## A Few Tricky Questions(now for all my math questions)

For the discussion of math. Duh.

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### A Few Tricky Questions(now for all my math questions)

I was just wondering if anyone had the answers to a few Mathematical Problems I was considering. This mainly my own musings, and I'm limited in that I'm only knowledge up to Single Variable Calculus, and after that I know a little Differential Equations and Linear Algebra. Basically I'm looking for some continuous function that passes through the following sets of numbers, preferably something Analytic:

0, sin(1), sin(sin(2), sin(sin(sin(3), sin(sin(sin(sin(4),...

Also, would happened to know if this function is continuous:

x= sum from i=-infinity to infinity aix10i (sorry I had to write that out, I'm not sure how to write Sigma Notation online)

f(x)=sum from i=-infinity to infinity rd(ai+ai-1)x10i, with rd(ai+ai-1) being equivalent to adding all the digits of (ai+ai-1)(for example, rd(10)=1, rd(11)=2, rd(1.32)=6, so on and so forth).

I'd appreciate anyone's input on the matter, thank you for your time.
___________________________________________________________

Previous problems already pretty well answered, but I've decided to save space and use this as my math questions thread.
Last edited by HeWhoCouldn'tThinkOfAUsername on Wed Oct 05, 2011 1:59 am UTC, edited 2 times in total.

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### Re: A Few Tricky Questions

sinx(x) {using normal function powers, i.e. sin2(x) = sin(sin(x))}
gmalivuk wrote:
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### Re: A Few Tricky Questions

I'm confused, what is the function for say a non-integer x?

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### Re: A Few Tricky Questions

HeWhoCouldn'tThinkOfAUsername wrote:I was just wondering if anyone had the answers to a few Mathematical Problems I was considering. This mainly my own musings, and I'm limited in that I'm only knowledge up to Single Variable Calculus, and after that I know a little Differential Equations and Linear Algebra. Basically I'm looking for some continuous function that passes through the following sets of numbers, preferably something Analytic:

0, sin(1), sin(sin(2), sin(sin(sin(3), sin(sin(sin(sin(4),...

The constant function f(x)=0 passes through all those numbers on the x-axis.

Or did you mean a function that passes through the points (0, 0), (1, sin(1)), (2, sin(sin(2))), etc.? If this is what you meant, then there definitely is an analytic function passing through those points. Given any countable subset of ℂ with no limit points, and any arbitrary complex values for each point in the set, there is always an analytic function through those points.

HeWhoCouldn'tThinkOfAUsername wrote:Also, would happened to know if this function is continuous:

x= sum from i=-infinity to infinity aix10i (sorry I had to write that out, I'm not sure how to write Sigma Notation online)

f(x)=sum from i=-infinity to infinity rd(ai+ai-1)x10i, with rd(ai+ai-1) being equivalent to adding all the digits of (ai+ai-1)(for example, rd(10)=1, rd(11)=2, rd(1.32)=6, so on and so forth).

I'd appreciate anyone's input on the matter, thank you for your time.

Definitely not continuous. Not even well-defined if you’re trying to make it a function from ℝ to ℝ. Even if you include a point at infinity, you still need to specify which expansion you’ll use for numbers with both a terminating and a non-terminating expansion. Every x whose decimal expansion is non-terminating will evaluate to infinity under your function. Specifically, all irrational numbers, as well as all rational numbers with least-terms denominator a multiple of any prime besides 2 and 5, have only a non-terminating expansion. Only terminating decimal expansions have finite values under your function.
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### Re: A Few Tricky Questions

HeWhoCouldn'tThinkOfAUsername wrote:I'm confused, what is the function for say a non-integer x?

Yeah, I didn't think it through fully at the time I posted that (I thought, "repeated function = power", then "sin2x = (sin x)2, so it can be extended to non-integers.")
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King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.
Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.

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### Re: A Few Tricky Questions

Qaanol wrote:Or did you mean a function that passes through the points (0, 0), (1, sin(1)), (2, sin(sin(2))), etc.? If this is what you meant, then there definitely is an analytic function passing through those points. Given any countable subset of ℂ with no limit points, and any arbitrary complex values for each point in the set, there is always an analytic function through those points.

I know, but what is the function? Moreover, I suppose a function cyclically returning to 1 for all Prime integers multiplied by some other function could equal it, as well as adding a function that returns to zero to any function there, or any other number of operations, I'm looking for the most "Fundamental", a function which has zero oscillation and passes through these points.

Qaanol wrote:Definitely not continuous. Not even well-defined if you’re trying to make it a function from ℝ to ℝ. Even if you include a point at infinity, you still need to specify which expansion you’ll use for numbers with both a terminating and a non-terminating expansion. Every x whose decimal expansion is non-terminating will evaluate to infinity under your function. Specifically, all irrational numbers, as well as all rational numbers with least-terms denominator a multiple of any prime besides 2 and 5, have only a non-terminating expansion. Only terminating decimal expansions have finite values under your function.

Look again at the function, you're think of rd(x), which is definitely not continuous, I'm talking about f(x)

x= sum from i=-infinity to infinity aix10i

f(x)=sum from i=-infinity to infinity rd(ai+ai-1)x10i, with rd(ai+ai-1) being equivalent to adding all the digits of (ai+ai-1)(for example, rd(10)=1, rd(11)=2, rd(1.32)=6, so on and so forth).

I'm not talking about if rd(ai+ai-1) is continuous, I'm talking about if f(x) is continuous. On that note, I should clarify ai is always a positive integer regardless of the value plugged in, as I'm adding up each value ai, with the i value specifying the digit of x that takes the value ai. Also to note, I should have defined further that for -x simply take -f(x).

On that note, doing some work I'll save the trouble of posting, I'm pretty certain this function as no derivative, which is all I have the ability to test for, still wondering if it's continuous but not differentiable(like the Weirstrass function). Further I'm curious as to this function's graph and properties, which if anyone has advice on how to find these would be awesome, but as long as these two main questions get answered I'm satisfied.

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### Re: A Few Tricky Questions

The function you want to go through iterated sine values will most likely have no simple representation in elementary functions. To see how hard it is, try finding an analytic function that agrees with the natural log for positive integers. If you just want a continuous function, you can take it piecewise with line segments. If you want it infinitely differentiable, you can use trickses with e-x[sup]2[/sup]. But to make it analytic is tough. You could try taking the sequence of polynomials with degree k that pass through the first k+1 of your chosen points. If the coefficients go to limits, you could check to see if the resulting power series does what you want.

You’re right, I did misread your f(rd) function. As it stands, it should be discontinuous at every number with a terminating decimal expansion.
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Qaanol

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### Re: A Few Tricky Questions

Hmmm, interesting. I'll give the power series a try, this'll be my first time attempting such a thing though.

On my function, I'm curious to how you got your answer, and would you have any advice on constructing a continuous function using the decimal expansion of x(i.e. a function working on the values of the digits of x).

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### Re: A Few Tricky Questions

Qaanol wrote:You’re right, I did misread your f(rd) function. As it stands, it should be discontinuous at every number with a terminating decimal expansion.

Indeed. As you've defined f, you have
f(1)=f(1.000...)=11, but
f(.999...9)=9.999...9, so limx->1- f(x) = 10 ≠ f(1) = 11. However, it is easy to see that if two numbers x and y agree up to the first n digits of their decimal expansion, then f(x) and f(y) agree up to the first n-1 digits. This means that f is continuous except at numbers with a terminating decimal expansion. If you defined a similar function in such a way that it naturally gave the same result for both expansions of numbers with non-terminating expansions, it would probably be continuous. In particular, if you defined f the same way but replaced rd(ai+ai-1) with a function g(a,b) such that g(a,0)=g(a-1,9) and g(9,9)=0, the modified f would be continuous. (An example of such a function would be g(a,b)=(a-b) % 10.)
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### Re: A Few Tricky Questions

Skeptical Scientist, thanks very much. I think this has become very clear now. So if I can find a way to make f(.999...)=f(1.000...) I can make this work. I'd be very interested if I could find a continuous function that is not a combination of previously known functions.

And I'm a little confused by your notation for an example of a continuous function working on a decimal expansion of x, particularly the %10 part, could you elaborate, maybe show me an example so I can see a little more clearly what you mean.
Last edited by HeWhoCouldn'tThinkOfAUsername on Thu Sep 29, 2011 1:02 am UTC, edited 1 time in total.

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### Re: A Few Tricky Questions

HeWhoCouldn'tThinkOfAUsername wrote:Skeptical Scientist, thanks very much. I think this has become very clear now. So if I can find a way to make f(.999...)=f(1.000...) I can make this work. I'd be very interested if I could find a continuous function that is not a combination of previously known functions.

And I'm a little confused by your notation for an example of a continuous function working on a decimal expansion of x, particularly the %10 part, could you elaborate, maybe show me an example so I can see a little more clearly what you mean.

The way you have it, when you add two adjacent digits and then add the digits of that number with your rd operation, you are actually just taking the value ai+ai-1+1 (mod 10). SkepSci is suggesting that you leave off the (+1) and just use ai+ai-1 (mod 10).
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### Re: A Few Tricky Questions

Oh, alright I see what you mean now, I guess I hadn't realized how rd(blah) was just blah+1 (mod 10).

So then, sum from -infinity to infinity ai+ai-1 (mod 10) is a continuous function, that should keep me busy. I'll play with that for a bit and see what if anything I find interesting with it, thank you for the help Qaanol and Skeptical Scientist. Cheers.

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### Re: A Few Tricky Questions

Er, I mistyped what SkepSci said. He used a minus sign not a plus sign. As in, ai-ai-1 (mod 10).
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### Re: A Few Tricky Questions

Alright, so same function but ai-ai-1? Duly noted.

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### Re: A Few Tricky Questions

HeWhoCouldn'tThinkOfAUsername wrote:Oh, alright I see what you mean now, I guess I hadn't realized how rd(blah) was just blah+1 (mod 10).

Not quite. blah (mod 10) is just the last digit of the integer blah, which is not the same as the sum of the digits (or the digits plus 1). It's just that ai+ai-1 is always the sum of two single-digit numbers, so it's always between 0 and 18, which means the 10s digit is either 0 or 1.

So then, sum from -infinity to infinity ai+ai-1 (mod 10) is a continuous function, that should keep me busy. I'll play with that for a bit and see what if anything I find interesting with it, thank you for the help Qaanol and Skeptical Scientist. Cheers.

Yeah, I don't think that what I said is true. If g(9,9)=g(0,0), then all of the digits after the nonrepeating version terminates (or the repeating version starts repeating) will be the same, and if g(a-1,9)=g(a,0), then the digit where the nonrepeating version terminates will be the same, but the digit right before might still be different unless you have g(b,a)=g(b,a-1). But then this rules out any interesting functions for g, so you can't actually define a function this way. (My example was not continuous after all.)

You can build some interesting functions using similar ideas, however. For example, take f to be the following function:
For x=\sum_{i=-\infty}^N a_i 3^i with 0≤ai≤2, (i.e., the base 3 representation of x), take n to be maximal such that an=1 (or n=-∞ if there is no such digit). For i>n, define bi = ai/2; for i=n, define bi = 1; and for i>n, define bi = 0. Then f(x)=\sum_{i=-\infty}^N b_i 2^i is continuous, since (for example) the two different base three representations of 1 give the same results: f(1.000...3) = 1, and f(0.222...3) = 0.111...2=1. (In words, you take the base-3 representation of a number, remove every digit after the first 1, replace all of the 2s with 1s, and interpret the result in base 2.)

In fact, this function is notable enough to have its own Wikipedia article.
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### Re: A Few Tricky Questions(now for all my math questions

A quick question, in a function I created I found something interesting.

Let us say rd(x) is a function that maps x to all the values of the digits summed, then summed again, repeated until some one digit positive integer a( 0<or=a<10, with a being an integer value). If we do this for all the square numbers, zero excluded, we find that the list we generate is cyclical, in a pattern 1,4,9,7,7,9,4,1,9, at least up to the 19th square and I believe to infinity. Furthermore, if we apply this function for 2^n, starting at n=0, we get the pattern 1,2,4,8,7,5. This seems rather curious to me, and I was curious if anyone could see if there was a reason why these cyclical answers are coming out of this function, I'd be glad to know.

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### Re: A Few Tricky Questions(now for all my math questions

HeWhoCouldn'tThinkOfAUsername wrote:A quick question, in a function I created I found something interesting.

Let us say rd(x) is a function that maps x to all the values of the digits summed, then summed again, repeated until some one digit positive integer a( 0<or=a<10, with a being an integer value). If we do this for all the square numbers, zero excluded, we find that the list we generate is cyclical, in a pattern 1,4,9,7,7,9,4,1,9, at least up to the 19th square and I believe to infinity. Furthermore, if we apply this function for 2^n, starting at n=0, we get the pattern 1,2,4,8,7,5. This seems rather curious to me, and I was curious if anyone could see if there was a reason why these cyclical answers are coming out of this function, I'd be glad to know.

This is because if n is a positive integer, then rd(n) is just the value between 1 and 9 which is equal to n, mod 9.

Proof:
Spoiler:
To demonstrate this, since rd(n)=n if 1≤n≤9, it suffices to show that for n≥1, rd(n)=rd(n+9). This is certainly true for 1≤n≤9. Suppose it fails for the first time at N≥10. Let ds(x) be the sum of the digits of x (so rd(x) is obtained by repeatedly applying ds until the value obtained is a single digit). Then N+9 is either of the form a9 where N was of the form a0 for some string of digits a, or is of the form a(d+1)0k(d'-1), where N was of the form ad9kd' where d'≠0 and d≠9 for some string of digits a and single digits d and d'. In the first case, rd(N+9)=rd(ds(N+9))=rd(ds(N)+9)=rd(ds(N))=rd(N), as ds(N)<N for N≥10, which means rd(ds(N)+9)=rd(ds(N))=rd(N) by the choice of N.

Then as the value of n2 mod 9 depends only on the value of n mod 9, the sequence rd(n2) must repeat every 9 terms. Similarly, the sequence rd(2n) repeats every 6 terms, because 2 is a unit in Z/9Z, and the group of units has order 6. (The numbers 1,2,4,5,7,8 that repeat in the pattern are precisely the numbers less than 9 which are relatively prime to 9, which is no coincidence.)
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### Re: A Few Tricky Questions(now for all my math questions)

Now, this is a much more advanced question that I probably won't understand the answer to, but I am curious, was recently researching the Riemann Zeta Function, and I was curious about something. The Riemann Zeta Function can approximate any Holomorphic Function arbitrarily well. It is possible to construct a function such that any part of the function approximates a section of any holomorphic function(just restating Zeta Function Universality), but is it possible to construct a function such that any section of all holomorphic functions are approximated by some section of the function, and furthermore can we make it so that this function is Infinitely Differentiable(i.e. Holomorphic)? Can we identify this function knowing this property alone?

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### Re: A Few Tricky Questions(now for all my math questions)

HeWhoCouldn'tThinkOfAUsername wrote:Now, this is a much more advanced question that I probably won't understand the answer to, but I am curious, was recently researching the Riemann Zeta Function, and I was curious about something. The Riemann Zeta Function can approximate any Holomorphic Function arbitrarily well. It is possible to construct a function such that any part of the function approximates a section of any holomorphic function(just restating Zeta Function Universality), but is it possible to construct a function such that any section of all holomorphic functions are approximated by some section of the function, and furthermore can we make it so that this function is Infinitely Differentiable(i.e. Holomorphic)? Can we identify this function knowing this property alone?

I can't really see how the formal statement here

http://en.wikipedia.org/wiki/Zeta_function_universality

doesn't imply that the zeta function satisfies your hypothesis. Of course, the function has to be non zero on the approximated region, and the region that its approximating on can't have a hole (or else the function could have a root in the hole, and then Cauchy integral theorem would cause some pretty big problems), but aside from that it says if you: pick an e>0; pick a (nice) region; pick a (nice) function f that is non zero on that region; then there is a t such that |zeta(z+it)- f(z)|< e for all z in the region. Isn't that what you're asking for?
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### Re: A Few Tricky Questions(now for all my math questions)

sinx(x) approaches zero as x approaches infinity, this is due to the fact that |sin(x)|<1 for all x, and |sin(x)|<|x| for x>0
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### Re: A Few Tricky Questions(now for all my math questions)

tomtom2357 wrote:sinx(x) approaches zero as x approaches infinity, this is due to the fact that |sin(x)|<1 for all x, and |sin(x)|<|x| for x>0

I don't think just having |f(x)| < 1 for all x and |f(x)| < |x| for x>0 alone implies that (f(x))^x goes to 0 as x goes to infinity. Or am I missing something? Let f(x) = (x-1)/x for x>1 and f(x) = x/2 for 0<x<=1. This satisfies the two requirements but the limit as x goes to infinity is 1/e.
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### Re: A Few Tricky Questions(now for all my math questions)

sinx(x) doesn't have a limit, because it varies between 1 and 'zero' as x-> infinity.

As soon as you pick a value of x that is 0 mod pi, you get 1.
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### Re: A Few Tricky Questions(now for all my math questions)

For x a positive integer and sinn(x) using the function-iteration notation for sin(sin(…(sin(x))…)), tomtom2357 is correct that sinx(x) must go to 0. First, since x is positive, we have 0 <= sin(x) < x. Next, since π is irrational and x is a positive integer, we know 0 < sin(x) < 1.

Let sn(x) = sinn(x), so s0(x) = x, s1(x) = sin(x), s2(x) = sin(sin(x)), and so forth. We are looking at the orbit of x under the sine function. Note that the sine function is a non-expansive mapping on (-1, 1), with fixed point at 0. It follows that for every x and n, we must have |sn(x)| < sn-1(1).

Now consider the sequence an = sn(n). That is, a1 = sin(1), a2 = sin(sin(2)), and so forth. We have just established that |an| < sn-1(1). Thus we have:
a1 = sin(1) < 1
a2 = sin(sin(2)) < sin(1)
a3 = sin(sin(sin(3))) < sin(sin(1))
etc.

Since the orbit of 1 approaches 0 under the sine function, it follows that the limit of sinn(n) equals zero as n grows without bound.
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### Re: A Few Tricky Questions(now for all my math questions)

Dason wrote:
tomtom2357 wrote:sinx(x) approaches zero as x approaches infinity, this is due to the fact that |sin(x)|<1 for all x, and |sin(x)|<|x| for x>0

I don't think just having |f(x)| < 1 for all x and |f(x)| < |x| for x>0 alone implies that (f(x))^x goes to 0 as x goes to infinity. Or am I missing something? Let f(x) = (x-1)/x for x>1 and f(x) = x/2 for 0<x<=1. This satisfies the two requirements but the limit as x goes to infinity is 1/e.

For large x, f(x)=1-1/x is just a bit smaller than 1. Now you apply it again, so f(f(x)) is a bit smaller than 1/2, f(f(f(x))) is a bit smaller than 1/4 and so on, it converges to 0.
fx is not meant as an exponent here, but as the function applied x times in a row.

But you are right that the two conditions are not sufficient. Take f(x)=x-eps/(1-x)^2 for x<1, f(x)=1-eps/(x-1)^3 for x>=1 with a small positive eps.
Now compare fx(x) and fx+1(x+1):

fx+1(x+1)=fx-1(f(f(x+1))) = fx(1-eps/x^3) = fx-1(1-eps/x^3 - 1/(eps*x^6))
fx(x) = fx-1(x-eps/(x-1)^3)

For large x, (1-eps/x^3 - 1/(eps*x^6)) > (x-eps/(x-1)^3)
As f(x) is monotonically increasing for x<1, "<" can be taken through the (x-1)-times application of the function.
It follows that fx+1(x+1) > fx(x) for large x.
The series ax=fx(x) is increasing and limited (for example to the range [0,1], so it converges - but not to 0. I think it converges to 1, otherwise I would be really surprised.

Edit: I noticed that I need an additional proof that "for large x", the whole thing is still well-defined >0. Hmm... may be a bit more tricky. Maybe next year (20:16 here) .

I think an additional requirement of f'(x)<1 would be enough.
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### Re: A Few Tricky Questions(now for all my math questions)

I think you guys were confusing sin with cosine when you said it oscillates between 0 and 1, sin(0)=0. Also, a function that is continuous for all numbers except the ones with terminating decimals is an interesting function indeed.
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