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Bob wrote:

hi Jack,

HW all correct, well done!

So you can apply the method but don't know why it works. For the moment let's just be grateful that you can do this. I'll have another think and see if I can find another way to justify what you're doing. Watch this space.

Bob

Hicies87 wrote:

Can you provide clarification on what you meant by "limits works" and why the concept is confusing to you, particularly with the use of the picture?

What did I mean by "limits works" I think it was this part : "I basically understand somehow how limits works that if this delta y and delta x is going to zero". I tried to say how do I understood the term : limits, So I understood it in a way that delta x and delta y are going to 0 but will never reach this value so it is basically a very very very small number.

When I look at the picture let's say the second one it says that if I want to find a slope of a point then it would be 0/0 but this is incorrect. So instead there are limits (lim) which are used to find that slope using a very small number. So if delta y and delta x are very small then why do they give a normal number (normal I mean not a small number like 0,00000001).

I just can't understand that delta x which will be eventually 0,0000000000000001 or smaller and delta y which is also small but somethimes delta y is smaller than delta x or delta y is bigger than delta x, so somehow they give a normal number and doesn't decrease.

So I don't understand why the number doesn't decrease when I take even smaller numbers when if they were that small like 0/0 then is would be incorrect. So logically taking even smaller number should give me smaller result but instead I have a constant number.

From the first picture it shows that the difference is really small and shrinks even further.

I don't know how to better describe my confusion. It is wierd for me that somehow I get a normal number what if delta y is smaller than delta x then it indicates that delta y is closer to 0 so getting smaller numbers shouls give me 0 but it doesn't. I'm just stuck with this imagination that when delta y and delta x are going to 0 and both of them are very small then why do I have a non 0 and a normal number.

Let's say y = x^2. So for x1 = 1 and for x2 = 1.1 then y1 = 1 and y2 = 1.21 here delta y is bigger than delta x so I get a normal number ? Even if the number gets smaller why doesn't the value change ?

Next example is : y = x/2. So for x1 = 1 and for x2 = 1.1 then y1 = 0.5 and y2 = 0.55 so now delta x is bigger than delta y and I still have a normal number ? And not 0 ? delta y is getting closer to 0 than delta x. I don't get it.

Thanks for the reply and sorry for not responding.

Bob wrote:

hi Jack,

HW all correct, well done!

So you can apply the method but don't know why it works. For the moment let's just be grateful that you can do this. I'll have another think and see if I can find another way to justify what you're doing. Watch this space.

Bob

Sure thing ! I will wait.

Of course I am not sure if what I said make any sense or if you understood it I basically understand somehow how limits works that if this delta y and delta x is going to zero then both delta y and delta x are a very very small numbers close to 0 but they are not 0. The fact is that both small numbers are divided by each other and both of them are trying to reach this "0". It is so confusing seeing also this picture :

Like why delta "y" is bigger than delta "x" what if delta "y" is smaller than delta"x" then does it mean that the number delta "y" is getting closer to 0 so it should be 0/small number. It is so wierd for me I don't know how to explain it. Tell me if what I said is wierd or not understandable, I'll try to make it more clear.

Bob wrote:

As you can see the total distance gets closer and closer to 2. Using MS Excel and accuracy set to 9 decimal places the values all came out as 2 after 32 rows. The 'gap' between the true answer and 2 has got so small that it's beyond the accuracy level of the software to show it.

Is it reasonable to say that eventually the frog reaches 2 metres? We would get to the point where, if you put a target tape at 2 metres, you wouldn't be able to see the frog hadn't got there.

That's what is behind the idea of a limit. We can get as close as you want to the actual gradient by getting closer and closer using B getting closer to A. In practice it works so it is widely accpeted by mathematicians as a good method.

Note: Some texts use Δx for the little bit in the x direction and some use 'h'. I think the first notation was used by Newton and the second by Leibniz. I'm happy to use either.

I can see your point in the post, however isn't my limit a it different ?

I also see that the frog is in fact reaching very closely to these 2 meters, so that I can say he reached 2 meters. The only thing is that my example is a little bit different. In which for the frog there was only one limit (distance he reaches) and in my example there is "x" and "y" limit and they divide each othery "y/x" also both of them have limit to 0.

I just couldn't get it that both limits are going to 0 but I get a normal number. Both of them reaches so close to 0. But dividing them gives me a normal number. It is just unintuitive. I see that the "y" is bigger than "x" but this thought that both of them reaches 0 really messes my mind. Sorry for being so stubborn.

Bob wrote:

=> => => => => => => => => => => => Here's some homework for you to try:

Work out the differentiated function by working from first principles here:

Bob

I can somehow calculate it, it's just a matter of both limits which is "y" and "x" which are going to 0 and from that I get a number.

Bob wrote:

Take care with the arithmetic.

A = ( 1 , 1 ) B = ( 1.1, 1.0201)

Bob

Ah yes you are right.

But this still shows that

Bob wrote:

hi Jack,

Loads of books will tell you you cannot find the value of 0 divided by 0.

Let's explore that first.

If I see the calculation 48 / 6 ; one way to answer this is to ask "What must I multiply by 6 to get the answer 48?"

This works because x and ÷ are opposite operations. This works for all divisions except one.

You cannot evaluate 0 / 0 by asking "What must I multiply 0 by to get 0?" Because any number times 0 will make 0.

This is the fundamental problem with trying to get the gradient at a point. If you pick a point A and write down its coordinates (x,y) then write down the coordinates of B where B is (x+0, y+o) you will hit the above difficulty when you try to do:

gradient = (difference in y coords) / ( difference in x coords) = ( y+0 - y ) / ( x + 0 - x) = 0 / 0

The differentiation from first principles method is a way of getting around this problem.

Let's look at this with an example.

Function: y = x^2 A = ( 3 , 9 )

Let's try B = ( 3.5 , 12.25 )

The gradient of AB is ( 12.25 - 9 ) / ( 3.5 - 3) = 3.25 / 0.5 = 6.5

Now try B = ( 3.1 , 9.61 ) G = ( 9.61 - 9 ) / ( 3.1 - 3 ) = 0.61 / 0.1 = 6.1

Now try B = ( 3.001, 9.006001) G = ( 9.006001 - 9 ) / ( 3.001 - 3) = 0.006001 / 0.001 = 6.001

You can try some more values for B for yourself. eg, B = ( 3.000001, ...........

What you will notice is that as B gets closer to A the value of the gradient, G, gets closer to 6.

You can also try B coordinates left of A such as ( 2.9, 8.41 ) You'll find the G values are under 6 but get bigger towards 6 as B moves closer to A.

Conclusion: The value of G at ( 3, 9 ) is 6.

Here's some algebra to justify this. Let B = ( x + h , x^2 + 2xh + h^2)

G = ( x^2 + 2xh + h^2 - x^2) / ( x + h - x ) = ( 2xh + h^2 ) / h

At this stage, letting h = 0 will get us nowhere as we'll end up with 0 / 0. The trick is to simplfy first.

G = 2xh / h + h^2 / h = 2x + h.

Now we can safely let h become zero as we are left with 2x. So we can conclude f'(x) = 2x

Please let me know if this makes sense so far.

Bob

I see the results and I tried myself to make B a lot smaller like B = (3.0000000000001,8.0000000000001).

but it sounds wierd.

Looking at it like this :

seeing that both = goes to 0 and = goes to 0. I know what do you mean but it is so wierd for me. I heard something about ratio of "y" that is bigger than "x".But when I used this example :

in a point "x = 1" so I had this result :so y = 1 and y = 1.01 here it looks like the difference is small but for "x" it is 1.1-1 = 0.1 and for "y" the difference is 1.01 - 1 = 0.01 hmmm maybe this is not a good example. Because here it looks like the difference for

is going to be closer to 0 than forSorry for this stupid question, it sounds wierd to not understand this

Bob wrote:

hi Jack

I made this picture some years ago for someone learning about functions.

https://i.imgur.com/gQXbI23.gif

A number goes into the box; the function makes a new number come out. (You can also have more than one input but let's not go there at the moment!)

Mathematicians often abbreviate the function box to f(x), meaning f is the box, x in the input.

And if you know f(7) = 5 for example, then 7 is the input and 5 is the output.So in your example, f(x) = 4x -2 you are being told that the rule for what to do with the input is "multiply the input by 4 and subtract 2; output the answer".

So f(7) = 4 x 7 - 2 = 26

f(-2) = 4 x -2 - 2 = -10

If the questioner wants a second function they'll use another letter so the two don't get muddled.

eg. g(x) = 8x so g(5) = 40

You say you have two ways of showing differentiation. Actually only one of these is correct:

This comes about by a procedure known as differentiating from first principles. This is the procedure:

I'll put my coordinates (x,y) in square brackets [x,y] so you don't get muddled with the function brackets.

1. Pick a point on the curve [ x, f(x) ]

2. Pick a second point close to the first [ x+h, (f(x+h) ]

3. Work out the gradient of the line joining the points:

4. Let h tend to zero and simplify the expression. Because the line joining the points gets closer and closer to the gradient at the first point, we conclude this is the gradient at the point. One way of writing the gradiant function is to put a ' after the f = f'

The second expression:

is just an expression for the gradient of a line joining [ a, f(a) ] to [ b, f(b) ]

It won't be the gradient at a point until you have applied the first principles rule.

Would you like some examples of how the 'first principles rule' works in practice?

Bob

Hello Bob !

Sorry for not answering for so long.

I think that I understand the difference between

Something like that ?

PS.

Could you also tell me how this lim works in f'(x) ? I've seen this picture :

I don't get it because

is going to 0 and is also going to zero, but somehow I got a normal value for f'(x).**Jack Omar**- Replies: 13

Hello !

I am having problem understanding the functions or symbols I am not sure.

For example let's say

The similar problem appeared in derivative for example :

and the same symbol but with different formula so now I can have . I don't get it. So when I have this : , f'(x) is a derivative of a function, but then I have the same symbol but with different formula : , doesn't this change how I calculate the derivative f'(x) I just now calculate the derivative using the second formula.I see similarities in both situations but I don't get it, when I am calculating the f(x) or when I have a result of f(x) or f'(x). Can someone guide me please

Or Am I wrong zetafunc ?

zetafunc wrote:

Jack Omar wrote:In the link about derivatives it is shown that it is 0/0 but I should take a really small difference and then shrink it to 0 so it is the same as 0/0 ???

I don't get that part with shrinking to 0, which is pretty much the same as 0/0.Hi Jack Omar,

Careful -- taking the limit as something approaches 0 is not quite the same as 0/0! We're only interested in what happens as getsreally closeto 0, not what happens at 0. This is the key part.The question we want to answer is: how do we describe the 'slope' of a function at any given point on its graph? Well, as the pictures you've added show, we can

(a) Take the first point on the curve, find another point on the curve, then draw a straight line through them...

That gives an estimate of what the slope at a given point is. But it's not quite right -- and as you move the points closer together on your graph, that 'estimate' becomes more and more accurate. (Likewise, moving the points further apart makes that 'estimate' less accurate.) That's what taking the limit as

(b) ...then calculate the slope of the line you've just drawn.approaches0 is intending to achieve, i.e. what happens when you bring those two points really, really close together?Post #1 wrote:I think I made a mistake in last equation but I don't know why.

Looks OK to me -- isn't the last line the same as the first one?

Post #3 wrote:No -- the division by is for the previous step. They are saying that:Second question isn't there an error in the result ? There is 2x + Δx, then it is devided by Δx, 2x÷Δx + Δx÷Δx then 2x÷Δx + 1 ? Shrinking it into Δx to 0 I get 2x÷0 + 1 ?

I saw that you asked some questions about integration as well -- but let me know if this makes sense first and then we can move on.

Hello

Thank you so much for the answer !

I do have problem still with understanding

andI've tried to do some examples.

For I choosed the points then and then the difference of is equal and for is equal So basically is going faster to 0 than . I know that if it is like that : then those small numbers will give normal number. But is getting smaller tna . It's hard to explain it. If I had different equation and then will give me 7. I don't know how to tell my confusion. I'll try again if it's not clear because my english isn't that good.

Or is it a no ?

Because I have a question for the links you've provided

ganesh wrote:

Hi Jack Omar,

The topic in the first post is ' I don't understand integrals (definite integrals and indefinite integr.)

Thats why is gave the links.

O okey

But is it possible to answer to my question I gave ? I am really uncertain about my answers.

Thanks alot !

ganesh wrote:

Hello

I've asked in my last post about derivatives.

What do you think of what I said in my last post about derivatives.

In the link about derivatives it is shown that it is 0/0 but I should take a really small difference and then shrink it to 0 so it is the same as 0/0 ???

I don't get that part with shrinking to 0, which is pretty much the same as 0/0.

And could you take a look at my calculations

Thanks alot

**Jack Omar**- Replies: 2

Hello ! I am Jack Omar, I live in Poland.

Usually I spend my time playing video games, listening to many musics and studying electronics I've met some difficulties in basic things. It's very difficult to say it, but I have annoyed many people with my questions, so I thought that maybe in this forum I'll try things slowly. Oh and I forgot to mention that I love Star Wars universe

I hope I won't annoy here anybody, I'll try my best !

Have a nice days everybody

Hello ganesh,

Thank you for the links, it was really helpful but I do have some questions.

I want to also say that english is not my native language so, I am really sorry if my post was really messy, and I want to apologize if I can be hard to talk with, because sometimes I tend to ask the same question again.

My question is relates to this picture :

And this picture :

It is said that there is no difference 0÷0. But in derivatives I take small a very small difference and then shrink it to zero. But if I shrink it to 0 then y is also 0 as I know ? So why does it work ?

Second question isn't there an error in the result ? There is 2x + Δx, then it is devided by Δx, 2x÷Δx + Δx÷Δx then 2x÷Δx + 1 ? Shrinking it into Δx to 0 I get 2x÷0 + 1 ?

I'll start later asking questions for integrals, but I wanted to start slowly if its not a problem.

Have a nice day guys

**Jack Omar**- Replies: 11

I am struggeling with understanding what is indefinite and definite integrals.

What is the problem ?

Derivative is a slope of a function, indefinite integral is the reverse of the derivative, but what does it mean reverse of the derivative ? Does it have any purpose ? What do I mean is that Derivative has a meaning that it is a slope of a function, and the anti derivative doesn't have this meaning, it's something is it a original function or what ?

I've heard that anti-derivative mean anything. Why ? Derivative means that it is a slope of a function and AntiDerivative means nothing ? I don't get it

Derivative has even a proof that to calculate derivative I have to use limit :

Like in the picture. So for example when I have f(x)=x2 and I have two points x1=3,x2=4.

Then :

I think I made a mistake in last equation but I don't know why.

But to the point, how do I prove that indefinite integrals is an inverse operation of derivative and to calculate it I must really reverse the derivative.

Last question about definite integrals I read this website : https://www.askamathematician.com/2011/04/q-why-is-the-integralantiderivative-the-area-under-a-function/

And I was a bit lost because to calculate the area I must take a derivative from a function that is already a derivative function. Like what ?

I tried to calculate it but I don't get it.

I don't get it why we use f'(c)*(B-A) = f(b)-f(a) is the area ? I've changed the equation for the derivative and now it doesn't have any sense because now what is this f'(c) in that equation ?

I've changed the equation so now I don't know what is this f'(c) in f'(c)*(B-A) = f(b)-f(a). Or even if I know then f(b)-f(a)÷(B-A) * (B-A) =f(b)-f(a) So f(b)-f(a) =f(b)-f(a) .

It doesn't have anysense also why integral of f'(c)*(B-A) = f(b)-f(a) is equal f(c)(B-A) = F(b)-F(a) why does (B-A) stays the same ? I know that the intervals doesn't have to be numbers it can be also different function.

I've read so many textbooks and websites and I still don't get it.

Help ;<

**Overall I don't get it why if I want to calculate the area I must derivate a derivated function, does calculating the area I need only derivative ?**

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