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#1 Re: Help Me ! » Conversion of recurring decimal to fraction algebraically » Yesterday 20:48:48

The OP has not suggested what digits come before the 251 repeats, so I'll make up an example.  It is always possible to get a fraction.

example.  Convert 0.3251251251 recurring into a fraction.

step 1.  Split off the non recurring part.

0.3 + 0.0251251251...….  note 0.3 = 3/10, so that part is now a fraction.

step 2.  Call the recurring part a/b, where a and b are integers and multiply the recurring part by a suitable power of ten.  There are three recurring digits so 10^3 is suitable.  For n recurring digits use 10^n.

a/b = 0.0251251251....    1000a/b = 25.1251251251251.....

step 3.  Subtract to eliminate the recurring part.

999a/b = 25.1

step 4.  Multiply by a suitable power of ten to make this a ratio of integers.  There was one non-recurring digit so 10^1 is suitable.  If there had been m non-recurring digits then use 10^m.

9990a/b = 251  => a/b = 251/9990

step 5. Combine the two fractions to complete the conversion.

3/10 + 251/9990 = (2997 +251)/9990 = 3248/9990

Bob

#2 Re: Maths Is Fun - Suggestions and Comments » How do I delete topics I've made (if possible)? » Yesterday 02:58:11

hi Benjamin,

I think the answer to your question is that members cannot do this but administrators can.  But, I've just done a search through your posts and I think they are worth keeping.  smile

Bob

#3 Re: Help Me ! » [ASK] Addition of a Fraction » Yesterday 02:51:09

Actually it is possible to do the steps just with fractions:

The second term in each bracket cancels with the first term in the next.

Bob

#4 Re: Help Me ! » [ASK] Addition of a Fraction » 2019-11-07 20:06:37

hi

The denominators are 'triangle numbers' and there's a formula for those.

n(n+1)/2

Then you can use partial fractions to split each term into two fractions, one positive and one negative.

Most terms then cancel out.

Bob

#5 Re: Introductions » I am a Kaboobly Doist » 2019-11-04 21:22:02

hi Agnishom

I remember when you first posted.  How time flies!

It's great to read your update.  You've done really well with your studies!

Bob

#6 Re: Help Me ! » [ASK] Minimum Dimension of a Map » 2019-11-04 21:20:16

hi Monox D. I-Fly

It looks to me that this question is about comparing the 1 metre width of the map with the 200 Km in reality.  You're right that the conversion isn't a whole number but I don't think you're meant to worry about that.

If the line is AB, 200 Km long in reality, we need to fit that line on the map.  Worst case is if its parallel with an edge. (diagonally is easier to fit)

If we made it 1 metre is equivalent to 200 Km the A and B would be right on the edge.  Most people wouldn't want that, so they've made it clearer by implying that you just need to make the width 210 Km.  Then we know the line will fit without A or B being on the very edge.

So the scale to make this happen is 1 metre : 210 Km.  Convert the second distance into metres and you've got your answer.

Bob

#7 Re: Guestbook » Applied Mathematics » 2019-11-03 03:39:36

hi seercalf

I don't but you'll find these pages are as good as, and in some ways better than, a book:

https://www.mathsisfun.com

Bob

#8 Re: Help Me ! » Analytic Geometry- PLEASE HELP! » 2019-11-03 03:36:57

hi DoeADeer

Welcome to the forum.

Thanks for posting anyway.  It was an interesting problem to have a go at.  First part was fairly quick to do.  I found the second part tricky as it took many steps.

My method was

(1) Write down the equation for the perpendicular bisector of CB.
(2) Calculate the equation for the perpendicular bisector of AB.
(3) Use these to find the centre D of the circumcircle.
(4) Get the vector AD and use this to get the coordinates of A'. (AD = DA')
(5) Show those coordinates satisfy the equation of the hyperbola.

Do you have a simpler method?  I'd like to see it if you do.

Bob

#9 Re: Help Me ! » The Last Few Terms in an Infinite Series » 2019-11-02 06:43:25

hi veeceeone

Oh great!  It will be such fun having you as a member.

How can people discuss this with each other if they cannot understand it?

But what have we been doing?

As for the second part; the 'validity' of any theory is tested by "does it work"?

Richard Feynman said: "If you think you understand quantum mechanics, you don't understand quantum mechanics."

And, slightly off the topic of understanding but even more fun,  from Lewis Carroll (real name Charles Dodgson … he was a mathematician as well as writing Alice in Wonderland):

"Alice laughed: "There's no use trying," she said; "one can't believe impossible things."
"I daresay you haven't had much practice," said the Queen. "When I was younger, I always did it for half an hour a day. Why, sometimes I've believed as many as six impossible things before breakfast."

Bob

#10 Re: Help Me ! » [ASK] Determinant of a Matrix with Polynomial Elements » 2019-11-02 01:29:12

hi Monox D. I-Fly

I'm not seeing your image; just a fail to load error.

When I copied the link and tried to 'go there' I got a 'lack permission to view page' error.

Bob

#11 Re: Help Me ! » The Last Few Terms in an Infinite Series » 2019-11-02 01:21:25

hi veeceeone

I doubt that anyone can 'understand' infinity.  You cannot see it or be aware of it with any other sense.  But it does have its uses.

Before I start on that a word about mathematics.  Maths doesn't exist in the above sense either … but you can use it for modelling.  There's a whole load of theory used to develop arithmetic for example.  This theory works on paper and could just be a study in its own right.  But most people use it when the model can be applied such as working out your bank balance; counting your sheep; deciding if you've bought enough marshmallows for everyone at your party to have three each.

But there are also situations where the rules of arithmetic don't work: eg.  when you add one pile of sand to another pile of sand the result is one pile of sand (1+1=1). eg.  show me half a piece of chalk.

You might have learnt lots of Euclidean geometry, but take care trying to use Pythagoras' theorem on the surface of a sphere.

So there's a branch of maths that considers what happens if you repeat something for ever.  Obviously we cannot really do this but we can ask what if?

There's a story about a frog that can jump 1/2 a metre on the first jump, then 1/4 m then 1/8 m and so on.  If it carries on jumping for ever how far will it jump altogether?

Put a line to indicate the frog's starting point and another exactly 1 metre away.  The frog covers half the distance on it's first jump; then half what is left on the second; then half of what's left on the third jump and so on.  It will get closer and closer to the 1 metre line as time goes by.  Will it reach the line?  Well no, because there's always a tiny bit left to cover.   But real frogs cannot jump so precisely and the line we made will have some thickness so eventually you won't be able to tell that the frog hasn't arrived on the line. So it is reasonable in practice to ignore the tiny distance left, and accept that the frog does get there in the real world.  In mathematics we make it sound more rigorous by saying that the limit as n tends to infinity of the series 0.5 + 0.25 + 0.125 + ……. is 1.

There's as much validity in that as assuming that lines have no thickness or that you can actually measure the three angles of a triangle, add them up and get 180.  I've tried this with a class and we got answers ranging from 178 to 182.

Now to the sums.

Let's say that S has n terms. (I understand that n is  ∞.)

In the theory of infinite sums I afraid the above is wrong on two counts.  The series has no last term.  And mathematicians don't allow infinity to be a number.

So, when you subtract one series from another the subtraction process goes on for ever.  So there is no last term left over.  The parts of the two series that go on for ever 'cancel' each other out completely and so you're left with some algebra you can deal manage without worrying about infinity.

If you look here:https://www.mathsisfun.com/algebra/sequ … etric.html

you'll find the same trick used to find the sum of n terms of a geometric series and then the sum to infinity of such a series.

Your investigative result is correct (well done!). It is a special case of the above sum formula with a = 1

Bob

#12 Re: Introductions » veeceeone an Introduction » 2019-10-31 06:00:46

hi veeceeone

Welcome to the forum.  Thanks for the kind words about the site.

It seems like you are asking for help so post your question in the Help Me section.  Click the Index tab.

Bob

#13 Re: Help Me ! » [ASK] Probability of Getting the Main Doorprize » 2019-10-27 21:11:54

hi Monox D. I-Fly

As the 'event' is the main prize I don't think it matters about any other prizes.  We should have been told that 240 tickets have been allocated.  This is not a trivial point as one member controls 15 tickets.  Is it allowed for one person to have more than one ticket or must we assume that Mr Aziz has 15 in his family and they are all members?

But we don't know all this so I'll assume that 240 tickets have been given out and he has 15 of them   One ticket will get the main prize, so he has 15/240 chances of getting it.

Bob

#14 Re: Help Me ! » Linear inequalities » 2019-10-27 21:06:31

Rules in maths are designed to make maths useful in practical situations.

If you owe money then it's easy to understand why $3 owed is worse than $2 owed, so mathematicians have decided to use < in the sense of 'further to the left' on a number line.  Thus -3 < -2

It's defined that way for the convenience of any situation where 'better' means further to the right.  Another example would be with temperatures.  -3 is colder than -2 so again -3 < -2 makes sense.

If you have two positive points on a number line, A and B with A to the left of B; then A < B

But when you multiply or divide by a negative then -B is now to the left of -A so -A > -B

This doesn't happen if you just add something or subtract something ( A + x < B + x) as the new numbers are still in the same left to right order.

Bob

#15 Re: Help Me ! » How to estimate a derivative from a graph » 2019-10-27 20:56:53

When x = -1 is substituted into the quadratic formula the 'a' term is ax^2 = a times -1 times -1.  So the minus signs cancel out.

If the equation isn't a quadratic then what happens next depends on what you know about the formula.  If for example you know it's a cubic then you'd have to start with

y = ax^3 + bx^2 + cx + d.

You could only find all four (a,b,c and d) if you know four sets of coordinates. 

Bob

#16 Re: Help Me ! » Standard Error or Standard Deviation ? » 2019-10-27 20:49:53

hi math9maniac

If you know all the values in a population (say height of people) then you can calculate the standard deviation as a measure of spread.  It's the square root of the deviations from the mean squared.

But often it wouldn't be practical to get every value so it is necessary to investigate a sample and estimate the sd. from those values.

Let's say you have a sample of size n.  If you could take sample after sample and get the mean of each sample, these means would themselves form a distribution with a mean and sd.  It can be shown that the mean of these sample means is likely to be the mean of the real population (more samples means a better estimate of this) and the sd (s) of the sample means is the standard error.  Using the formula

where s is the standard error and sigma is the population sd, you can then estimate the population sd.

There is a wiki page on this here:

https://en.m.wikipedia.org/wiki/Standard_error

There is also some notes on this here:

https://www.mathsisfun.com/data/standar … mulas.html

but the proof is not given there.

I'm a bit rusty on this but could look it up if you want.

Bob

#17 Re: Help Me ! » How can i modify the fundamental error? » 2019-10-22 22:25:36

hi IEd

Welcome to the forum.

You can use the usual rules to integrate the function and substitute values but that doesn't make your answer correct.  For any integral you should always make a sketch of the function so you can check for places where the function changes sign and where the function goes to infinity.

This function is 'infinite' at x = 0 and the area under the graph cannot be determined.  I suppose it might tend to a limit but try here:
https://www.wolframalpha.com/input/?i=i … -2+to+%2B1 and you'll see that doesn't happen here.

So I don't think any tricks exist to help.

Even if you said x not zero you'd still encounter a problem if the area tends to infinity.

One way to examine this would be to create a program or use a spreadsheet to calculate the area of rectangles that are approximately vertical strips from the graph.  What happens to the size of these as you approach x = 0.  (Probably enough to just do right hand side values).

Here's my attempt:

I created row headings and then entered x = 0.1 into cell A2
I made B2 = A2/10 so that I am dividing the region between that x value and zero into 10 strips.

I made A3 = A2 minus $B$2.  The dollar signs mean that B2 will always be used rather than advancing the reference when I copy it

So I could then copy the formula down the column.

Then I calculated the y values and the area of each strip. 

So it was easy to reduce the A2 value towards zero and see what happens.  Here's a screen shot when x = 0.001 with the formula view alongside so you can see how each line is calculated.

zqZpQRM.gif

This is using MS Excel.  If you have that software you can try this yourself. As x gets smaller the area values tend to infinity as expected.

This is because, for this function, 1/x^2 gets bigger more rapidly than dx gets smaller.

Bob

#18 Re: Help Me ! » voltage derivative » 2019-10-22 19:46:13

All derivatives are defined as the limit of the chord gradient between points A and B on the curve as B approaches A.

If the gradient changes instantly to a different value it cannot be determined at that point as the limit is different on one side of the point to the other. 

If you compare the 'official' answer  with yours, you'll see that your lines 'overlap' each other so that, for a given value of x, there are two values of dy/dx.  That cannot be correct.

Here's a written description of what the gradient is 'doing'.

It is positive and constant for a while; then it switches to zero for a while; then it is negative and constant for a while; then zero again; then this four stage description repeats.

Bob

#19 Re: Help Me ! » Set theory » 2019-10-22 19:37:20

hi 666 bro

Introduce a second set B that does not intersect A. 

A intersect B = null set, so the null set must be in A and in B.

This might seem to be just playing with non existent sets but it's safer to assume this is true than trying to re-write the whole of set theory.

Bob

#20 Re: Help Me ! » How to estimate a derivative from a graph » 2019-10-22 19:27:50

The curve  goes through (1,3) , (3,-1) and (-1,-1). 

Substituting these into y = ax^2 + bx + c  you get

3  = a + b + c                                (1)
-1 = 9x^2 + 3b + c                        (2)
-1 = a -b + c                                  (3)

If you subtract (2) from (1) you'll eliminate c
Repeat subtracting (3) from (1)

So you've got a pair of simultaneous equations and can work out a and b, thence c

Bob

#21 Re: Help Me ! » How to estimate a derivative from a graph » 2019-10-21 22:14:07

By looking at the curve I can see it goes through (1,3) , (3,-1) and (-1,-1). 

I'm assuming it is a quadratic as the gradient graph is a straight line.

Substituting these into y = ax^2 + bx + c and solving for a, b and c you'll get:

y = -x^2 +2x +2.  You can check this with a function grapher.

so dy/dx = -2x + 2

So the equation of the gradient graph is y = -2x + 2 as your second graph shows.

Bob

#22 Re: Help Me ! » instantaneous velocity at a point » 2019-10-19 21:01:44

hi

That value is correct 2 x cos(2) using calculus.

I think for this question and your other one, you will find this derivative plotter helpful:

https://www.mathsisfun.com/calculus/der … otter.html

Bob

#23 Re: Help Me ! » Visualizing proof » 2019-10-16 04:12:39

hi

Here's the proof that I was taught many years ago.

ztwLV2t.gif

Draw a circle radius R and centre C.  Mark points A and B as shown and D so that AD is perpendicular to CB.

Let angle ACB be x in radians.

sin(x) = AD / R

arc length AB / whole circumference = arcAB / 2.pi.R = angle of arc / whole circle angle = x / 2.pi

cancelling the 2.pi gives arcAB / R = x

So sin(x) / x = AD / arcAB

implies sin(x) < x but as x approaches zero AD approaches arcAB so sin(x) / x tends to 1.

Bob

#24 Re: Help Me ! » Statistics and Economics » 2019-10-15 19:54:39

Many years ago when I was doing A-level maths I studied pure maths and mechanics. Many years later I was teaching the subject.  And statistics had been added to the curriculum.  So I had to self teach so I could teach my classes.  Actually that worked quite well because I knew what had been difficult for me and how I'd overcome this. 

So I do know about regression lines for example.  Post if you need help.

Bob

#25 Re: Help Me ! » Using limits with complex numbers » 2019-10-15 00:45:30

hi 666 bro

Strictly speaking, the Argand diagram is not a graph.  As it has 'x' and 'y' coordinates I can see why you might think it was.  But it is just a way to represent complex numbers and there is lots of useful maths that stems from it.

A function such as y = 2x + 3 has an input (x) and an output (y).  So you can show the effect of the function by plotting a graph.

A function like z = x^2 + y^2 cannot be represented in 2 dimensions, so, if you want a visual representation you have to try and represent the z axis, perpendicular to the other two … usually rising up from the x-y plane.  Tricky to show clearly because we are only 3 dimensional beings (as far as I can tell smile )

So if that was the  function of complex numbers of the form x + iy and z was limited to real values that's how you'd have to graph it.  But z could also be complex in which case we run out of dimensions to show what is happening. Nevertheless there's a whole area of maths that does consider such functions. 

Bob

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