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

You are not logged in.

- Topics: Active | Unanswered

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Problem # n+7

If x ∈ Real Numbers, What is the minimum value of x^x?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,664

Minimum?

0^0 = undefined

0.1^0.1 = 0.794

0.5^0.5 = 0.707

0.7^0.7 = 0.779

Somewhere between 0.1 and 0.7 ... I will leave the rest up to someone else

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The 'proper' way to solve this would be to find turning points by differentiating.

y=x^x.

dy/dx = x^x * ln x (I make no guarantee that that is right )

Turning points are horizontal, so we need to solve x^x * ln x = 0.

x^x * ln x = 0 when either x^x or ln x = 0.

Do ln x first, as it's easier.

ln x = 0

Take to the eth power: x=e^0=1. So 1 is a turning point, but as x^x=1, and there are values of x^x less than 1, this is clearly the wrong one.

So we are left with x^x=0. The solution to this is the minimum value of x^x. And, because of a loophole in your question, I can answer that the minimum value of x^x is 0. I don't know what value of x gives that, but you didn't ask for it .

Why did the vector cross the road?

It wanted to be normal.

Offline

**wcy****Member**- Registered: 2005-08-04
- Posts: 117

for quadratic equations, you can make use of the symmetry of the graphs to find the minimum point, which is halfway in between the two x-intercepts

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Am I allowed to slightly modify my question?

If yes, the question should read

'For what value of x is x^x the minimum, x>0?'

If, as I had said earlier, x∈ Real numbers,

my question would have no answer as

(-1 x 10^n) -1 would give a negative odd number for a

large value of n (n ∈ Natural Numbers). And this number raised to itself would give a negative number which would be much lesser than zero!

Therefore, the smallest value would be -∞

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If you are allowed negative values, then they would have to be odd integers or reciprocals of odd integers, otherwise x^x would be imaginary or positive. The lowest value of x^x would be when x= -1/(2n+1), such that 1/(2n+1) is the closest possible to the value of x that gives the lowest value of x^x. This is -1/3, that returns a value of -1.44224927...

Large negative numbers would return values that were extremely close to 0.

Back to your original question, I brute-forced it in Excel for a while, and it came up with x≈0.3678, meaning x^x≈0.692200633.

Why did the vector cross the road?

It wanted to be normal.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Oops, I made a mistake.

If, as I had said earlier, x∈ Real numbers,

my question would have no answer as

(-1 x 10^n) -1 would give a negative odd number for a

large value of n (n ∈ Natural Numbers). And this number raised to itself would give a negative number which would be much lesser than zero!

Therefore, the smallest value would be -∞

I forgot while posting that one that a^-n = 1/a^n

Therefore, the minmum value of x^x is not -∞

I am sorry I wasn't concentrating 100%

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Mathsy, you are right.

But, this can be done with Calculus.

Let x^x = y

x logx = log y

Differentiating both side,

xLogx(1) + x(1/x) = d(logy)

logx + 1 = d(logy)

When this is equal to zero,

Logx + 1 = 0

Log x = -1

x = e^-1 = 1/e

This is because logy is minimum when y is minmum

*Last edited by ganesh (2005-08-11 22:29:35)*

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**wcy****Member**- Registered: 2005-08-04
- Posts: 117

how to differentiate log x ?

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

d/dx (logx) = 1/x and ∫(1/x)dx = logx

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Problem # n+8

Can you find two numbers composed of only ones which give the same result by addition and multiplication?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I use 10 ones to make these: -- --

2+2=2x2. | |

Alright, I know that's not -- --

allowed, but I'm stumped | |

otherwise. -- --

Why did the vector cross the road?

It wanted to be normal.

Offline

**NIH****Member**- Registered: 2005-06-14
- Posts: 33

wcy wrote:

how to differentiate log x ?

Out of interest, you can derive it from the derivative of the exponential function:

y = log x <=> x = e^y.

So dx/dy = e^y = x.

Hence dy/dx = 1 / dx/xy = 1/x.

2 + 2 = 5, for large values of 2.

Offline

**NIH****Member**- Registered: 2005-06-14
- Posts: 33

ganesh wrote:

Problem # n+8

Can you find two numbers composed of only ones which give the same result by addition and multiplication?

How about

and . Works in any number base!2 + 2 = 5, for large values of 2.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

You are correct, NIH !

Problem # n+8

Find the next term of the series

3, 14, 39, 84, 155, _____

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,664

Well, at least this one goes UP (rather than those up-down-left-right-next-time-add-subtract-multiply-divide types!)

3+11=14, 14+25=39, 39+45=84, 84+71=155 ... hmm ... * walks off mystified *

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

3 14 39 84 155

11 25 45 71

14 20 26

6 6 ---> 6/3!=1 ---> (1)n³+...

3-1³ 14-2³ 39-3³ 84-4³ 155-5³

2 6 12 20 30

4 6 8 10

2 2 2 ---> 2/2!=1 ---> n³+(1)n²+...

2-1² 6-2² 12-3² 20-4² 30-5²

1 2 3 4 5

1 1 1 1 ---> 1/1!=1 ---> n³+n²+n+...

1-1 2-2 3-3 4-4 5-5

0 0 0 0 0 ---> 0/0!=0 ---> n³+n²+n+0n° = n³+n²+n

So the next number would be 6³+6²+6=258

Yay for brute force!

Why did the vector cross the road?

It wanted to be normal.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

For the one where you have two numbers that add and multiply to give the same thing, I got bored so worked out that if you are given a value of z, then you can work out values of x and y such that x+y=xy=z like so:

x = √(z(z/4-1))+z/2

y = z/(√(z(z/4-1))+z/2)

This will work for any value of z, but if 0<z<4 then x and y will be complex.

Why did the vector cross the road?

It wanted to be normal.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

mathsyperson wrote:

So the next number would be 6³+6²+6=258

Mathsy, thats very good!

Problem #n+9

From a well shuffled pack of 52 cards, four cards are drawn one after the other. What's the probability that the four of them would belong to different suits?

*Last edited by ganesh (2005-08-15 16:12:15)*

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Why did the vector cross the road?

It wanted to be normal.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Correct, Mathsy!

Problem # n+10

Mathsyperson, Kylekatarn and wcy are a team for solving problems. The possibility of one solving a problem is 45%, the other is 50% and the third is 55%. What is the probability that a problem would be solved?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

This is all in my head, so there might be a mistake somewhere...

Person 1 has a 45% chance of doing it, meaning he has a 55% chance of *not* doing it, which is 11/20.

Similarly, person 2's chance is 1/2 and person 3's chance is 9/20.

So the chance that the puzzle won't get solved is 11/20x1/2x9/20=99/800, meaning the chance that it will is 701/800.

Convert back into a percentage: (701/8)% =87.625%

It's probably less than that though, because we would all solve easy problems and all not solve difficult problems. This assumes that we solve problems independant of each other.

Why did the vector cross the road?

It wanted to be normal.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Problem # n+11

A circle of diameter 'a' is cut from a square metal sheet of side 'a'. From the four corner cuttings, four circles are cut. Whats the maximum area of the four circles put together?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,374

Problem #k

What is the total number of squares, of all sizes, in a Chess board?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

here it is... just the answer

*Last edited by kylekatarn (2005-08-22 16:41:39)*

Offline