Mathematics desk
< October 10	<< Sep \| October \| Nov >>	October 12 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 11[edit]

Does there exist a function that...?[edit]

It maps R+ to R+, and has f(f(x))<f'(x) for every x>0. I guess it is not possible, but what is the trick?--124.172.170.233 (talk) 07:20, 11 October 2013 (UTC)[reply]

Am I missing something? What's wrong with f(x)=2x? AlexTiefling (talk) 09:28, 11 October 2013 (UTC)[reply]

There's a ' on the right f.Phoenixia1177 (talk) 10:04, 11 October 2013 (UTC)[reply]

f(x)=-{1 \over -{1 \over x}}\,\!

perhaps? -- Toshio Yamaguchi 10:23, 11 October 2013 (UTC)[reply]

Isn't that just x?Phoenixia1177 (talk) 10:27, 11 October 2013 (UTC)[reply]

How is the derivative of that x? The inner derivative of the denominator is

1 \over x^{2}

and then doing the outer derivative you get

1 \over {x^{4}}

, unless am I missing something here? -- Toshio Yamaguchi 10:44, 11 October 2013 (UTC)[reply]

Perhaps I should have added some parentheses. -- Toshio Yamaguchi 10:46, 11 October 2013 (UTC)[reply]

Like this:

f(x)=-{1 \over (-{1 \over x})}\,\!

-- Toshio Yamaguchi 10:57, 11 October 2013 (UTC)[reply]

Perhaps I should try to graph that as its becoming a bit confusing. -- Toshio Yamaguchi 11:04, 11 October 2013 (UTC)[reply]

f(x) simplifies to f(x) = x. You've made a mistake calculating the derivative:

f'(x)=-{1 \over (-{1 \over x})^{2}}*{1 \over x^{2}}=x^{2}*{1 \over x^{2}}=1,

as expected.--80.109.106.49 (talk) 11:07, 11 October 2013 (UTC)[reply]

I don't what the proper notation for the function I have in mind would be. Consider the following. Take say

x=2

. Calculate

y=-{1 \over x}

. Then calculate

z=-{1 \over y}

. I am pretty sure z is not 2. -- Toshio Yamaguchi 11:17, 11 October 2013 (UTC)[reply]

It could be expressed as a function composition:

f(x)=-{1 \over x}\,\!

g(x){=}f\circ f

-- Toshio Yamaguchi 11:30, 11 October 2013 (UTC)[reply]

Your notation is correct for what you're trying to express. However, it does simplify to x. The negatives cancel out, and dividing by a fraction is the same as multiplying but the reciprocal. So it becomes 1*x = x.--80.109.106.49 (talk) 11:34, 11 October 2013 (UTC)[reply]

Yes, I can see it now. Sorry, don't know what I thought when writing that, I was trying to follow some weird thought that actually led nowhere. -- Toshio Yamaguchi 11:39, 11 October 2013 (UTC)[reply]

By an argument using Rolle's theorem, f must be monotonic. Clearly it cannot be decreasing, so it must be strictly increasing. If it were bounded, the derivative would converge to 0, so it must be unbounded. Thus the derivative must be unbounded. So eventually, f(x) - x > 1. By the mean value theorem, ff(x) = f(x) + f'(c)( f(x) - x), for some c greater than x. So ff(x) > f(x) + f'(x)( f(x) - x) > f'(x). Contradiction.--80.109.106.49 (talk) 10:28, 11 October 2013 (UTC)[reply]

I guess I'm assuming f'(x) is monotonic at the end. We can fix that. Eventually, f(x) - x > 2. Since

\liminf f'(x)=\infty

, choose x such that for all c > x, f'(c) > f'(x)/2. Note that this does not require f'(x) be continuous. Now we have ff(x) = f(x) + f'(c)( f(x) - x) > f'(x)/2 *2 = f'(x).--80.109.106.49 (talk) 10:43, 11 October 2013 (UTC)[reply]

I disagree that, just because in the bounded case f' goes to zero, f must be unbounded; even though f(x) keeps getting larger and hence greater than f', it could be that f(f(x)) keeps getting smaller and always stays smaller than f'.

How about this example?: Let f(x) = k₁x for 0 < x < 1/k₁, f(x) = k₂x for 1/k₁ ≤ x < 1/k₂ (with k₂ < k₁), .... Here {k_i} is a decreasing sequence that goes asymptotically to zero. In the range 1/J < x < 1/K we have f(x) = Kx and f(f(x)) = K²x and f'(x) = K, so f'(x) > f(f(x)). This function has an infinite number of kink points in which the left derivative f' is not defined, but I wonder if this can be fixed by smoothing out the kink somehow. Duoduoduo (talk) 16:13, 11 October 2013 (UTC)[reply]

every base is base 10 -- isn't calling it that kind of dumb?[edit]

in the sense that "every base is base 10", isn't calling it that kind of dumb? Wouldn't it be much more descriptive to call base 10 'base 9', call octal 'base 7', call binary 'base 1' and call hexedecimal 'base f' (base 15)?

I mean what kind of sense does it make to use two positions to name our base - a 1 and a 0? 212.96.61.236 (talk) 22:30, 11 October 2013 (UTC)[reply]

Nothing 'dumb' about it at all, given that (unless indicated to the contrary), written numbers are normally given in base ten: so '10' is clearly intended to mean 'ten', rather than 'the digits 1 and 0 in some unknown base'. AndyTheGrump (talk) 22:42, 11 October 2013 (UTC)[reply]

isn't calling it that kind of dumb? — Perhaps it is... but -after all- we're not geniuses, we're just mathematicians... :-) — 79.113.213.168 (talk) 00:47, 12 October 2013 (UTC)[reply]

How is "every base base 10"? Bubba73 ^{You talkin' to me?} 01:09, 12 October 2013 (UTC)[reply]

The OP's point is that if you write the base in the number system itself then it always becomes "10" = 1×base + 0. But base ten is implied so when we write "base 10" we mean ten and when we write "base 16" we mean sixteen. PrimeHunter (talk) 01:42, 12 October 2013 (UTC)[reply]

OK, now I understand. Bubba73 ^{You talkin' to me?} 01:56, 12 October 2013 (UTC)[reply]

Our system is called base 10 rather than base 9 because when we write multi-digit numbers, the position of the digits indicates multiples of powers of 10. Like 4562 means

4\times 10^{3}+5\times 10^{2}+6\times 10^{1}+2\times 10^{0}

. Each term is a constant times an exponential function with base 10. Base 9 would be representing numbers in terms of powers of 9. Staecker (talk) 01:30, 12 October 2013 (UTC)[reply]

Also, we have 10 digits, not 9. As for why other bases are named in base 10 (why we would say base 12, 12 in decimal) is because we use decimal usually. Notation is all that number bases are, the point of notation is communication- it is not foundational, so it's not circular to denote the base in decimal, it's convenient.Phoenixia1177 (talk) 05:37, 12 October 2013 (UTC)[reply]

If we have ten digits, why do you have to spell 'ten' 1-0, instead of being able to write the tenth digit? As far as I can tell we have 9 and a positional null. 212.96.61.236 (talk) 13:57, 12 October 2013 (UTC)[reply]

0 counts as a digit. so in base ten we have ten digits to hold place value, 0-9. compare to binary, 2 digits: 0 and 1. likewise, the number two in binary is '10' .. so perhaps you're somehow conflating the issue of digits with counting, since we normally learn to start at 1, only later learning that 0 is actually the first digit if we're ordering them 0-9. 76.17.125.137 (talk) 05:51, 17 October 2013 (UTC)[reply]

Sure, you can make up your own definitions if you like. Doesn't change the fact that you're wrong, of course...Sebastian Garth (talk) 16:29, 12 October 2013 (UTC)[reply]

Thank you! 212.96.61.236 (talk) 00:30, 13 October 2013 (UTC)[reply]

We call it this way because we use base 10. If we were a base-12 civilization we'd probably be referring to decimal as "base ᘔ". A base-6 civilization would most likely call decimal "base 14". And so on. We would all refer to our default bases (decimal, dozenal and senary here respectively) as "base 10", understanding the base "10" is written in to be implied. The base you are using by default is implied. Double sharp (talk) 03:31, 15 October 2013 (UTC)[reply]

Write 'base ten' or 'base twelve' or 'base six' or 'base sixty' or whatever. As the base number is always written '10', writing 'base-10' is useless for communication purposes. Kind of dumb? Yes! Bo Jacoby (talk) 19:03, 15 October 2013 (UTC).[reply]

No. There is a convention that when you write "base X", X is always expressed in base ten. PrimeHunter (talk) 12:09, 16 October 2013 (UTC)[reply]

Yes, and he who does not know this dumb convention is lost. Bo Jacoby (talk) 22:40, 17 October 2013 (UTC).[reply]