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#1 Re: Help Me ! » Reviewing area of polygons » Yesterday 18:45:02

Q5 has recently and many times earlier been answered in other posts.  Use the search to find these.

Bob

#2 Re: Dark Discussions at Cafe Infinity » Giving up smoking/drinking » 2019-05-19 19:35:50

hi Alg Num Theory

Best of luck with both resolutions!  I don't smoke and never have done so but I have noticed one thing amongst friends who have tried to give up.  They manage so many days and then give in to the urge and have a smoke.  Then they feel they have failed and so go back to smoking as before.  I'd say that's the wrong attitude to the 'failure'.  If you've gone let's say a week, then in exam terms that's a score of 90+%.  Terrific achievement and definitely worth celebrating, not feeling a failure.  So if you do have the odd sly ciggy don't despair … look on it as a success that you've got so far …. and set a new target to go at least as long before the next one.  By gradually making the 'non-smoking periods' longer and longer, you will still be improving your health and saving money … it may just take longer before the "n-s p" becomes infinitely long. smile

Keep us posted.

Bob

#3 Re: Dark Discussions at Cafe Infinity » 25 Tips to Beat Anxiety and Depression » 2019-05-19 19:27:46

hi Mathegocart

Before my time I think.  I've checked and it's no longer available.  So let's leave this.

Bob

#4 Re: Help Me ! » Probability Distribution » 2019-05-19 19:22:25

hi Zeeshan 01

The question doesn't make clear if you are expected to fit a formula to this data, such as Poisson.  Also it says nothing about the time scale under which the data was collected. If, for example, we had been told " 'x' shows the number of machine breakdowns during each month collected over the last several years" , we could then estimate the probability of getting more than 2 breakdowns in the next week.

As we're not told this and not asked for such answers, I'd take the simplest interpretation and just calculate total f = T and the probability that X > n by adding a third row to the table.  Then for example P(X>6) = (4 + 3 + 1)/T

[X only takes discrete values so a continuous function is not appropriate I think]

Bob

If you have been given an example of a question like this, then post it and it may make things clearer.

#5 Re: Maths Is Fun - Suggestions and Comments » Minor Upgrade - may or may not go smoothly » 2019-05-15 21:44:49

hi Phro

Just a bobbym search with no keyword specified gives me this:

HTTP 500 error

It would seem that you can get to any post if you search diligently for long enough.  Given that bobbym is  a mathematical god, maybe that's as it should be.  Enlightenment requires some effort. smile


Bob

#6 Re: Maths Is Fun - Suggestions and Comments » Minor Upgrade - may or may not go smoothly » 2019-05-15 19:37:30

The search function yielded no results for me but the following did:

User list gave his profile, which offers a link to topics or posts.  Posts also failed here, but topics worked and gave a list starting in 2009 and running through to his last.  10 pages worth.  Is that a complete list … I'm not sure.  I tried a few at random and his posts are still there.  But only threads started by him.  If you know the thread you can find the rest.

addendum: I've just checked my PM list and all my conversations with bobbym are still intact.

Bob

#7 Re: This is Cool » Integral of 1/sqrt(1-x^2) » 2019-05-12 19:01:02

hi Anthony Lahmann

Welcome to the forum. 

According to Wolfram Alpha, this is sin-1(x) and the second is -sin-1(x).  So there's a sign error in there somewhere.  I'll try to track it down later.

Bob

#8 Re: Dark Discussions at Cafe Infinity » 2015 UK General Election » 2019-05-12 18:56:06

At the bottom of each post there are four options: report, delete, edit and quote.  Delete and Edit are only available to the poster and moderators.  If you click Report,  it flags that post for moderators to look and take action.  That helps because a post may otherwise creep through un-noticed, particularly if it is followed by several more posts by others.

Options then are ban the spammer, edit the post, delete the post.

Bob

#9 Re: Maths Is Fun - Suggestions and Comments » Minor Upgrade - may or may not go smoothly » 2019-05-10 18:25:52

Seems ok for me.  Just deleted an inappropriate post and  it happened straight away.  Great!

Bob

#10 Re: Help Me ! » Simplify by multiplying by denominator before Implicit Differentiation » 2019-04-24 07:46:53

LATER EDIT:

I think I have found the flaw in this post.  This is the corrected version.

The test of any differentiation process is "Does it give the correct gradient function?"  So I thought I would investigate this thoroughly.

The function is:

It can also be expressed:

and also re-arranged to make y the subject:

I have tried all three using the equation grapher and all give the same result:

gmj5h9E.gif

So I then described the gradient function by experimenting with this graph.  The gradient function has these properties:

It tends to negative infinity as x approaches -1 from either side.
There is a local maximum at about -2.4 and a local minimum at about +0.4.
The gradient tends to 1 as x tends to infinity and negative infinity.

Here are the three versions of a possible gradient function:

By quotient rule:

By product rule:

By direct differentiation:

woL8BpC.gif

This has the right properties and so I claim is the correct answer.

My difficulty lies with the way I was using the equation grapher.

So I have used algebra instead and managed to show that all three are in fact equivalent.

By quotient rule:

Substituting

the quotient version becomes

By product rule:

So all three versions are, in fact, the same.

Bob

#11 Re: Help Me ! » Areas of Polygons » 2019-04-24 00:35:27

B6Gj6ox.gif

It's a regular octagon so you can work out the central angles such as GAC. That triangle is isosceles so you can work out ACH or use the right angle.

Bob

#12 Re: Help Me ! » Simplify by multiplying by denominator before Implicit Differentiation » 2019-04-23 19:35:47

hi ioann

Welcome to the forum.

I can find no fault with your working.

The test of whether differentiation has been done correctly must surely be "Does the gradient function have the right shape when compared with the start equation?"

So I put the two functions into the MIF equation grapher and got this result:EDIT the gradient graph here is wrong.  See later post.

9N2LvDL.gif

The gradient curve is negative in the right places, zero at the right point, and positive in the right places.  Also the values increase and decrease at the right times.  This doesn't totally prove the differentiation is 100% correct but I'd say it is strong evidence.

LATER EDIT.  I thought I would try the quotient rule as well.  I got:

How does this compare with your answer?

Then I plotted these and the gradient curve isn't anywhere near close to what is required. ??? So I need to work on this some more.

EVEN LATER EDIT:  This is weird.  I've checked both results using Wolfram Alpha.  Both are confirmed and they are different!  The quotient answer isn't correct as it fails the basic "Does this look like the right gradient graph?" test.  ?????

Bob

#13 Re: Help Me ! » Areas of Polygons » 2019-04-22 22:23:35

AHC is a right angled triangle, so use trig.

Have you calculated angle ACH or HAC yet?

B

#14 Re: Help Me ! » Co-prime numbers » 2019-04-22 22:16:34

hi Amartyanil

If you make x the subject:

x = mc/n

We know x is in Z and m,n have no common factors other than 1, so n must divide c, let's say c = dn where d is in Z


So c is the product of two integers, ie is composite.

Bob

#15 Re: Help Me ! » Find the greatest number of boxes which can be packed in the crate » 2019-04-22 20:34:14

OK, I have a proof. And a simpler packing arrangement.

I'll introduce some 'notation' to make it clearer.

With the crate in front of our position so we're facing one side, I'll call the left-right direction x, the back-front direction y, and the up-down direction z.

And to indicate which way round a box should be placed, I'll give its dimensions x first, then y then z.

Start by placing boxes 40x20x10.  You can make two entire walls of boxes like this taking up 80cm of the x direction.  This packs 50 x 2 boxes.

But no more can be placed with 40 in the x direction because 100 / 40 goes twice with a remainder of 20cm.

So now place boxes 20x40x10.  This fills across the front and we can make two columns of these packing another 10 x 2 boxes.  Then we hit the remainder problem again … there's a 20cm gap at the back in the y direction.  So no more boxes can be placed so that the 40 goes in the y direction.

Only choice left is to pack them with the 40 in the z direction: 10x20x40.  You can get another 2x2 like this and then there's only a 20cm gap left so no more can be placed with the 40 in the z direction.  That's all three directions filled to the maximum extent so maximum number of boxes is 50 + 50 + 10 + 10 + 2 + 2  = 124.

QED.

Bob

#16 Re: Help Me ! » Find the greatest number of boxes which can be packed in the crate » 2019-04-22 19:24:09

hi peterbill

If the boxes were liquid, so that they could occupy any volume, regardless of shape, then you'd be correct.  But the boxes are rigid so it's not just a case of comparing volumes; you must also find a packing arrangement that works.  Two of the dimensions (20 and 10) divide into 100cm but the third (40cm) doesn't.  So you cannot just pack all the boxes the same way round without overflowing the space.

I started by filling a space 80 x 100 x 100 with 100 boxes and then tried fitting more by placing them the other way round. I only got to 120 and the remaining space doesn't have a 40 cm length to fit any more.  I'm thinking that a more complicated packing arrangement is needed but I haven't found it yet.  And I don't think there's a standard procedure for finding one.  I think it's just a case of trial and improvement.

LATER EDIT: Here's a way of packing 124:

0Ojyp1K.gif

There's a 20x20x20 space that I cannot fill.  That doesn't prove that 125 is impossible.  Not sure if that proof can be formed.  Thinking ……………...

Bob

#17 Re: Help Me ! » Areas of Polygons » 2019-04-22 00:58:08

hi dAnta

Yes, that's what I made it!

From what I can remember the pool looks like this:

L9l2lrt.jpg

Here's a view from above.  It is not accurately drawn. 

B6Gj6ox.gif

To get the area of the pool surface you can work out the area of the octagon (no reason why you shouldn't put the two halves together) and the area of the rectangle; then add them together.

Start by calculating the angle ACH.  The sides (such as CG) are given so you can work out AH and AC using trig.  Then you can work out the area of triangle ACG and times by 8 for the whole octagon.  The rectangle length (CF) is two sections long and the width is twice AC.

The pool is a prism so it's volume will be area of the top times by the height of the sides.

It's not clear if 'surface area' means just the water surface or the area of the whole prism.  You'll already know the first.  I'd work out the second as well just to make sure.  You'll need to double the top area so as to include the bottom and add on the sides.  These are all rectangles and there are 12 of them.

Bob

#18 Re: Help Me ! » Area of Polygons » 2019-04-21 09:46:41

When you get an answer, think about the triangle and whether this seems right.  You've ended up with a height and adjacent nearly equal (1.5 and 1.4). But the angles are 54 and 36 so these two sides should be quite different.  How has this happened?

The angle is 54 so you should be calculating tan(54) x 1.5  This will give a result bigger than 1.5

And yes, find the area by half base x height.

Bob

#19 Re: Help Me ! » Complex complex number problem » 2019-04-20 08:10:15

Ok.  I think I have it.

If z = 1/2 + √3/2 i then 'n' isn't any value.  Specially it is 6 because only this complex number raised to the power 6 (or any higher multiple of 6) will actually give 1.  Use the angle made with the X axis ... It's π/3 so needs six of these to make 2π.

So, best answer I can come up with: n must be a multiple of 6.

I think I can sleep on that.

Bob

#20 Re: Help Me ! » Complex complex number problem » 2019-04-20 07:49:55

Let's say w^m = 1 where w = 1/z

By the same argument w is a root of 1, so w = a + ib where a^2 + b^2 = 1, and z + 1/z = 2a again.

Seems to me n can be any integer.

Is any of this correct?  Now I'm worried my whole reasoning is faulty.  I guess that's why I originally hoped someone else would do this question for us.

Still thinking ?

Bob

#21 Re: Help Me ! » Complex complex number problem » 2019-04-20 07:36:11

I've realised that I've been assuming that n > 0.

Let's stick with that for the moment. 

The first equation fixes z as any of the nth roots of 1 and that will work for any n > 0.

The second equation becomes ( as a + ib + a - ib = 2a ) .... (2a)^n = 1 which implies a = 1/2 and therefore b = √3/2 .  This equation also has these solutions for all n > 0.

Now to consider other values of n.

If n = 0, and as anything to the power zero is 1, we still have solutions.

If n < o.      What ?

The first equation can be re written as (1/z)^positive integer = 1 let's say (1/z)^m = 1 where m >0

Have to go back to pencil and paper to explore some more. Continued in next post.

B

#22 Re: Help Me ! » Complex complex number problem » 2019-04-20 06:15:45

Oh, that's interesting.

Let's say z = a + ib.

If z lies on that unit circle then a^2 + b^2 = 1

1/z = 1/(a + ib) = (a - ib)/((a + ib)(a - ib)) =  (a - ib)/(a^2 + b^2) = a - ib

So, yes, once equation 1 is given, 1/z is the conjugate of z.

Bob

#23 Re: Help Me ! » Complex complex number problem » 2019-04-20 04:15:56

Hi,

I've been hoping someone else would make a suggestion here.  The first equation alone gives the nth roots of 1.  In an Argand diagram, draw a unit circle around the origin and mark off points every 2π/n radians

Obviously you need more ... I'll keep thinking.

Bob

#24 Re: Help Me ! » Area of Polygons » 2019-04-20 04:10:30

That's ok so far.  If you split one triangle down the middle, you make a 90, 54, 36 triangle so you can use trig .  In this triangle the opposite to angle 54 is the height you need and the adjacent is half of a side.  Once you have the height you can calculate the area and times by 5 for the pentagon.

Bob

#25 Re: Help Me ! » Areas of Polygons » 2019-04-19 20:39:52

Hi dAnta,

When you have calculated a length or angle, it's a good idea to ask yourself 'is this answer reasonable?'. 24 is far too large for that triangle.  The method is ok up to the point where you make adj the subject of the equation.  It should be adj = 7 / tan(60)

Bob

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