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#1 Re: Help Me ! » Differentiation » 2013-04-09 02:52:30

Try: Quotient Formula:  D(f/g) = (g*Df - f*Dg)/(g²) where  f(x)=sqrt(1-x²)^(½) and g(x) = 1-x

Also for the 1/2 power:  D[f]^n = n*[f]^(n-1) * Df. 

I hope this helps. smile

#2 Re: Help Me ! » exponents » 2013-02-18 12:08:00

Yes, you do think.  And you are quite good at it! smile

#3 Re: Help Me ! » exponents » 2013-02-17 11:52:30

Thanks stefy!

I might have worried all night about what that meant.  I'll check the site out and see what it is about.

Bless you!  smile

#4 Re: Help Me ! » Properties of addition » 2013-02-17 11:49:20


And here is a little different way to look at commutativity and associativity.

Commutativity can be viewed more as a language problem than an axiom problem.  Let me
explain.  Take the example of adding four and five.  We typically write and speak in a linear array
of letters or syllables.  This forces us to say one of the numbers first and the other second.  So
we say "four plus five" or "five plus four" and write 4+5 or 5+4.  But we want these two
expressions to mean the same thing; that is, we want them to get us to "9."

As a function (binary operator in this case) we have a function "+" which maps (4,5) to 9 and
also (5,4) to 9.  If we had a way to "say" the four and five SIMULTANEOUSLY, then we would
have but one way to "speak" or "write" about the addition of four and five.  Thus we wouldn't
have to "invoke" a commutativity rule.

Actually we could define addition of 4 and 5 as a mapping from the SET {4,5} to 9.  Since the
set is unordered this gives us only one way to express the combination of the 4 and 5 to get
the 9.  Then "4+5" and "5+4" could be viewed as ways of saying that {4,5} maps to 9 or that
these are ways of expressing the result of mapping {4,5} to 9.  But then the notation +{4,5}=9
is a bit awkward compared to 4+5=9 or 5+4=9.  On the other hand as the sets get larger, for
example {3,4,5}, +{3,4,5}=12 is not so bad.  And here with the three numbers using this
approach we don't have to get into associativity either.

If we define 3, 4, and 5 by 3={ooo}, 4={oooo} and 5={ooooo} (multisets) then adding these
three sets together is just in essence dumping them altogether to get {oooooooooooo}.  How
we dump them together (all at once, one at a time, etc.) is immaterial.  We get the same result.

So commutativity and associativity can be view more as a language problem than an axiom
problem.  On the other hand, the additive and multiplicative identity and inverse axioms are
axiom problems in the sense that they do not just involve ways of "saying the same thing".
They inject the notion of "existence" into the system.

All this being said, viewing the way our systems of numbers have evolved, it is perhaps easier
to just introduce commutativity and associativity as axioms and be done with it.

Have a very blessed day! smile

#5 Re: Help Me ! » exponents » 2013-02-17 10:49:10

Hi niharika_kumar,

You are quite welcome! Seeing patterns is often the name of the game in solving math problems.
And if you have access to a good math program you can often write a program to test for patterns
associated with a given problem. smile

Hi cooljackiec,  Not only am I not levans,  I have no clue as to what AoPS is.  What is it?

And thanks bobbym for the kind comment.  I do try to be polite and not hurt anyone's feelings
even if I think what they are saying is incorrect.  After all,  I have been wrong before.  As I
recall once in 1947 I was wrong about ... smile

#6 Re: Help Me ! » exponents » 2013-02-13 11:44:03

Or another way to look at it:

0th power of 12357 ends in 1.
1st power ends in 7.   2nd power ends in 9.   3rd power ends in 3.   4th power ends in 1
5th power ends in 7    6th power ends in 9    7th power ends in 3.    8th power ends in 1
9         ...            7    10      ...               9   11       ...              3    12    ...                  1, etc.

There are only four digits that occur in the powers and these repeat in the sequence 1, 7, 9, 3, 1, ...
Divide the power (655 in this case) by four and see what the remainder is.  It is three.  So the 655th
power ends in the same digit that the 3rd power does, namely 3. 

But this is a bit confusing since we get 3 for the remainder and for the last digit.
Try raising to the 656 power.  Then dividing 656 by 4 we get a remainder of zero.  So the 656
power ends in 1 same as the 0th power.

For the 654th power dividing 654 by 4 has remainder 2.  So the 654th power ends in 9 the same
as the second power.

For the 653rd power dividing 653 by 4 has remainder 1 so the units digit of the 653rd power is 7 the
same as the 1st, 5th, 9th, ... powers.

Be very blesses! smile

#7 Re: This is Cool » 4-D cube! » 2013-02-10 09:40:55

Hi stefy! smile

Inside/outside whatever we wish to call it.  A Klein bottle only has one side, right? (similar to a Moebius strip).

Have a quite blessed day!

#8 Re: This is Cool » 4-D cube! » 2013-02-09 12:20:16

Maybe we could use a "Klein bottle" for the robot and put the load on the side. smile

#9 Re: This is Cool » 4-D cube! » 2013-02-07 14:54:24

Looks like a good prototype for a load carrying robot!  But where would one put the load?

#10 Re: Dark Discussions at Cafe Infinity » What does it take ... » 2013-02-07 14:50:35

Hi Al-Allo,

Webster's dictionary defines it as "a specialist or expert in mathematics."
There are several hundred areas of mathematics where one could spend the rest of their
life doing research in mathematics.  PhD's in math typically know MUCH LESS than
1% of the mathematics known worldwide today.  I'd say that anyone who is willing to call
themselves a mathematician probably is one.

Now if you are talking about a "working" statistician, they go through a battery of tests that
would make diamond wilt.  It's not an easy road, but usually quite lucrative.  They are
mathematicians, but usually only expert in a small area of mathematics.

And today some of the most active mathematicians use computers to bounce their ideas
off of often.  Wonderful tools for exploring patterns, testing formulas, etc.

And age doesn't necessarily have a lot to do with it.  Gauss could have probably been called
a mathematician before he was a teenager.  He taught himself to add columns of numbers
by the age two and a half.  I was still trying to get my walking down about that age.

Have a super day! smile

#11 Re: Help Me ! » complex numbers » 2013-02-05 11:14:21

Hi cooljackiec!

Three ways to get the result:
1)  Convert z = 1+ i√3 into polar (exponential) form  2e       so the 6th power is
                           i60° 6     6  i360°         i0°
                      (2e      )  = 2 e        = 64e    = 64(cos0°+isin0°) = 64( 1 + 0 ) = 64.
     re    =  (r,θ) are polar forms for complex numbers.
     The rcisθ = r(cosθ+isinθ) is really a "hybrid" in the sense that it is given in terms of
      r and θ, but when the trig functions are evaluated it gives the rectangular form x+iy.
      So rcisθ is handy for converting complex numbers from polar to rectangular form.

2)  First get z² = 2(-1+i√(3)).  Then square z² to get z^4 = -8(1+i√(3)).  Then multiply
     z^4 by z^2 to get the 64.

3)  Multiply (x - (1+i√3))*(x - (1-i√3)) = x^2-2x+4.  Then divide x^6 by x^2-2x+4 to
     get a remainder of 0*x+64.  Substitute z for x in 0*x+64 to get the result 64 using the
     Extended Remainder Theorem.

The Extended Remainder Theorem:  Given  P(x)/D(x) = Q(x) + R(x)/D(x).  For any x for which
                                                     D(x)=0, P(x)=R(x).

Simple proof:  Multiply by D(x) to obtain P(x) = Q(x)D(x) + R(x).  Then since we are using
only x's that make D(x)=0 this reduces to P(x)=R(x).

Dividing x^6 by x^2-2x+4 yields Q(x)=x^4+2x^3-8x-16 with R(z)=0*z+64  Since z=1+i√3 is
one of the two values that makes D(x)=0, P(z)=R(z)=0*z+64=64.  [Note (1-i√3)^6 = 64 also.]

I find 1) to be easiest and 2) to be the hardest.  The reason I find 3) easier than 2) is that
the division can be done quickly using the Generalization of Synthetic Division (GenSynD)
algorithm.  An article on how to do GenSynD can be found on the site that comes up when
googling "Math: The Original Four Letter Word."

Have a superduper day! smile

#12 Re: Help Me ! » squares » 2013-01-29 09:03:22

Considering just positive integers, there are 6 factorizations of 495 into two factors:
1*495, 3*165, 5*99, 9*55, 11*45 and 15*33.  Each of these corresponds to one of
the six (x,y) pairs bobbym listed in post #4.  For example: 1*495=248^2-247^2 and
24^2-9^2 = (24+9)(24-9) = 33*15.

In general for an odd composite number each of its unique factorizations (other than a perfect
square factorization) corresponds to a difference of squares.  For 9 = 1*9 we get 5^2-4^2
= (5+4)(5-4) = 9*1 but 3*3 has no difference of squares representation unless we allow zero:
3^2-0^2 = (3+0)(3-0) = 3*3.

But we were talking about POSITIVE integers.

If M is an odd composite number and M=n*m where n and m are different, we get


  as a difference of  squares factorization.

If I recall correctly this was involved in one of Fermat's methods of factoring odd composites.

Have a grrreeeeaaaaaaat day! smile

#13 Re: Help Me ! » Just to be sure.... (problem) » 2013-01-27 15:02:03

Hi Al-Allo!

Try  x for Patrick, 3x for Aaron  4(3x)=12x for Shad.  (Patrick is the slowest reader, Aaron next and
Shad the fastest.)  So if Patrick reads 4pages in 1hr, then ...

Can you get it from there? smile

#14 Re: This is Cool » tau=2*pi » 2013-01-22 10:20:26

Wasn't there a guy that bc (before computers) calculated pi out to 700 places by hand.  Then ac (after
computers) it was discovered that he made a mistake in the 200th digit and all the rest was trash
since each digit depended on the previous digit.  Probably spent the great bulk of his 15 year project
on the last 500 digits.  Mercy me!  But 200 digits of pi ain't bad, eh?  smile

#15 Re: Help Me ! » Binomial expression » 2013-01-18 14:13:50

How about something like:

A polynomial in x is the result of a finite number of additions and/or multiplications of -1 and x.
  Example:  3x² - 2x + 1 = (-1*-1 + -1*-1 + -1*-1)*x*x + (-1 + -1)*x + (-1*-1)
  Example:  0 = -1*-1 + -1
  Example:  -x^3 = -1*x*x*x

Resulting polynomials involve integral coefficients and non-negative integral exponents on x.

If one traces back each defined term until nothing but undefined terms are present, then terms
like "expression," "independent variable," "form," "variable," "denominator," and "function" would
create a succession of definitions terminating in a terribly convoluted definition of polynomial in
terms of just undefined terms (not unlike "well formed formulas" in logic).

Typically the best communication is obtained by tailoring the discussion to the intended audience.
Too much detail or not enough detail (especially in proofs) makes it difficult to follow.  And sometimes
when two (or more) people think they have finally arrived at a good understanding of what they have
been discussing, they later find out that they really had not grasped what the other was trying to get
across.   Each used perhaps the same words, but in the back of their mind had quite different ideas
as to what the words meant.  This can be especially troublesome when trying to "flesh out" a new
concept or new area of mathematics.

And therein lies much of the FUN in doing math.  Communicating with each other and trying to
figure out what in the world is going on!  Two minds (and the more the merrier) are "better than
one."  What one says usually sparks different thoughts in another's mind.  And back and forth
the exchange goes quite often culminating in some interesting stuff.  It's probably quite closely
akin to graffiti.  smile

#16 Re: Help Me ! » Binomial expression » 2013-01-17 16:32:46

Sometimes mathematicians play "fast and loose" with definitions.  We probably will not find total
agreement about what a "polynomial" is.  Some books will define "a polynomial in x" as opposed
to just a polynomial.  And what about a "polynomial in two variables?"  If y=2+x is x+y a binomial?

And what does "quotient" refer to?  If we consider 7÷2 as a fraction, is 7/2 a quotient?  Or is it
3.5?  Or is the quotient 3 (with remainder 1)?

And what is a fraction?  Which among the following are fractions?

     x, 2/x,  x/y,  2/3, 2÷3, x÷y, 2*(1/x), x/1, x/3, xy where y=1/z, z where z=1/y, etc.

     Let x=1/y,  y=1/3 and z=1/x.   Which are fractions?  x, y, 1/x, 1/y, y/3, xy, xz, yz, 1/yz, etc.

What is an arithmetic fraction, an algebraic fraction, etc.?

And in geometry is an equilateral triangle also isoseles?  There has been disagreement on whether
to make the set of equilateral triangles a subclass of isoseles triangles or a separate category of

The best we find at times is a "local" definition where an author defines a term precisely for his
following discussion.  And often other mathematicians may find fault with this.

Mathematics is a LANGUAGE and has not been (nor ever is likely to be) "nailed down" so as to be
without ambiguity or disagreement even for the most common and "simple" concepts.

Sometimes we just have to "roll with the punches" and at times ask for clarification of what the
author intends (as is often the case in this forum).

Often pushing for the exact meaning of all the terms we encounter may result in returning to the
undefined terms of a system.  But then the expression we obtain for a "higher level" concept's
definition may be so long and involved with the elementary undefined terms as to be basically

As an example consider the "Sheffer stroke" or "Dagger" in T/F two valued logic.  Each of these
can be used to define the usual AND, OR, if..then, if and only if, NOT, exclusive OR.  But the
expressions for some of these are quite long and complicated using just the one stroke or dagger.
It is interesting that one "operation" can be used to define all the usual stuff, but it is way too
unwieldly to want to use it.  As humans we work better with the AND, OR, etc.

As another example consider binary vs hexadecimal.  Working with hexadecimal is much easier
for us humans than working with binary, especially when numbers are fairly large.  The strings
of 1's and 0's just get too long and difficult to deal with.

1/2(grateDAY )! smile

#17 Re: Puzzles and Games » post prime numbers! » 2013-01-13 08:47:24

1234567891 and 123456791  easy to remember primes for testing some things in number theory.
Also  31, 331, 3331, 33331, 333331, 3333331, 33333331 are all prime.

#18 Re: This is Cool » The biggest number? » 2013-01-13 08:32:51

where 3!! means (3!)!, 3!!! means ((3!)!)! etc.
Note that (2!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!)^(2!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!) = 4 so
there is a bit of a leap from using 2 to using 3 in these expressions.

#20 Re: Help Me ! » Golden Key - Riemann Zeta Function » 2013-01-08 07:26:14

Hi therussequilibrium!  smile

You didn't waste my time.  I enjoyed looking up and reading about the Riemann Zeta function.
Quite interesting (and mostly over my head).  Gauss at age 15 came up with the n/ln(n) formula!
He was quite the genius, eh?

Have a blessed day!

#21 Re: Exercises » Index Laws Practice » 2013-01-08 07:16:45

Hi Toast!  smile  Nice set of problems!

Here is a variation of law 2 that is easy to use and always ends up with a non-negative exponent.
It takes care of all three cases:  m>n, m=n, m<n.  It is especially nice when the problem involves
negative exponents.

(p is the opposite of the smaller of m and n)  For example recalling that a^0 = 1 we get
( Whether this is true for a=0 has been thoroughly explored in other threads of this forum.)

       m     n          p       a^(m+p)/a^(n+p)

       2      5         -2     a^0 / a^3  = 1/(a^3)
       3     -2          2     a^5 / a^0  =  a^5   
      -5     -3          5     a^0 / a^2  = 1/(a^2)
       4     -7          7     a^11 / a^0 = a^11
       3      3         -3     a^0 / a^0  = 1
      -4     -4          4     a^0 / a^0  = 1

Of course the last two cases here would obviously be 1 from the start.

This law takes care of all cases for positive, negative or zero exponents m and n, leaving the
answer with a non-negative exponent for a in the numerator or denominator as most books
require for the answer.

If p is anything else but -min(m,n) the equality is still true, but it will not be in "simplified" form.

Play around with it a bit and I think you will find it is quite handy and easy to use. smile

#22 Re: Maths Is Fun - Suggestions and Comments » Binary Converter - Leading Zeros » 2013-01-07 06:20:07

Thanks!  Works great!  Handles enough digits to do my bank account in binary smile

How hard would it be to allow the user to input a base of their own choosing?

Have a great day!

#23 Re: Help Me ! » binomial thereom » 2013-01-06 18:45:00

Hey cooljackiec!

You might google "multinomial coefficients" and check out some of the sites to see how to easily
obtain the coefficient of the ab^2c^3 term in (a+2b+3c)^6.  I'm guessing that that is where bobbym
got his answer.

#24 Re: Help Me ! » Complex no. » 2013-01-06 18:35:53

(B)  When a=c we obtain ax²+bx+a= 0 which has roots


And multiplying these two roots together gives b²-(b²-4a²) = 4a²/4a² = 1.

So the two roots are reciprocals of each other,  BUT the Discriminant is not negative unless
b²<4a².  So |b|<|2a| is required for the roots to be complex (not real roots).

So the first condition would be true if it were |b|<|2a| instead of |b|<|a|. smile

From this problem we see that there are infinitely many complex number pairs that are complex
conjugates AND reciprocals at the same time.  But there is only ONE pair of complex numbers that
are both OPPOSITES AND RECIPROCALS at the same time.  And that would be ...

#25 Re: Help Me ! » Prime numbers. » 2013-01-06 10:16:30

Hi Al-Allo! smile

Given a positive integer M>1 let PM denote the set of all prime factors of M.  As in your example
if M=12 then PM={2,2,3} and if M=20 then PM={2,2,5}.  As a further example consider M=105 so
PM={3,5,7}.  The first two (for 12 and 20) are multisets because they have more than one copy of
a prime factor (two 2's in each).  {3,5,7} is a typeset because it has only one copy of each prime

The union of two sets whether multisets or typesets is obtained by choosing each type of
prime seen and scanning all the sets for the maximum number occurring in any
ONE of the sets.

Then using "u" for union we get
I:     If M=12, N=20 then PMuPN = {2,2,3}u{2,2,5} = {2,2,3,5}.
II:    If M=12, N=18, L=30 then PMuPNuPL = {2,2,3}u{2,3,3}u{2,3,5} = {2,2,3,3,5}.
III:   If M=56, N=9, L=250 then PMuMPuPL = {2,2,2,7}u{3,3}u{2,5,5,5} = {2,2,2,3,3,5,5,5,7}.

The lcm of any two, three or more numbers is the PRODUCT of all the elements of this UNION.
For example I:    lcm(12,20) = X{2,2,3,5} = 2x2x3x5 = 60
                  II:   lcm(12,18,30) = X{2,2,3,3,5} = 2x2x3x3x5 = 180
                 III:  lcm(56,9,250) = X{2,2,2,3,3,5,5,5,7} =  2x2x2x3x3x5x5x5x7 = 63000

These products produce the smallest positive integer that each of the given numbers divides into
with NO REMAINDER because one can choose from the union a set looking like each of the sets
obtained from the numbers M, N, etc.

Example:  In the third example above:   the union:  {2,2,2,3,3,5,5,5,7}
                                                                      M:  {2,2,2,             7}
                                                                      N:  {2      3,3           }
                                                                      L:   {2,          5,5,5   }

If we replace the commas in these sets with "x" then we get lcm/M =  2x2x2x3x3x5x5x5x7
                                                                                                     2x2x2        x         7
which is 3x3x5x5x5=9x125=1125 which is an integer. 
Doing likewise for N and L we get  lcm/N=3500 and lcm/L=252.

If we leave out just one factor from the union, then at least one of M, N and L will NOT
divide into the product of the union with zero remainder.

NOTE:  In a similar fashion the highest common factor hcf (or greatest common divisor gcd) of two
or more numbers is the PRODUCT of the INTERSECTION of the prime factor sets:

                   hcf(M,N,L) = X(PMnPNnPL)   ( and lcm(M,N,L) = X(PMuPNuPL) as above)

where "n" represents intersection which is based on choosing minimums (instead of maximums)
from the scanning of each of the sets.  Be careful though.  If at least one of the sets has NONE of
a prime that OCCURS IN ANOTHER  set, then the minimum is ZERO for that type of prime.

For the three given examples: I:  PMnPN = {2,2}
                                           II:  PMnPNnPL = {2,3}  no 5's since 5 is missing from another set.
                                          III:  PMnPNnPL  = { }    empty since each type of prime is missing
                                                                              from at least one of the sets.
( NOTE:  X{ } = 1 by definition since X{a,b,c} means 1xaxbxc.  Always start with a 1 factor.)

So hcf(12,20) = X{2,2} = 4
     hcf(12,18,30) = X{2,3} = 6
     hcf(56,9,250) = X{ } = 1. 

The hcf is the largest integer with divides into each of the given numbers with no remainder.

I hope this will help you, though it may be more information than you wanted.  smile

P.S. The edit was just to get some spelling corrected and to make the wording a little clearer.

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