Math Is Fun Forum

A = ( 1 , 1 )  B = ( 1.01, 1.0201)

G = ( 1,0201 - 1 ) / ( 1.01 - 1 ) = 0.0201 / 0.01 = 2.01 / 1 = 2.01

Sorry I missed a zero in my previous post.

There's an old story about a frog which I'll adapt for our purposes here.

Let's imagine a frog that can make a mighty leap of one metre, but is so exhausted by the attempt that the next leap is only half a metre, and the next one quarter of a metre and so on. Each leap is half the one before.  How far can the frog travel?

First leap             = 1 metre
First plus second  = 1.5 m
Three leaps         = 1.75 m
Four leaos            = 1.825 m
Five leaps             = 1.9375 m
.....
I set this up on a spreadsheet using formulas to generate the values quickly.  Here's the formulas I set up.
1            =1
=A1*2    =B1+1/A2
=A2*2    =B2+1/A3
=A3*2    =B3+1/A4
=A4*2    =B4+1/A5
=A5*2    =B5+1/A6
=A6*2    =B6+1/A7
=A7*2    =B7+1/A8
=A8*2    =B8+1/A9
=A9*2    =B9+1/A10
=A10*2    =B10+1/A11
=A11*2    =B11+1/A12
=A12*2    =B12+1/A13
=A13*2    =B13+1/A14
=A14*2    =B14+1/A15
=A15*2    =B15+1/A16
=A16*2    =B16+1/A17
=A17*2    =B17+1/A18
=A18*2    =B18+1/A19
=A19*2    =B19+1/A20

And here's some output:

1    1
2    1.5
4    1.75
8    1.875
16    1.9375
32    1.96875
64    1.984375
128    1.9921875
256    1.99609375
512    1.998046875
1024    1.999023438
2048    1.999511719
4096    1.999755859
8192    1.99987793
16384    1.999938965
32768    1.999969482
65536    1.999984741
131072    1.999992371
262144    1.999996185
524288    1.999998093

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.

Here's some homework for you to try:

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

Bob

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

hi Yan,

Welcome to the forum.

Bob

hi Jack

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

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

hi imcute

I'm flattered that you think I (or anyone else) could make sense of this as posted.  I think the following is the minimum you need to back up your code:

(1) What you are trying to code in words.

(2) The algorithm(s) you are using (or a flow diagram, for example).

(3) Some comment lines to make it clear what each section of code does.

(4) The language you are coding in.

(5) A list of the variables, stating whether each is an integer, decimal, string etc.

(6) A clear statement of what output you were expecting and the output you are getting.

Probably more but that will do for starters.

If I was your teacher, I'd know what the exercise was and I'd be able to question you about what you have submitted. But 119 lines of code without any of the above more or less means, just spend a week trying to figure out what this is and correct it. Eeekkk!

Bob

hi Winatte

Welcome to the forum.

The formula for the sum of squares ( 1 + 4 + 9 + 16 + ... + n^2) is

Sum = n(n+1)(2n+1)/6

Substitute n = 1, 2, 3 etc to show that it works.

There are also similar formulas for the sum of cubes and higher powers but they get increasingly complicated.

Q1 is just 5 + 5 + 5 + 5 ... 'n' times = 5n.  I don't understand what " an3 + bn2 + cn" means for this.

We can make more progress with this over several posts so please reply.

Bob

hi undertaker

I'm assuming you mean this problem:

The size of the sectors is not relevent.  Let's say you had a slice one sixth of the cake and sub-divided as in the picture into two equal parts then <outer part> = <inner part> so 6 times <outer part> = 6 times <inner part>

= whole cake divided in two. 

The area of the whole cake is pi r squared so we want the two parts to have area half pi r squared.

The inner is a smaller circle, so let's call it's radius 's'.  Then pi s squared = half pi r squared.

Cancel the pi on each side and you have s = Square root (half r squared) = root(2) . r

Hope that helps

Bob

Do you mean if the student(s) doesn't 'get it' and doesn't want to try?

I once had to take a special needs class doing a coursework task; the object being to investigate then arrive at an algebraic formula.  I brought in some sweets of different types and presented the task as involving the sweets as a prop.  It went down well and they quickly got the idea making up packs of mixed sweets according to the 'rule'. They were able to write their 'conclusion' in terms of a 'formula' using the names of the sweets instead of x and y.

The final step was to persuade them to re-write this, abbreviating each sweet name to it's start letter. Result: one perfectly respectable algebraic conclusion.

So, I suppose my answer is 'find a way to present the work using their language/interests/culture'.

Bob

hi Mister_JWO

What you normally try will identify vertical asymptotes. But there may be other lines that the graph tends towards.

There's an example here http://www.mathisfunforum.com/viewtopic … 36#p427436 post 7.

As x tends to infinity the curve approaches the line y = 1.

For your function as x tends to infinity the bx and c terms become negligible so that the ax^2 dominates.

So the function behaves like y = √(ax^2) . Hopefully you can finish this off by simplifying that.

You can also try using the function grapher to 'see' what a curve does.  You'd have to choose a, b and c values.

https://www.mathsisfun.com/data/function-grapher.php

Bob

From time to time I check out the new members list to weed out any who have only joined to post adverts or spam.  The list shows adiministrators how many posts a member had made.  This number is usually wrong such as showing one post when there are none.  It may be that the counting for this has a bug or it may be that the forum software automatically removes posts that break the rules.

Good if it's working properly but not good if it is taking away genuine posts. At the very least it would help if I knew what's being removed, then maybe I could work out why.  If you are able to recreate your missing post please put it here so I can try to help.  This has happened to another member recently so I'd like to get to the reason.

Bob

The standard way to do intregration of absolute functions is to split the function into two;
one for when |expression| > 0 and one for when |expression| < 0. 
So you end up with a two part answer.

https://photomath.com/en/calculus/integ … -functions

Bob

hi DrLiangMath

Your videos are really great but, as you add more, it is becoming harder and harder to find a particular one or type.

I suggest you give some thought to how you might index them so an interested person can find a particular one.

Maybe you could divide them by similar topics ?

Best wishes,

Bob