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#1 Re: Help Me ! » Help me please! Area of Polygons » 2023-01-28 05:39:07

Bob

hi aquafina09

These look like Compu High questions. Whatever,  the method is the same for all of them.

If a polygon is regular that means all its sides are equal and all its angles.

So if you fnd the centre of the shape (call it point O) and draw lines radiating out to the vertices, A, B, C, D, E,  etc
then OAB, OBC, OCD, ODE etc are all identical isosceles triangles. So, if you can calculate the area of one of these and then times by how many there are, you get the total area of the whole polygon.

Follow these steps:

(1) Let n be the number of sides and s be the length of one side.

(2) Angle AOB = 360/n so you can calculate the angle at the centre.

(3) Let M be the midpoint of AB.

(4) As AOB is isosceles and the angle sum of any triangle is 180 you can calculate OAM.

OAM = (180 - AOB) / 2

(5) AM = s/2

(6) Use trig to calculate the height of the triangle AOB.

h =  s/2 tan(OAM)

(7) Area of any triangle is half base times height.  In our triangle

area AOB = half x s x [s/2 tan(OAM)]

(8) times by n

Bob

#2 Re: Help Me ! » I need help understanding fuctions. » 2023-01-28 05:26:00

Bob

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

#3 Re: Help Me ! » I need help understanding fuctions. » 2023-01-22 23:52:28

Bob

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

#4 Re: Help Me ! » I need help understanding fuctions. » 2023-01-21 20:29:32

Bob

Take care with the arithmetic.

A = ( 1 , 1 )  B = ( 1.1, 1.0201)

Bob

#5 Re: Help Me ! » I need help understanding fuctions. » 2023-01-21 07:24:08

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

#6 Re: Guestbook » Pi is fun » 2023-01-19 21:32:44

Bob

Hi blu£

Welcome to the forum.

Hope you don't mind but I corrected the Pi formula in your posts above.

It doesn't matter whether a person likes pi.  Pi goes on existing anyway.  Even alien beings with different methods of arithmetic will have the chance to know about pi.

My favourite bit of maths is:

There's a lot of more advanced maths in that.  It would take me a few posts to explain it but I will if you wish. smile

Best wishes,

Bob

#8 Re: Help Me ! » Intersecting a graph with a line in 3d » 2023-01-18 21:20:12

Bob

hi mathdrop

It looks like you've got

The line joining the origin to (1,1,1) is

So every point on the line has x = y = z

So it will intersect the shape at

Hope that helps,

Bob

#9 Re: Help Me ! » I need help understanding fuctions. » 2023-01-16 22:48:19

Bob

hi Jack

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

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

#10 Re: Help Me ! » Math probability » 2023-01-14 21:05:40

Bob

hi Alberto

Welcome to the forum.

We haven't got the net diagram so I need to check some things.

A triangular prism net will have three rectangular faces ad two trianglular ones.

The triangles are equilateral so the probabilities of any triangular face should be equal and similarly for the rectangular faces.. So a roll could yield a rectangular result or a triangular one. It is necessary to know whcih letters are on which faces.  I'll say that the probability of landing on a triangle is t and on a rectangle is r.

As one of these events must happen we can say that 2t + 3r = 1

To get the results given the faces must have M, A, T, H, E on them.  I'll further assume that MATHE may be made from 5 rolls with the letters coming up in that order. ( Not sure if it makes any difference if I allow the letters to be rolled in any order ... I'll try it that way too and see)

So once I know where the letters are I can get an expression for the probability of the two words and hence make an inequality statement.  That should allow me to determine t>r or r>t and I can proceed from there.

Over to you for the essential extra details.

Bob

#11 Re: Help Me ! » Werid linear error term on self made ln function » 2023-01-02 02:10:38

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! dizzy

Bob

#12 Re: Help Me ! » Werid linear error term on self made ln function » 2022-12-30 22:48:17

Bob

Hi imcute

I have added url /url code to your link address to make it easier to jump to.

That's a lot of code to analyse.  Please help by explaining what the ode is for; and the algorithm you are attempting to code.

Thanks,

Bob

Ps replit is safe.

#13 Re: Formulas » Simplifying summation. I am clueless » 2022-12-29 19:51:47

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

#14 Re: Help Me ! » Linear Algebra » 2022-12-12 23:48:12

Bob

hi Jake204598

welcome to the forum.

Looking at Q1 it looks to me that T is

But what are T2, T3 etc?  I see no definition for these.

Q2. Yes. But showing it from the rules for determinant expansion could get messy.  Have you been given any rules that might make this easier?

Bob

#15 Re: Help Me ! » geometry! one question :) » 2022-12-08 23:24:27

Bob

hi undertaker

I'm assuming you mean this problem:

yQ2zLKX.png

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

#16 Re: Exercises » interval » 2022-12-01 20:48:19

Bob

hi tony123

It's worth trialling different k values using the grapher here:
https://www.mathsisfun.com/data/function-grapher.php

You'll find that x = 1/4 is always the vertical line of symmetry, so one limit will be found by finding the equation when the graph goes through (1,0)

And as k becomes more positive the curve shifts up and eventually fails to cross the x axis at all. That should give you the other limit.

Bob

#17 Re: Help Me ! » a ball/torus resting in the tip of a pyramid » 2022-11-25 22:32:53

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

#18 Re: Help Me ! » multiplying matrices » 2022-11-25 22:21:17

Bob

Just treat, say, 4 by 6 matrix as 6 separate 4 by 1s and do the multiplication of each column like that.

So for example if you have a 3 by 4 to multiply a 4 by 6 you treat the 6 columns like in the above post and then re-assemble the answer. It will be a 3 by 6 matrix.

Bob

ps. You may be able to help another member who is having difficulty making a new post. He keeps adding his new posts to existing threads.  When you view the forum on a phone can you see the post numbers?  And if you switch to the help index can you see the 'make a new post' option?

#19 Re: Help Me ! » Asymptotes of irrational curve » 2022-11-19 20:33:20

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

#20 Re: Help Me ! » Help me! I have work problem to solve » 2022-11-15 06:38:49

Bob

hi Knew43,

Welcome to the forum.

If you had one set {A,B,C,D,E} and you had to choose any 3 this would be a combinations problem.

{5x4x3)/(3x2x1)

If you had to choose them and order matters (so that ABC and BCA are different) then the problem becomes a permutions one. 5x4x3

Where do these calculations come from. Let's deal with the permutations first.

There are 5 choices for the first selection; then 4 choices then 3. This gives 5x4x3 altogther.

If it doesn't matter which order we select them, for example, ABC, ACB, BAC, BCA, CAB and CBA have all been counted so divide by 3x2x1 to get the number of combinations.

For your problem we are making 3 independent selections; so deal with each separately and then times them together.

So choose one from {A,B,C,D,E,F} That's 6C1 ie 6.

Same for the next two selections so the number of ways to make these three selections is 6 x 6 x 6.

If each set has only 5 members then do a similar calculation with 5s.

Bob

#21 Re: Maths Is Fun - Suggestions and Comments » Deleted post » 2022-11-14 22:41:03

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

#22 Re: Computer Math » A very inefficient way of approximating the golden ratio » 2022-11-14 22:31:56

Bob

hi Dries

Welcome as a member!

The golden ratio has been identified as having the very worst rational approximations which means it is least well approximated by rationals.

It has long been known that plant seed development tends to follow in Fibonacci numbers and these directly lead to the golden ratio.  An expalnation for this and in the appendix a proof of the number theory statement above can be found here: https://onlinelibrary.wiley.com/doi/10. … 04.01185.x

I suspect that your discovery is related to this property although I don't think I could prove it.  There is also lots of interesting facts (enough to last you a year!) on the wiki page:
https://en.wikipedia.org/wiki/Golden_ra … quare_root

Bob

#23 Re: Help Me ! » How do you integrate the ||->v|| function?(arrow on top of v) » 2022-11-14 00:46:15

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

#24 Re: Computer Math » A very inefficient way of approximating the golden ratio » 2022-11-14 00:41:46

Bob

hi MrDries

Welcome to the forum.  I seem to remember that the golden ratio does have some interesting properties that may explain this but I'll have to do some research first.  Have you considered becoming a member of the forum?

Bob

#25 Re: Maths Teaching Resources » How to Improve Math Problem-solving Skills? - Free Math YouTube Videos » 2022-11-10 21:55:51

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

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