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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,879

Can anyone tell me what's special about this number?

It is indeed a very special number...I shall tell you later.....

The number is

approximately

1.444667861............ and it goes on and on after the decimal.......

Ganesh

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,552

Cool, that sounds like fun, ganesh.

Something to do with pi, or the golden mean?

I will have to put my thinking cap on.

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

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

What's the golden mean?

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

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

The golden mean is a ratio that the 'perfect rectangle' has sides of. This is the rectangle that most people agree looks nicest.

Also, if you stick a square on the long side of the perfect rectangle, then it retains its proportions, just switches from portrait to landscape or vice versa. If the perfect rectangle has a width of 1, then its length would be the golden mean.

The golden mean can be approximated by dividing one number in the fibonacci sequence by the one before it, with the estimate growing more accurate as you get to higher fibonacci numbers.

The golden mean is represented by the greek letter Phi.

Finally, the golden mean is worked out by (√5+1)/2.

*Last edited by mathsyperson (2005-06-29 04:13:24)*

Why did the vector cross the road?

It wanted to be normal.

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

Interesting. You learn something new every day!

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,552

I try to use the golden mean when I design things - it makes it look "natural".

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,879

e, the natural logarithm base is approximately 2.7182818284......

1.444667861 (approximately) is the value of eth root of e.

This is a very special number for two reasons.

(1) The maximum value of xth root of x for any value of x is obtained when x=e. This value, i.e. xth root of x is approximately 1.444667861.

(2) We all know 1 raised to itself ad infinitum is 1.

2 raised to itself ad infinitum is divergent, that is infinity.

But, there's a value between 1 and 2 for which the number raised to itself ad infinitum is a finite value.

The maximum such value is 1.444667861 (approximately) such that the number raised to itself ad infinitum remains finite. And this finite value=e.

Please feel free to mail me if you have any doubts.

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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You know, I think I had this number on my calculator years ago.

I was playing with the "n"th root of "n" and noticed:

The square root of 2 is 1.41421

The cube root of 3 is 1.44224

The fourth root of 4 is 1.41421

The fifth root of 5 is 1.37972

So, the cube root of 3 is the biggest, and they just get smaller after

So I tried numbers between 2 and 3 and found that the largest I could get was:

The 2.718th root of 2.718 is 1.44467

But that next bit has me confused:

1^1^1^1... = 1 YES

2^2^2^2... = ∞ YES

But even

1.001^1.001^1.001^1.001... = ∞ ???

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,879

No, gentleman.....

1.01^1.01^1.01^1.01.......... is not infinity.

For any value less than approximately 1.444667861, the resultant is NOT infinity...

I have got a proof for this....If you want, I shall e-mail it to you....

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,552

I beg your pardon, you are quite right, I had my formulas the wrong way around.

In fact, as an example:

Start 1.1 ^ 1.1 = 1.110534241

Then 1.1 ^ 1.110534241 = 1.1116498

Then 1.1 ^ 1.1116498 = 1.111768002

Then 1.1 ^1.111768002 = 1.111780527

After a few more runs the number reaches 1.111782011 with no changes (not at that level of accuracy)

I played a little in Excel (when I should have been doing stuff on the website!) and found:

When I tried 1.4, after 45 lines it settles to 1.886663306

But when I tried 1.5 it hit 1,499,263,005,586,480 after just 12 lines. The next line was just "#NUM!"

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,879

Yes, my friend....

I discovered this more than 15 years back,

but very recently, I wrote a proof for this

The proof is on my website,

in the bottom of 'the number 'e''

page....

that is another elegant property of the number e....

i shall tell you one more...

into how many pieces has a number to be divided in order to get the maximum

product?

when the parts are closes to 'e' !

for example, when 10 is divided into two equal parts, the product is 25,

when diivided into three equal parts, the products is roughly 37.03...

but when divided into four parts (each part is 2.5, closest to 'e'),

the product is the maximum,,that is, 39.0625

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,552

I played a bit more (what fun a spreadsheet can be) and found:

1.45 blew up after 43 lines

1.44 stabilized at 2.393811748

1.445 blew up

1.444667861 reached 2.718197788 on line 65,536

1.444667865 blew up after 51,277 lines

It's very touchy!

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**angel****Guest**

hey i'm angel aka aisha if still on e me bk k

**candy****Guest**

hi

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,552

So, let's choose, I don't know, 15:

15 in 2 parts is 7.5, and 7.5*7.5=56.25

15 in 3 parts is 5, and 5*5*5=125 (its higher)

15 in 4 parts is 3.75, and 3.75*3.75*3.75*3.75 = 197.7539063 (its higher)

15 in 5 parts is 3, and 3*3*3*3*3 = 243 (its higher)

15 in 6 parts is 2.5 ... = 244.140625 (its higher)

15 in 7 parts is 2.142857143 ... = 207.468675 (goes down again)

Yep, that is interesting!

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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Oh, hi angel and candy!

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**mathsyperson****Moderator**- Registered: 2005-06-22
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e^((pi*√67)/3)=an approximate value for the number of feet in a mile.

Yay! Pointlessness!

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,879

Consider a room, 10m x 10m x 10m

A spider starting from one vertex, has to reach

the diagonally opposite vertex....

What's the minimum distance the spider would have to crawl?

Character is who you are when no one is looking.

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

Say, it starts at the floor. It'd have to cross the diagonal of the floor, so to find the length of the diagonal I need to use pythagoras' theorm (a²+b²=h²).

a² is ten squared, and b² is ten squared. This makes 20². h² = 20², so find the square root of both sides and we get h=4.472135954........

So that'd be it, I think?

That was my best go.

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

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**justlookingforthemoment****Moderator**- Registered: 2005-05-26
- Posts: 2,161

Does 10² + 10² = 20² though?

10² + 10² = 20² is the same as 100 + 100 = 400 which doesn't work out

10 square metres + 10 square metres = 20 square metres but

10² + 10² = (14.142)²

10² + 10² = 200 so

10² + 10² = (√200)²

10² + 10² = (14.142)² to three decimal places

So the length of the diagonal is actually 14.142 metres...

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**mathsyperson****Moderator**- Registered: 2005-06-22
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I think Rora must have done √20 by accident...

If I'm interpreting this right, the spider needs to walk 10m forward, right and up to reach the diagonally opposite corner. If this is the case, then it would take 30m if it stuck to the edges, but that wouldn't be an optimal solution.

Breaking the cube down into its net, we can see that the spider essentially needs to get to the opposite corner of a rectangle 20m by 10m. Using Pythagoras it needs to travel √(20²+10²)=√500=22.36m to the nearest cm. Clever spider!

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,879

The question was the shortest distance a spider crawls from one vertex to a diagonally opposite vertex;

the solution is reached thus;

the spide crawls from one vertex to the mid point of the two dimensional opposite vertex;

therefore the distance travelled would be square root of 10^2 + 5^2

Then, the spider crawls a similar distance to reach the diagonally opposite vertex;

Hence, the distance travelled would be 2 multiplied by square root of 10^2 + 5^2

That is, 2 multiplied by square root of 125, i.e. 2 multiplied by 5 multiplied by square root of 5

around 22.3 meteres.....

This is the shortest distance

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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That is cool. Two different approaches, and same answer.

I think it is because if the cube were flattened out (as mathsyperson said "Breaking the cube down into its net") it would look like:

Π

ΠΠΠΠ

Π

And the spider actually has to go from the corner of one square to the far corner of the next square, so is crossing like this:

.-----.

| / |

| / |

|-----|

| / |

| / |

.-----.

Which is either 2 movements of (10,5) or one lot of (20,10)

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,879

Two dices are marked with numbers from zero to nine on each face.

The two dices are meant to display all the dates in a month from 01 to 31.

Remember, single digit dates would have to be shown as 01, 02, 03 etc.

That is, one dice should show zero and the other 1 or 2 or 3 and so on.

The two dices can be interchanged i.e. placed in any order.

What would be the numbers written on the six faces of the two dices???

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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ganesh wrote:

Two dices are marked with numbers from zero to nine on each face....What would be the numbers written on the six faces of the two dices???

Umm.. confused.

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