if
[latex]f(x) = \frac{x^2 - 1}{x - 1}[/latex]
then f(1) is not defined (division by zero), yet as x moves arbitrarily close to 1, f(x) correspondingly approaches 2.
...
f(x) can be made arbitrarily close to the limit of 2 just by making x sufficiently close to 1.

In other words, [latex]\lim_{x \to 1} \frac{x^2-1}{x-1} = 2 [/latex]

This can also be calculated algebraically, as [latex]\frac{x^2-1}{x-1} = \frac{(x+1)(x-1)}{x-1} = x+1[/latex] for all real numbers [latex]x\neq 1[/latex].

Now since x+1 is continuous in x at 1, we can now plug in 1 for x, thus [latex]\lim_{x \to 1} \frac{x^2-1}{x-1} = 1+1 = 2[/latex].

So you can still substitute the value, but at the right time - after simplification, in these cases. And (as pointed out) the function needs to be continuous up to the limit.

Blackpanther95

bnt wrote: »

I'm not looking at the Limit as a "thing", so I don't know how to answer a question like "what is it?". It's a bit of mathematical notation that says "this variable tends to a value", but we can't just substitute the value straight away, since that leads to an undesirable result.

Wikipedia uses a non-Calculus example to illustrate why we need a Limit:

So you can still substitute the value, but at the right time - after simplification, in these cases. And (as pointed out) the function needs to be continuous up to the limit.

I won't explain what the limit is here, but if you make it a separate thread, i bet you'll get 100's of responses. But i guess the way you think of the limit is correct (intuitively) even if the definition and formalism of the limit makes calculus rigorous. I forget his name but a mathematician once gave 100 ways to think of the derivative, each valid and correct (and distinct). The limit is the most rigorous method, hence it is most widely used