# Math Is Fun Forum

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

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## #1 Re: Puzzles and Games » Connotations. » 2009-03-21 14:10:20

JaneFairfax wrote:

1. andesite

2. subgroups

3. benthic

1. pumice

2. normal

3. abyss

## #2 Re: Dark Discussions at Cafe Infinity » Tsk tsk for British parents » 2009-03-21 12:56:58

It's not how long they're online for, but what they do when they're online that's important. Are they watching stuff they're not supposed to watch, for example?

## #3 Re: Puzzles and Games » x » 2009-03-21 12:34:19

I get it now. Thanks, TheDude.

Incidentally, are you all (apart from Ricky) only assuming that x is positive? Can't x be negative as well? After all, if you multiply a number four times, you do get a positive number. Ricky's method may allow for negative numbers but I don't really know what you can do with the method.

From all the numbers I've tried, I'm inclined to believe that there is no negative solution. I'm just wondering if you can definitely prove that no negative solutions exist.

1 - basalt

2 - cosets

3 - suboceanic

1 - granite

2 - Laplace

3 - submarines

1 - rocks

2 - measure

3 - acoustics

## #8 Re: Puzzles and Games » x » 2009-03-15 09:02:26

TheDude wrote:

We can also see that 3 < x = a/b < 4.

How?

## #9 Re: Help Me ! » Real Analysis Help » 2009-03-07 07:39:57

We can prove that

Suppse it is not. Then there exist

in I such that
.

Now

and
.

Also

is either
or
.

Thus we have either

or
. This contradicts the fact that both f and g are increasing on I.

Therefore

is increasing on I.

You can prove that

is also increasing on the interval I by the same method.

Oops, sorry.

## #11 Re: Puzzles and Games » Complete the series.... » 2009-02-15 03:45:10

Ooh! It looks all so simple once you know the answer that it makes you feel kinda stupid wondering why you never thought of it before.

Thanks.

Yoda rocks!

## #16 Re: Puzzles and Games » Complete the series.... » 2009-02-15 02:37:03

JaneFairfax, how did you get the answer to (xxvii)?

## #17 Re: Help Me ! » Real Analysis Help please » 2009-02-15 02:24:58

sumpm1 wrote:

1. A function :R->R has a PROPER RELATIVE MAXIMUM VALUE at c if there exists d>0 such that f(x)<f(c) for all x that satisfy 0<|x-c|<d. Prove that the set of points at which  has a proper relative maximum value is countable.

This is one of those questions which are intuitively obvious but present such a challenge to prove. I can only suggest using the metric-space properties of the real numbers. Try proving that the set of points at which  has a proper relative maximum value is nowhere dense in R. Countability should follow from the fact that all nowhere-dense subsets of R are countable. (Are they? I think so anyway. )

## #18 Re: Help Me ! » Real Analysis Help » 2009-02-15 02:09:19

1(b)

Hence the LHS is minimized

Use the same trick for 1(c), rewriting

## #19 Re: Help Me ! » Random Walks Question? » 2009-01-25 08:56:58

In theory, the betting game could go on forever, but in practice no game can go on forever. You can't possibly spend an eternity in the casino. Besides, you'll miss your bus.

A more practical phrasing of the question might be, say, what is the probability that you have £50 within 50 bets? Assuming you only have time for 50 bets before your bus arrives.

## #21 Re: Jai Ganesh's Puzzles » Brain Teasers » 2009-01-25 07:39:27

JaneFairfax wrote:

## #23 Re: Jokes » Jokes galore...... » 2009-01-22 13:01:05

JaneFairfax wrote:

Q: While the musician was at the supermarket, what were her children doing?
A: Playing Haydn seek.

I thought they were playing fluteball.

## #24 Re: Help Me ! » (1+i)^i » 2009-01-22 12:37:59

for all integers n. The principal value is when n = 0.

## #25 Re: Help Me ! » Unbounded function » 2009-01-22 11:31:37

mathsyperson wrote:

I *think* this result can be generalised so that instead of f' and f'', you have f[sup](m)[/sup] and f[sup](n)[/sup], where m and n are respectively odd and even.
You'd also need f to be differentiable max(m,n) times rather than just twice.

Haven't worked out a solid proof yet though.

Intuitively, f', f'' > 0 means the function is convex and strictly increasing; such a curve should intuitively go to infinity as x goes to infinity. I just wanted a concrete analytical proof of what was intuitive to me.

In fact f' and f'' don't have to be positive on the whole real line, they just have to be positive on the interval [a,∞) for some real number a. This can even be easily shown - just replace 0 by a in the your proof above.