Math Is Fun Forum

Hi learn2teach,

Please, not Uranus.  It is the same here, with the same pressures, the media jumping on any chance to bash teachers and kids hating math.

But, that's not a reason for giving up on trying.  Your equation example, and bobbym's quadratic both show that the students were just trying to apply a 'parrot fashion' technique, not thinking about the problem and what it is actually asking.  I've had that too. 

eg.  Kid: "And so the man earns £30 000 a week."  Me: " Doesn't that seem a bit high for cleaning windows?  Do you think there might be a decimal point in there somewhere?"

It sounded to me like your own tests were close to what you'd done in class ... and hence they could do them; ..... and then they get someone else's test, where the questions are a bit different, and their brains cease up.  That's why I advocate trying to get them to think.  You won't find it easy;   I've had kids saying 'Just tell me what to do; I  don't want to know why!'  But I've got more determination than they have, so gradually I wear down their resistence.

Can you afford lollipops?  That's to reward good thinking.  "And today's lollipop for good thinking goes to Bill for ......."

Or little rubber stamps that print a happy message like "Skate boarding Joe (+picture) says well done" or "A+ from the Smart Cat".  I've had tough, street-wise, 16 year olds begging for a stamp.

Or postcards that you send home to their parents, saying how great their kid was today.

If thinking is not what they're used to then it's an uphill battle.  But that's what you're paid for isn't it?

Hope you find a way forward, best wishes,

Bob

hi kerryadelfred

Go to the help section, choose a suitable topic title for your post and ask your question.

Best wishes,

Bob

Hi lindah,

Glad to be of help.

Hi bobbym,

Bags I get to drive the Gran Torino!

Bob

hi lindah

My first thought was that I ought to recognise this as a text book distribution and I could not see which one.

But your suggestion of listing the possibilities is a good one.  Sometimes the number of possibilities is just too big to list them all but, for this question quite small.  You should have done it!  If you want the satisfaction of solving this yourself, then read as little or as much as you need from what follows.

It's not 5! however.  Because one D is just like the other and similarly, the  nonDs are the same,  you have to divide by 2! and 3!

so

Whoops, I uploaded the images in the order I wanted but the software has reversed them.

I've listed them below in image 2.  (D = defective,   G = good)

And image 1 is the distribution, showing X, Frequency of X, and Probability of X.

You can get the mean by either

or by

Hope that helps

Bob

hi irrational

Here's a diagram to show that the decimal number system behaves the same under increasing magnifications ( x 10 each time).

Bob

Hi,

I was hoping you would not notice that I had not shown the complete proof.  But you have called my bluff, so now I've had to sit and think about it.  And my wife wanted me to mow the lawn ... which helped a lot because I could make a plan while I was doing this.

Let's go back to post 2.  When I read your question, my first reaction was 'surely this cannot be true .... let's find a number that cannot be written as a sum of non-repeating Fibonacci numbers.'

So I began this list and quickly realised they always could.

1 = 1
2 = 1 + 1
3 = 2 + 1
4 = 2 + 1 + 1
5 = 3 + 2
6 = 3 + 2 + 1
7 = 3 + 2 + 1 + 1
8 = 5 + 3
9 = 5 + 3 + 1
10 = 5 + 3 + 2
11 = 5 + 3 + 2 + 1
12 = 5 + 3 + 2 + 1 + 1
and then there's no more small numbers but I can use 8 so
13 = 8 + 5
....
20 = 8 + 5 + 3 + 2 + 1 + 1
and again I've run out of small numbers but I can use 13 so
21 = 13 + 8
....
33 = 13 + 8 + 5 + 3 + 2 + 1 + 1
and I've run out of small numbers but I can use 21 so
34 = 21 + 13

If a , b and c are three Fibonacci numbers in sequence, a < b < c so that a + b = c
I noticed that any number up to c - 1 could be done without using b at all.

eg If c = 34, then anything up to and including 33 can be done without needing to use 21.

This is what I needed to make the induction step.

Suppose the next Fibonacci number after a, b, c is d so that b + c = d.

I also need to invent some notation to cover the sum of some Fibonacci numbers so:-

Definition:  Let SF(a) mean the sum of some Fibonacci numbers up to and possibly including a. (not totalled but rather just written out so SF(5) could be 5 + 3 + 2 and SF(21) could be 21 + 13 + 8 + 5 + 3 + 2 + 1

SF(a) may be just a single Fibonacci number and may be all up to a ie. 1 + 1 + 2 + 3 + ..... + a

Induction Assumption.  That all numbers up to c - 1 can be done with some SF(a) and c = b + c.

Now consider numbers between c + 1 and d -1.  Let's say N is such a number.

d = b + c  = b + a + b

So N < b + a + b                          =>         N - b  <  a + b

But the assumption says numbers under a + b can be done with SF(a)

So N can be done with b + SF(a).

An example will make this clearer.

Say a = 13, b = 21, c = 34 and d = 55

Try N = 50

N - b = 29

29 = SF(13) = 13 + 8 +  5 + 3

therefore N = 21 + 13 + 8 + 5 + 3

d itself is of course b + c

This shows that if all numbers up to c can be done, then all numbers up to d can also be done.

The initial step needs the initial values of a, b, and c ( 1, 1, 2) so you need to show that these three (two?) can be done.

Hope that helps,

General method. Try it with numbers first to see what's going on.

Bob

Thinking ....................

Bob

Hi Akpolome,

OK, now I know where we're starting from.

The 1/2 at the start can simply be ignored until you write your answer as it's just a multiplier.

So what about the sin(2x)  ?

If you wanted to differentiate cos(2x) you'd have to use the 'chain rule' as it's a 'function of a function' .

ie cos u where u = 2x

So you have to differentiate cos u (=-sin u) and multiply by u differentiated with respect to x (= 2)

So d (cos2x) /dx = -sin u . 2 = -2sin(2x).

Now back to the integration, ask what must I have differentiated to get (1/2). sin (2x)

It must have a cos and must start minus so that the result ends up positive and it shouldn't have a 2 in it, so better half at the start  and finally another 1/2 for the multiplier.

...............................   - (1/2) . (1/2) . cos (2x) + C  = -(1/4)cos(2x) + C

and that's the answer.

When I have to do these I usually start with a guess at the function ( cos something) and then adjust my answer so that when it is differentiated it gives the desired result.

Hope that helps,

Bob

Hi,

Fibonacci question.

When I get a problem that seems tough I look for examples first so I can see what is happening.

Can any number really be the sum of Fibonacci numbers used, at most, once each?  Let's try.

FB = 1, 1, 2, 3, 5, 8, 13, 21, 34, .......

1 = 1
2 = 1 + 1
3 = 2 + 1
4 = 2 + 1 + 1
5 = 3 + 2
6 = 3 + 2 + 1
7 = 3 + 2 + 1 + 1
8 = 5 + 3
9 = 5 + 3 + 1
10 = 5 + 3 + 2
11 = 5 + 3 + 2 + 1
12 = 5 + 3 + 2 + 1 + 1
and then there's no more small numbers but I can use 8 so
13 = 8 + 5
....
20 = 8 + 5 + 3 + 2 + 1 + 1
and again I've run out of small numbers but I can use 13 so
21 = 13 + 8
....
33 = 13 + 8 + 5 + 3 + 2 + 1 + 1
and I've run out of small numbers but I can use 21 so
34 = 21 + 13
and so on.   

So there seems to be a general rule emerging here.

Can you show that any number can be written as the sum of two smaller numbers without using the same FB number twice?

If the theorem is true for the two smaller numbers then its true for their sum.  This is what you've got to show.

Other problem.

I'm still thinking about this one.  Weak induction means you assume it's true for n, and show this leads to it being true for n+1
Strong induction means you may assume it's true for 1,2,3 ... up to n and show this leads to it being true for n+1.

Bob

You're welcome.

Bob

Hi hannahr,

Do you know 'Pascals Triangle'.  This can be used to generate terms in binomial expansions

eg.   

The expression

is the formula for the coefficients in this expansion.

So you can use the Pascal rule to do the induction step.

Consider the expression

This is the general term for the next row down in the triangle.

split off some bits from the factorials

separate into two fractions

cancel

multiply out

This shows that a term in the triangle is made by adding the two terms in the row above.

If they are each integers, then their sum will be too.

So if any row of the triangle consists of all integers, then the row below will be too.

Finally you need an initial step.

Show that

is true when a = 1 and b = 1 and you have an inductive proof.

Hope that helps.  Post again if you want more on this.

Bob

Hi

I'm unclear about the exact form of your integral.  Are you able to use LaTex commands such as

math]\frac{numerator}{denominator}[/math]  (missing [ to stop this being parsed.)

to show which expressions are above the line and which below.

Or add brackets ???