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#1 Re: Coder's Corner » Technical Coding » 2019-03-16 20:08:46

hi Zeeshan 01

You haven't said what you want from this post.  Also what coding language?

Bob

#2 Re: Help Me ! » Please help me with a proof » 2019-03-13 01:35:06

hi mrpace

Depends on the sign of a and b.

if

If a and b have same sign (both positive or both negative)

This will be true for all a and b ∈ {reals}

If a and b have opposite signs the inequality switches:

I'll leave you to finish this off.

Bob

#3 Re: Exercises » Calculating Quartiles » 2019-03-11 21:03:27

hi Zeeshan 01

I agree with your answers.

Why this formula?  Let's start with the median. 

It'll make more sense if the number of data items is 7.  Let's say:

2, 4, 6, 8, 10, 12, 14

The median is the middle item in ordered list so the median is 8. 

If the list had 8 items, say 2, 4, 6, 8, 10, 12, 14, 16  then there's no middle number.  So statisticians have chosen a rule for this: Choose the number half way between the 4th and the 5th numbers ie 0.5(8+10) = 9.  They call this the 4.5th number in the list, just to give it a name.  It's just a name for it so don't get too bothered by it.

The quartiles are defined to be the numbers one quarter and three quarters of the way along the list.

So for n=7, list = {2, 4, 6, 8, 10, 12, 14} the lower quartile is the 2nd number which is 4, and the upper quartile is 6th number which is 12.

This works nicely when n is a multiple of 4 minus 1.  You can easily find actual numbers in the list 1/4, 1/2 and 3/4 of the way along the list.

But what should you do when n isn't a multiple of 4 minus 1 ?  To understand the formulas look at how it works when n=7

(7+1)/4  , (7+1)/2  , 3(7+1)/4 give 2nd, 4th and 6th numbers in the list so that's the formula that has been chosen for any value of n.

(n+1)/4, (n+1)/2 and 3(n+1)/4

Of course, in a list where n is not a multiple of 4 minus 1, this leads to apparently silly positions as you have found 1.5th etc.  For small lists it isn't really a useful statistic anyway but if n is large it gives a good estimate of the values 1/4, 1/2 and 3/4 of the way along the list.

The situation where I have found it is really useful is when trying to compare two sets of data.  Let's say you have the exam results for two classes and you want to consider which class is better.  Probably you'd work out the class averages for this.  But it's also useful to know how spread out the results are.

Just comparing the range of marks can be misleading because there might be outlying values that distort the overall comparisons.  If you calculate the inter quartile range (upper quartile minus lower quartile) you remove the bottom and top quarters of the data and just look at those in the middle 50%.  That statistic is a better way to compare the data.  But this calculation has to work whatever 'n' is, which is why a formula is needed.

Bob

#4 Re: Maths Is Fun - Suggestions and Comments » Does mathisfunforum block certain IPs from access? » 2019-03-06 22:32:21

hi pcallahan

The technical aspects of the forum works are a bit of a mystery to me.  If the problem persists we'll have to get MathsIsFun himself on it.

I've had a look at the IP addresses and there's no block on either of yours.  It is possible to ban one to prevent a spammer from just changing to a fresh username.  But there's no ban on yours.

A while back I created  a page which was an index to a series of geometry pages.  So it had lots of hyperlinks.  I fell foul of some forum limit and had to set up a second page to complete the task.

The forum records the date an IP was last used, so it might help the investigation if you can remember these for each of the two IPs.

Bob

#5 Re: Computer Math » How much math is involved in computer science?? » 2019-02-28 06:33:56

hi Jacobcullen

Welcome to the forum.

I've taught computer science to A level and no calculus was required.  Working in number bases other than decimal would be useful eg. binary, hexadecimal maybe octal.  I had done logic (sometimes called propositional calculus) at university and this was very useful for understanding NAND gates and FLIPFLOPs for example.

Bob

#6 Re: Help Me ! » [ASK] Exponents and Roots » 2019-02-28 06:12:44

I have decided to assume that there are several errors in this question.  As the five possible answers are very similar it seems more likely that these lie in the start expression.

x^{5/6) leads to a cube root of 2 whereas we are hoping for a square root of 2.  It is possible to show that no rational multiple of the first can give the second so that 6 must be in error.  More likely a 4. This leads nicely to the (1+2root2) part of an answer.

56 is indeed a problem.  How can a numerator with an integer and a root 3 lead to just a root 3.  So I have altered the minus sign to a times sign.

9/2 leads to an 81 in the answer.  Could it be 9^(1/2) rather than 9 times 1/2 ?

And x^-2 at the denominator leads to an answer that is far too large.  Given that this is an exercise in fractional powers perhaps it should be x^(1/2)

So my suggestion is

This does lead to one of the given answers.  You can have a try or uncover my hidden working.



Bob

#7 Re: Help Me ! » [ASK] Exponents and Roots » 2019-02-27 02:24:27

hi Monox D. I-Fly

I too got stuck with this.  So I tried evaluating the expression and the five possible answers and could find none that worked.

x^(5/6) seems to be a sticking point for me as that leads to a cube root of 2 not a square root.  The answer must have a root of 3 in it.

If we're looking for a minimum typo to make this 'do-able' then losing the 7 would be a start … but that power 5/6 will have to change too … maybe 5/4 ???

I'll try exploring alternatives.

Bob

#8 Re: Exercises » Number of integer solutions » 2019-02-25 02:11:40

hi Amartyanil

If you have some centimetre squared paper then you could try plotting the straight line.  The answer will be easy to spot then and you'll see a pattern

Bob

#9 Re: Guestbook » Site » 2019-02-25 02:09:32

hi gaconvn1106

Welcome to the forum!

Bob

#10 Re: Help Me ! » anyone good with doing math on excel? » 2019-02-20 19:29:01

hi Wolfwood7388

Have a look at my post 12.  Is this what you're asking about?  In which case, are you able to say what needs more explanation.

=COUNTIF(L2:L10001,>0) this returns an error

I checked the syntax for this function and found this example:


=COUNTIF(B2:B5,">55")

Counts the number of cells with a value greater than 55 in cells B2 through B5. The result is 2.

So try adding the ""s

Bob

#11 Re: Help Me ! » Maximum and Minimum question, help desperately needed » 2019-02-13 21:29:42

hi Wch12267

Welcome to the forum.

I'm going to use x rather than o here to avoid confusion with zero.

root(1-x) cannot have a real value if x is positive as the square root of a negative doesn't exist in real numbers.

So I did a bit of trial and improvement.  I set up Excel with 11 lines of formulas, testing from -1 to 0 in steps -0.9, -0.8 and so on.  I could see a solution lay between -0.9 and -0.8 to I repeated the search between these and gradually 'homed in' on -0.866.  That looked to me like root(3)/2 so I tried that directly and the expression evaluated to 1.

Now Excel only works with 'so many' digits of accuracy so this isn't a proof of the solution, but it gave me enough to switch to an algebraic method.  What I'm about to show you has to be used with caution as it can lead to values that aren't solutions as well as values that are.  Let me show you why:

Suppose we have a simple equation like x = 5.  That's easy to 'solve' ; the answer is 5.

But if I square it: x^2 = 25 ; this has 2 solutions, x = 5 and x = -5.  The second isn't a solution of my original equation!  So if you ever use this technique, always check the values you end up with, to make sure they really do solve the original problem.

I'll replace the inequality with an equals.

square both sides

Simplify and rearrange

Square again:

The 'other' value can be disregarded as it is positive and I've already discounted that.

Bob

#12 Re: Help Me ! » geometry help » 2019-02-05 19:24:47

Hi Monox D. I-Fly,

If you stand close to the tree and stand the stick so that it mimics the 'lean' of the tree (if necessary as the tree may not be vertical) and not in the tree's shadow, then the tree and its shadow and the stick and its shadow make similar triangles.  So, by measuring the two shadows you can calculate the height of the tree.

Bob

#13 Re: Introductions » Kamov's introducing himself » 2019-01-27 06:59:19

hi Kamov50K

Welcome.  Or you could ask her to teach you smile

Bob

#14 Re: Help Me ! » Last step » 2019-01-27 06:56:39

x = 6, y = 2/9 is a point of inflexion.  Cannot find any others.

Bob

#15 Re: This is Cool » Ellipse Whirl » 2019-01-20 20:32:08

hi benice

Thanks beautiful!  I think I may be hypnotised now.  dizzy

Bob

#16 Re: Exercises » Integration » 2019-01-18 19:27:55

Thanks zetafunc.  It seems that the limits make all the difference.

Zeeshan 01: After a bit of research I think I can take you through this, but it'll take more than one post.  If you would like that please reply.  Or you could do like me and use a search engine.

Bob

#17 Re: Exercises » Integration » 2019-01-17 22:19:59

hi Zeeshan 01

Wow! Where did that come from?  Not out of a beginners book of integration, I hope.  I couldn't see how to do it, so I used https://www.wolframalpha.com/ to compute it.  It has a far from easy solution as you'll see.  Did you just make this one up yourself?  That's a hazard with trying that.  Not all functions are easily integrable.

Bob

#19 Re: Exercises » Derivative » 2019-01-13 20:36:56

hi Zeeshan 01

There are two stages to this problem.  I think you have done the first but I'll say anyway.

This is a product of two functions so use the product rule:

You can combine the trig functions into a single trig function using the compound angle formulas.  You want a cosine so I'll use

To make the derivative look like the right hand side here:

Hope that helps,

Bob

#20 Re: Exercises » help intercepts » 2019-01-12 20:36:15

hi yorkmanz

Welcome to the forum.

Looks like you've got these sorted ; I agree with all your answers.

Bob

#21 Re: Help Me ! » I don't how to solve this problem! » 2019-01-11 20:51:46

hi xxStellaxx

Welcome to the forum.

This question has been asked many times.  Post 18 of this thread:

http://www.mathisfunforum.com/viewtopic.php?id=18391

has a diagram that may help.

Bob

#22 Re: Help Me ! » Areas of Polygons » 2018-12-28 21:44:49

hi Strangerrr

The area of the triangle will be different as it depends on both the side length and the angle at the top.

n is the number of sides.  Here's an example using a regular 12 sided polygon.  The centre point is O.

LQS5lM7.gif

Let's say that A1A2 = side length = 7cm.

First I'll work out the angle A1-O-A2.  Calculate 360/12 = 30. 

The triangle A1-O-A2 is isosceles.  The vertex angle is 30 and the base is 7.

Split it in half; see yellow shaded right angled triangle.  This triangle has an angle of 15 and the 'opposite' is 3.5cm.

Use trigonometry to calculate the height.

Now calculate the area of the triangle A1-O-A2.

Now calculate the area of the whole polygon.

Hope this helps,

Bob

#24 Re: Help Me ! » Areas of Polygons » 2018-12-27 21:20:34

I agree with all those answers.

Provided the polygon is regular (with n sides) you can draw lines from the centre out to the vertices and make n triangles.  Once you have calculated the area of one triangle you can times by n to get the area of the polygon.

These triangles will always be isosceles.  The base is the side of the polygon.  But you'll also need the height of the triangle.

The angle at the top of each triangle is 360/n in size.  With an isosceles triangle you can always split the triangle in two down the middle, making a right angle at the bottom.  The top angle will be split into two angles of 180/n each.  The base is split into two parts of side/2 each.  You can get the height by either

or

Then the area of the polygon is

Bob

#25 Re: Maths Is Fun - Suggestions and Comments » Merry Christmas! » 2018-12-24 22:07:16

And a Very Happy Christmas and New Year to you!

Bob

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