Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2015-11-04 17:22:44

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Random questions I need help with

1.   Identify a function f(x) so that f(x). f(y) = f(x + y).

2.   Prove that a . a^1/2 .a^1/4 .a^1/8. ... ... [infinity] = a^2

Got to go. I'll bring the rest later.
Thanks


Only a friend tells you your face is dirty.

Offline

#2 2015-11-04 20:03:55

Bob
Administrator
Registered: 2010-06-20
Posts: 10,623

Re: Random questions I need help with

hi math9maniac

Q1.  logs have this property.  http://www.mathsisfun.com/algebra/expon … ithms.html

Q2.  To multiply numbers written as powers you add the powers:

So the whole expression 'hinges' on being able to write a single expression for

See here http://www.mathsisfun.com/algebra/seque … etric.html

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#3 2015-11-05 02:28:48

Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,996
Website

Re: Random questions I need help with

Logs don't have that property, exponentials do


'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

Offline

#4 2015-11-05 02:59:58

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,436
Website

Re: Random questions I need help with

Or as an even simpler alternative to Q1, the zero function trivially satisfies this property (i.e. f(x) = 0 for all x).

Offline

#5 2015-11-05 09:37:32

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

Ok. Thanks so much guys for these. Here are the others.

3.   Find two numbers such that their arithmetic mean is 15 and Geometric mean is 9 without using the identity (a + b)^2 = (a - b)^2 + 4ab.

4.   If tanA = 1/3 and tanB = 1/2, prove that sin 2(A + B) = 1.

5.   If A = {(x, y): y = a ^x, x € R } and B = {(x, y) : y = a^-x, x € R }, then what is A n B?
PS. : I have used € to represent the mathematical symbol " belongs to " or " is a member of "; n in place of "the intersection " symbol. I don't have these on my keyboard. Thanks for understanding.


Only a friend tells you your face is dirty.

Offline

#6 2015-11-05 20:03:44

Bob
Administrator
Registered: 2010-06-20
Posts: 10,623

Re: Random questions I need help with

Agnishom wrote:

Logs don't have that property, exponentials do

Whoops.  embarrassed smiley.  Thanks Agnishom.  Glad to see at least you are awake.

Q3.  So put (a+b)/2 = 15 and ab= +/- 3 and solve.

Q4. There's a formula that will give you a start:

Then you can convert to a sine  by setting 'opp' = numerator and 'adj' = denominator, and calculating 'hyp' using Pythagoras.

Finally:

Q5.  Compare the graphs by using http://www.mathsisfun.com/data/function … ?func1=2^x

Experiment with different numbers than a=2.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#7 2015-11-05 23:02:40

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,436
Website

Re: Random questions I need help with

bob bundy wrote:

Q3.  So put (a+b)/2 = 15 and ab= +/- 3 and solve.

I think that should be ab = 81?

Last edited by zetafunc (2015-11-05 23:02:57)

Offline

#8 2015-11-06 01:16:01

Bob
Administrator
Registered: 2010-06-20
Posts: 10,623

Re: Random questions I need help with

Thanks zetafunc.  I seemed to have caught inverse function disease.  I wonder if there is a cure.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#9 2015-11-06 05:23:06

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

Thanks so much.

Sorry to say that I still don't understand #1.


Only a friend tells you your face is dirty.

Offline

#10 2015-11-06 07:30:49

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,436
Website

Re: Random questions I need help with

Consider the function
. Does this satisfy
?

Other questions to think about:

How about
,
, or
for some
? Can you think of any similar functions that also satisfy this property?

If you wanted to be a bit cheap, you could also say that
is a valid solution.

Offline

#11 2015-11-07 00:08:50

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

Ok zetafunc. Thanks for reply. Please how does f(x) = 0 satisfy the condition. Or in other words, how is it a valid solution?


Only a friend tells you your face is dirty.

Offline

#12 2015-11-07 00:11:39

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

6.   Find the domain and range of the function
  f(x) =  (x + a)/(a + 1 - x)

where a is a positive integer.


Thanks


Only a friend tells you your face is dirty.

Offline

#13 2015-11-07 02:39:18

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,436
Website

Re: Random questions I need help with

math9maniac wrote:

Ok zetafunc. Thanks for reply. Please how does f(x) = 0 satisfy the condition. Or in other words, how is it a valid solution?

If f(x) = 0 for every value of x, then that means that whatever I put inside the brackets gives me zero, e.g. it's also true that f(2) = 0, f(0) = 0, f(y) = 0, f(x + y) = 0. It is then trivially true that f(x)f(y) = f(x + y), since both sides of that equation are 0. However, this is usually called a "trivial solution" -- so some question-writers like to ask you to find the "non-trivial solutions".

Offline

#14 2015-11-07 07:18:43

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

Ok. So what then are the non-trivial solutions?

Thanks


Only a friend tells you your face is dirty.

Offline

#15 2015-11-07 07:21:14

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

Can we also relate the function

f:x==> a^x
f(x) = a^x


Only a friend tells you your face is dirty.

Offline

#16 2015-11-07 08:43:11

Bob
Administrator
Registered: 2010-06-20
Posts: 10,623

Re: Random questions I need help with

hi math9maniac

My apologies for earlier slipups.  As a penance I set myself a task that I have now completed.

If you know the properties of the power function, then all you have to do is to show it fits this equation.  But what if you have not met it before or want to get it as an answer without knowing in advance that it works.  To find a way to derive the power function as the solution was my penance.  Here is my answer:

Set y = 0 then

Now set y = -x

now set y = x

and setting y = 2x

continuing this we see that

Now put x = 1

So the power function is a solution.  n is a positive integer  If I could show that n may also be a fraction or real number then I'd have shown that only the power function works.  This is a work in progress.  smile

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#17 2015-11-07 15:12:52

Bob
Administrator
Registered: 2010-06-20
Posts: 10,623

Re: Random questions I need help with

put x = 1/2

f(1/2).f(1/2) = f(1/2 + 1/2) = f(1) = a

so f(1/2) = square root(a)

Similarly f(1/3) = cube root(a) and so on.

Thus f(n/m) is a^(n/m)

All real numbers lie between two rationals

eg. 31/10 < pi < 32/10

Further these bounds can be narrowed

eg.  314/100 < pi < 315/100

So f(r) can be determined to any degree of accuracy where r is real by determining a pair of rationals q1 and q2 either side of r and computing f(q1) and f(q2).  Improved accuracy can always be achieved by narrowing the bounds q1 and q2.

Thus f(x) is determined as a power function for all x.  The case discussed in an earlier post is just when a = 0.

Thus the general solution if f(x) = a^x for some constant a.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#18 2015-11-08 07:35:56

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

Thanks so much Bob. I appreciate this.


Only a friend tells you your face is dirty.

Offline

#19 2015-11-08 10:54:40

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

Q7.   Three cars are there in a race. Car A is 3 times as likely to win as car B. Car B is twice as likely to win as car C. What is the probability of each car winning?


Only a friend tells you your face is dirty.

Offline

#20 2015-11-08 20:06:05

Bob
Administrator
Registered: 2010-06-20
Posts: 10,623

Re: Random questions I need help with

hi math9maniac

You could call the probability of C winning 'p'.  Then write the prob. for B and for A in terms of p.

If you then assume there are no other outcomes (like a draw), the three probabilities must add up to 1.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#21 2015-11-09 07:27:35

math9maniac
Member
From: Tema
Registered: 2015-03-30
Posts: 443

Re: Random questions I need help with

Wow, that seems quite easy but far from what I thought initially. Thanks a bunch.

I'll see you later for a piece of advice. Gracias


Only a friend tells you your face is dirty.

Offline

Board footer

Powered by FluxBB