# Infinite Series/Calculus III

1. Aug 3, 2005

### Edwin

Suppose you start with the number 0. If you Integrate 0 you get 1.
Next, if you integrate 1 to get x + C, but for simplicity, let C =
0, so you get just x. If you integrate the result again, that is,
if you Integrate x, you get (x^2)/2 + C2, but for simplicity, let C2
also = 0, so you get just (x^2)/2. Let all arbitrary constants Cn,
of the given sequence of polynomial antiderivitives = 0 for
simplicity's sake. If you integrate each result an infinite number
of times, you would get an infinite sequence of antiderivatives of
the number 0.

From this you could create an infinite series from these anti-
derivitives as follows:

1+x + (x^2)/2 + (x^3)/6 + (x^4)/24...etc.

The question I have is, is the infinite series ever convergent?

If so, for what values of x is the sequence convergent.

If the sequence is convergent for some values of x, and x=1 happens
to be one of those values, then what does the sum of this infinite
series converge to
if x = 1? The answer may, or may not, surprise you!

Inquisitively,

Edwin G. Schasteen

Last edited: Aug 3, 2005
2. Aug 3, 2005

### iNCREDiBLE

Have you tried the case for which x=1?

f(1) = 1 + 1 + (1^2)/2 + (1^3)/6 + (1^4)/24 + (1^5)/120 + (1^6)/720 = ?

3. Aug 3, 2005

### dlaszlo88

This fits as a simple power series. Here you can integrate which is like doing the Integral Test

4. Aug 3, 2005

### Maxos

Wow!

You have just discovered Taylor's series of $$e^x$$!!!

(Anyway, to demonstrate its convergence, you should use the formula with Lagrange's remainder, showing that the rest goes to 0)

By the way, that's included in Calculus II, in my University

Last edited: Aug 3, 2005
5. Aug 3, 2005

### lurflurf

This is a joke right? By calculus 3 this should be obvious.
I will procede by operator methods without particular concern about rigor.
You have defined a operator
$$Lf=\int_0^x f(t) dt$$
thus
(L^k)1=x^k/k!
and
DL=1
and
next you operated on 1 with a geometric series of L
(1+L+L^2+L^3+...)1=f(x)
we sum the geometric series
(1-L)^-1 1=f(x)
operating with (1-L) on both sides
1=(1-L)f(x)
operating with D on both sides
0=(D-DL)f(x)=(D-1)f(x)
so
f'(x)=f(x)
also f(0)=(1+L+L^2+L^3+...)1|{x=0}
f(0)=1
so
f(x)=exp(x)
and the series given converges everywhere.
f(1) is now obvious

Last edited: Aug 3, 2005
6. Aug 5, 2005

### Edwin

Thanks for the replies!

That's right! e^1.

We just started covering infinite series in calculus III, and this was one of the series that was taught to us. I just thought it was kind of cool that integrating 0 to get a term a1, and integrating a1 to get a term a2, and doing this consecutively through infinity to get the infinite sequence an, and then summing this sequence yields the infinite series for the function e^x.

Please forgive me if this question sounds a little dumb, but if one discovers a function, or algorithm, that successfully yields all prime numbers greater than a certain prime number, would this be that significant of a discovery?

Inquisitively,

Edwin G. Schasteen

Last edited: Aug 5, 2005
7. Aug 5, 2005

### Edwin

Question

Consider the function f(x) = x for x greater than or equal to 5.
If a given value of x, such that x is a prime number greater than or equal to 5, then f(x) is also a prime number that is greater than or equal to 5. The area under the curve f(x) = x from the prime number x=5 to the next prime number x=7 is an even integer that is divisible by the number 12 an integer number of times. Similarly, the area under the curve f(x) = x from the prime number x=7 to the prime number x=11 is an even integer that is divisible by the number 12 an integer number of times. For each number I've checked, it appears that the area under the curve from f(x) = x from x = a prime number, to x = the next prime number, is an even integer divisible by the number 12 an integer number of times, so long as x is greater than or equal to 5. Let
g(x) = the integral of f(x) from 5 to x.

g(x) = [(x^2)/2 - 25/2] for x greater than or equal to 5.

It seems that when g(x)/12 is an integer, that x may, or may not, be a prime number. I say may, or may not, because if you plug in the number x=995, g(x)/12 is an integer, but 995 certainly is not a prime number.

I'll do a little simplification,

Let P(x) = g(x)/12 = (1/24)*(x^2 - 25).

The question I have is as follows.
Given x is a prime number greater than 5, will P(x) = (1/24)*(x^2 - 25) always yield an integer?

Inquisitively,

Edwin G. Schasteen

8. Aug 5, 2005

### shmoe

Yes, it works if x=5 as well. Any prime not 2 or 3 can be written as 6k+1 or 6k-1 for some integer k. If it's 6k+1 you get:(1/24)*(36k^2+12k-24) which is an integer if (3k^2+k)/2 is. But 3k^2+k=(3k+1)k, and is therefore even whther k is even or odd. Same thing if your prime is 6k-1.

A similar idea will let you change the lower limit of 5 to any larger prime.

Stated another way, the square of any prime larger than 3 is congruent to 1 mod 24 (it's remainder is 1 when divided by 24).

9. Aug 8, 2005

### Edwin

Wow, that's Interesting

That is interesting. It might be interesting to explore the properties of the set of even numbers k-sub-n to see if there is any pattern within the primes that compose these even numbers that you mentioned. If a function and some conditions to discern the next even numbers that have a prime number 6k+1 or 6K-1 in a sequence, one might be one step closer to deriving an algorithm that can spit out the entire set of prime numbers. What do you think?

Inquisitively,

Edwin

Last edited: Aug 8, 2005
10. Aug 8, 2005

### shmoe

I think you might have a slight misunderstanding (the "sandwiching bit"). Let me reword slightly (this is equivalent to what I wrote before): if p is any prime greater than 3 then there is an integer k where p=6k+1 or p=6k+5. This k and choice of 1 or 5 is found by dividing p by 6, e.g. 23=6(3)+5, 97=6(16)+1. The 1 or 5 is dependant on the remainder of p when divided by 6.

You know that when you divide a number by 6 the remainder is 0, 1, 2, 3, 4, or 5. If it's 0, 2, or 4 then the number was even. If it's 3 then the number was divisible by 3. So if you divide a prime (>3) by 6 you are going to get 1 or 5, and in fact 1 or 5 are both equally likely (in a sense that can be made precise).

Now if 6k+5 is prime, there's no guarantee that 6(k+1)+1 is prime, so there's no guarantee of any "sandwiching". Actually if you have both 6k+5 and 6(k+1)+1 being prime, these are twin primes, and it's a long standing conjecture that there are infinitely many such pairs of twin primes.

More often than not primes occur "alone", rather than as twin primes. I don't know if you were specifically talking about twin primes with your sandwiching comment, but thought I'd give some more detail.

11. Aug 10, 2005

### Edwin

Correction

I initially misunderstood. You are correct in the concept mentioned. The conjecture of there being infinitely many prime pairs has not been proved.

I was thinking that one could construct a series

(a-subn+1/a-subn) similar to the sequence formed from the ratio of two consecutive Fibonacci numbers, and then map each element of the sequence to a unit position vector, originating from the origin by convention, with length = 1 so that the ratio of each consecutive prime numbers, twins or not, occupies a unique position on the unit circle defined by the position vector of length 1 with a unique angle defined by Arctan(Bn/An), where An is the nth prime, and Bn is the (n+1)th prime.

For example, the first few angles of the unit position vector would be:

Arctan(2/1)
Arctan(3/2)
Arctan(5/3)
Arctan(7/5)
Arctan(11/7) and so on...

I've done a few of them, and at first it seems like it may be converging on 45 degrees, but I'm not completely sure it does. The reason is that I tried a few angles for prime numbers in the 600 range, and it gave values like:

Arctan(661/659) = 45.08681172 degrees
Arctan(673/661) = 45.51539041 degrees
Arctan(677/673) = 45.16976478 degrees

But if you shift up to the 900 range,

Arctan(977/971) = 45.17647515
Arctan(983/977) = 45.1753947
Arctan(991/983) = 45.23220047
Arctan(997/991) = 45.17292436

Although I haven't explored this concept thoroughly yet, I'm not yet sure whether in the limit as n approaches infinity, the Arctan(Bn/An) converges on 45 degrees. If, in the limit as n approaches infinity, {Bn/An} = infinity, that is, if the sequence {Bn/An} is divergent, then the angle approaches 90 degrees. It doesn't seem like it's approaching 90 degrees, it seems like it is approaches 45 degrees, which would mean that in the limit as n approaches infinity, the sequence is approaching 1.

The way to ensure the length of the vector is always one, is as follows.

(a+bi)(a-bi) = a^2 + b^2

(a^2 + b^2)/(a^2 + b^2) = a^2/(a^2 + b^2) + b^2/(a^2 + b^2) = 1

x^2 + y^2 = a^2/(a^2 +b^2) + b^2/(a^2 + b^2) = 1.

It follows that x^2 = a^2/(a^2 + b^2) and y^2 = b^2/(a^2 + b^2)

Where a is the nth prime number, and b is the (n+1)th prime number.

The set of all points P(x.y) are the points on the unit circle that correspond to each unique ratio of consecutive primes {Bn/An}.

The angle of the position vector that points to a given point P(x,y) on the unit circle is given by

Arctan(y/x) = Arctan(b/a), which are the angles we were finding above.

What interests me is that each consecutive point on the circle can be connected by a secant line, but because the angles seem to oscillate in magnitude, it follows that many of the secant lines form triangles within the circle. The lengths of the secant lines also appear to get shorter and shorter with each iteration. The question I have is, if one were to measure the sequence of lengths of each secant line, what would the sum of the secant lines approach as the n approaches infinity? Is it a finite number? If so, it would be a constant that can be derived from the set of prime numbers, which may, or may not, be interesting.

What do you think?

Inquisitively,

Edwin G. Schasteen

Last edited: Aug 10, 2005
12. Aug 12, 2005

### shmoe

I'll use $$p_n$$ to denote the nth prime. It's known $$p_{n+1}=p_n+O(p_n^{9/10})$$ (exponents much better than 9/10 are known, but this will suffice). so we must have:

$$\lim_{n\rightarrow\infty}\frac{p_{n+1}}{p_n}=1$$

so the arctan of this sequence will converge to $$\pi/4$$ or 45 degrees if you prefer. (the O above is the usual "big-O" notation)

There will be no triangles with vertices on the circle formed from secant lines (though I'm not sure if this is what you meant). If $$\arctan{p_{n+1}/p_n}=\arctan{p_{m+1}/p_m}$$ then $$p_{n+1}/p_n=p_{m+1}/p_m$$ (arctan is increasing) so we must have n=m (by unique factorization), and you never hit the same point on the circle twice.

I'm not sure whether or not the sum of the secant lengths will converge or diverge. It's probably easier to ask this question about the sum of the lengths of the arcs rather than the secants, these series will converge or diverge simultaneously. You can then use some properties of arctan to simplify. Maybe the various bounds for $$p_n$$ can answer the question of convergence, I'm not sure, but that's a place to start.

I don't know that there's really much signifigance in putting these points on a circle. I think there's anything to be gained from looking at $$\arctan(p_{n+1}/p_n)$$ instead of $$p_{n+1}/p_{n}$$, especially since $\lim_{x\rightarrow 0}\arctan(x)/x=1$.

13. Aug 15, 2005

### Edwin

Factoring Composites Composed of Two Primes

Thanks for the information! Last Thursday, a friend of mine mentioned to me something called the RSA factoring challenge. Have you heard of that challenge? Apparently, if you factor a really large number, RSA laboratories will pay you \$200,000.
That sort of tweaked my interest in the direction of searching for methods to factor prime numbers. So far, this is what I was able to come up with over the weekend.

Consider the function

f(x) = C/x^2 + x^2/C

Where C is the composite composed of two primes to be factored.

Consider also, the function

g(x) = a constant c

There is one constant c, such that g(x) = c, where g(x) = f(x) at
values of x that are the two prime factors of C.

For example,

consider the following prime product:

55,837*99,577 = 5,560,080,949

The number to be factored is 5,560,080,949.
So, you plug the number to be factored into the function as follows:

f(x) = 5,560,080,949/x^2 + x^2/5,560,080,949

There exists a constant valued function g(x) that intersects the function f(x) at values of x that are the two prime factors of 5,560,080,949.

For the function f(x) = 5,560,080,949/x^2 + x^2/5,560,080,949:

The constant valued function g(x) = 2.34409 intersects the graph of the function f(x) = 5,560,080,949/x^2 + x^2/5,560,080,949 at values of x that are the two prime factors of the composite number 5,560,080,949.

Something else I noticed from this function is that it appears that the parabola type function is always centered at x = the square root of the composite number to be factored, and y = 2, ( (C)^(1/2), 2). It also appears that one of the prime factors of C is always less than the square root of C, and the other prime factor of C is always greater than the square root of C.

The five questions I have are as follows:

Suppose that the number C is a composite that factors exactly into two prime numbers.

Is it always true that a function g(x) = some constant will intersect the graph of f(x) = x^2/C + C/x^2 at values of x that are the two prime factors of C?

Will one of the prime factors of C always be less than the square root of C, and the other, always greater than the square root of C?

Is there an easy method, or any method, or is it possible to construct an algebraic method to determine the value of the constant c for the constant function g(x) for any given function f(x) = C/x^2 + x^2/C such that the graph of the constant function g(x) = c intersects the function f(x) = x^2/C + C/x^2 at the values of x that are the prime factors of C?

If no such algebraic method can be constructed, are there any other mathematical methods that can be used to approximate the value of c?

Is it possible to create non-algebraic methods to exactly determine c?

Inquisitively,

Edwin G. Schasteen

egschasteen05@yahoo.com

Last edited: Aug 15, 2005
14. Aug 16, 2005

### shmoe

Ok, let's write C=pq. Let's also assume that p and q are distinct.

This g(x)=constant business, there's no need to bother with it, but the answer is yes because f(p)=f(q). This is really what you've noticed about your function.

Note this is true for any real numbers z and y where zy=C, f(z)=f(y). This is why I don't think this f will be of any use at all for factoring C (this contains my answer to questions 3-5).

If n is any integer, then n will always have a prime divisor less than or equal to n^(1/2). Apply this to C=pq, if p and q are distinct.

15. Aug 16, 2005

### Edwin

Thanks!

Thanks! That makes perfect sense. Any even degree polynomial funtion with positive lead coefficient shifted to the right a given number of units will have values of x corresponding to the values of the c that are equal to two distinct prime numbers p and q of some composite number c. I also agree that, from what I can tell, this alone is not sufficient to factor C as the constant c is not uniquely determined by the function itself, that is, unless you already know the prime factors p and q, you can not determine with any degree of sucess which constant value for y will be the constant value that intersects f at values of x = p and x = q. Once again, thanks for the info!

Best Regards,

Edwin

16. Sep 4, 2005

### Edwin

Pattern Regarding Composites

The following are the details regarding a pattern
that I discussed with our math group relating to factoring of large composites composed of two prime numbers. Just wanted to run it by you for feedback.

Rp = 2 -> 2^2: 4
Rp = 3 -> 3^2: 9 8
Rp = 4 -> 4^2 16 15 12
Rp = 5 -> 5^2 25 24 21 16
Rp = 6 -> 6^2 36 35 32 27 20
Rp = 7 -> 7^2 49 48 45 40 33 24
Rp = 8 -> 8^2 64 63 60 55 48 39 28
Rp = 9 -> 9^2 81 80 77 72 65 56 45 32

The row (Rp) that a given composite (Cp) composed of
two prime numbers (a) and (b) is given by the equation
(Rp) = (a+b)/2. When each composite is subtracted from
the square of the row the composite belongs to,
(Rp)^2, the result is a perfect square. (Rp) might be
a unique number such that when a composite number (Cp)
composed of two prime numbers (a) and (b) is located
in the row (Rp), then if (Cp) is subtracted from
(Rp)^2, then a perfect square results only if (Cp)
exists in the row of the given (Rp). The smallest of
the two prime factors of (Cp) is be obtained by
subtracting a number from the row (Rp-1) that is
located at the top left corner of (Cp), and then
dividing by the number 2. For example:

Consider Rp = 4, notice that the composite number Cp =
15, which is composed of the two prime numbers a = 3
and b = 5, is located in row Rp = 4.
Row (Rp-1) = 3. Now the number located to the top left
corner of Cp = 15 in row (Rp-1) = 3 is the number 9.
Notice that (1/2)*(15 - 9) = (15 - 9)/2 = 6/2 = 3,
and that 3 = a; the smallest of the two prime factors
composing Cp = 15. The larger prime factor b = 15/3 =
5.

Consider Rp = 5, notice that the composite number Cp =
21, which is composed of the two prime numbers a = 3
and b = 7, is located in row Rp = 5.
Row (Rp-1) = 4. Now the number located to the top left
corner of Cp = 21 in row (Rp-1) = 4 is the number 15.
Notice that (1/2)*(21 - 15) = (21 - 15)/2 = 6/2 = 3,
and that 3 = a; the smallest of the two prime factors
composing Cp = 21. The larger prime factor b = 21/3 =
7.

Consider Rp = 8, notice that the composite number Cp =
55, which is composed of the two prime numbers a = 5
and b = 11, is located in row Rp = 8.
Row (Rp-1) = 7. Now the number located to the top left
corner of Cp = 55 in row (Rp-1) = 7 is the number 45.
Notice that (1/2)*(55 - 45) = (55-45)/2 = 10/2 = 5,
and that 5 = a; the smallest of the two prime factors
composing Cp = 55. The larger prime factor b = 55/5 =
11.

Notice the following pattern:

Consider the row Rp = 9, Rp^2 = 81.

81 - 80 = 1^2 = 1
81 - 77 = 2^2 = 4
81 - 72 = 3^2 = 9
81 - 65 = 4^2 = 16
81 - 56 = 5^2 = 25
81 - 45 = 6^2 = 36
81 - 32 = 7^2 = 49

Consider next, the row Rp = 8, Rp^2 = 64.

64 - 63 = 1^2 = 1
64 - 60 = 2^2 = 4
64 - 55 = 3^2 = 9...and so on.

Finally, consider the row Rp = 4, (chosen randomly).
Rp^2 = 16

16 - 15 = 1^2 = 1

It appears that any given composite composed of two
prime numbers is the element of a set Rp which is
defined by the equation Rp - some perfect square = Cp.
It appears that the set of all Rp form a group that
contains all products of prime numbers. This group
appears to also contain the sum of all pairs of prime
numbers divided by the number 2 as the number value of
the nth row, or set, Rp is equal to (a + b) / 2, such
that the set Rp contains the element a*b, where a and
b are smaller and larger prime numbers respectively.
The set of all Rp; therefore, is a group Gp that is
composed of an infinite number of ordered sets of Rp,
where p is an index that is greater than or equal to
2. Another way to interpret this is that value of Rp
has a domain restriction such that Rp >=2, as the
number 1 need not be included as part of the
pattern. Actually, the number Rp = 3 is the first number Rp
that has more than one element in its set, namely, Rp
= 3 has two elements in it's set: the numbers 9 and 8
respectively. What do you think? Is this accurate?

Inquisitively,

Edwin G. Schasteen

17. Sep 5, 2005

### shmoe

Let's denote the entry in your ith row and jth column by E(i,j), where the first row and column are indexed by 2 and 0 respectively and j<i-1. So E(2,0)=4, E(3,0)=9, E(3,1)=8, etc. Then in general E(i,j)=(i+j)(i-j), but this isn't taking into account leaving off any composites that have 3 or more prime factors. This and a little algebra will explain most of what you've observed with a minor problem and why the patterns you've seen so far will soon fail if you only consider entries with 2 prime divisors.

You'll get the smallest of the two factors that put this composite in this row, not necessarily a prime (e.g. with 80 in row 9 this process will yield 8). This is no suprise since you are computing (E(i,j)-E(i-1,j-1))/2=(i-j).

If you leave off composites with more than 2 prime factors, this pattern will eventually fail (e.g. row 12, since 105 will be missing in row 11).

You're looking at the list begining E(i,0)-E(i,1)==i^2-(i+1)(i-1)=1^2, E(i,0)-E(i,2)=2^2, E(i,0)-E(i,3)=3^2, etc... so you'll get the squares

If you were to leave off composites with more than 2 prime factors this pattern will also fail (in row 11 first, 105 will be "missing", so 16 will be "missing" from the sequence of squares you get)

You won't get any numbers of the form 2*p^k, where p is a prime other than 2. If you define E(i,j) as I did above, you'll get every composite except those of the form 2N, where N is odd. This is because i+j and i-j are either both even or both odd.

Some are missed for the reasons above, for example (5+7)/2=6 is not on this table.

18. Sep 5, 2005

### Edwin

I will have to review your reply in more detail to see if it is accurate.
Your last quote is partially inaccurate atleast.

Qoute: Some are missed for the reasons above, for example (5+7)/2=6 is not on this table.

Actually, (5+7)/2 = 6 is on the table.. 5*7 = 35, and 35 is in row 6.

(35 - 25)/2 = 10/2 = 5, and 5 is the smaller of the two prime factors.

I will check the rest of what you mentioned to see if it is accurate.
Thanks for your assistance!

Best Regards,

Gary

19. Sep 6, 2005

### shmoe

Maybe I have no clue as to what you're doing, but you said "This group appears to also contain the sum of all pairs of prime numbers divided by the number 2...", 6 is one of these numbers (the sum of two primes divided by two), yet I don't see 6 anywhere on your table, nor do I see how it could appear anywhere.

Or did you mean that the sum of any two (odd) primes divided by two would appear as an index of a row? That's obvious...

20. Sep 10, 2005

### Edwin

I appologize for the delay in responding. I currently do not have access to a computer at my home, so I have to drive down to the local library to gain access to the internet. Anyhow, I appologize for the lack of clarity regarding the concept I posted.

Smoe wrote:

Or did you mean that the sum of any two (odd) primes divided by two would appear as an index of a row? That's obvious...

Yes, that is what I meant. Sorry for the confusion about that.
The sum of two prime numbers divided by two appears as an index of the row, and the product of those two prime numbers exists in that row.

I just realized also that I did not show how the numbers in the rows are generated.

Every perfect square can be written as the sum of consecutive odd numbers:

Consider the 2nd row Rp = 3 from the chart below:

Rp = 2 -> 2^2: 4
Rp = 3 -> 3^2: 9 8
Rp = 4 -> 4^2: 16 15 12
Rp = 5 -> 5^2: 25 24 21 16
Rp = 6 -> 6^2: 36 35 32 27 20
Rp = 7 -> 7^2: 49 48 45 40 33 24
Rp = 8 -> 8^2: 64 63 60 55 48 39 28
Rp = 9 -> 9^2: 81 80 77 72 65 56 45 32

To generate the numbers 9 and 8, we rewrite the perfect square 3^2 as the sum of consecutive odd integers starting with the number 1:
1+3+5 = 9

To get the number 9, you add the entire sum of odd integers

1+3+5 = 9

To get the number 8, you add the last two odd integers 3 and 5.

3+5 = 8.

Consider the row Rp = 9:

To generate the numbers 81, 80, 77, 72, 65, 56, 45, 32, we rewrite the perfect square 9^2 as the sum of consecutive odd integers starting with the number 1 and ending at the number 17 (17 included).
1+3+5+7+9+11+13+15+17 = 81

To get the number 81, you add the entire sum of odd integers.

1+3+5+7+9+11+13+15+17 = 81

To get the number 80, you add all odd integers starting with the number 3 and ending at the number 17.

To get the number 77, you add all odd integers starting with the number 5 and ending at the number 17.

To get the number 72, you add all odd integers starting with the number 7 and ending at the number 17.

To get the number 65, you add all odd integers starting with the number 9 and ending at the number 17.

To get the number 56, you add all odd integers starting with the number 11 and ending at the number 17.

To get the number 45, you add all odd integers starting with the number 13
and ending at the number 17.

To get the number 32, you add all odd integers starting with the number 15 and ending at the number 17.

The last element of the row is generated by the largest two odd integers that that compose the perfect square Rp^2.

Any of the rows can be generated using the pattern above. What I suspect you will find is that product of any two prime numbers exists in the row Rp = the sum of the two prime numbers divided by two.

The number that is the sum of the two prime numbers divided by two, which determines the index as you mentioned above, is also an element of the set of numbers that compose the row Rp, and so does the perfect square.
For example, the set of all numbers that compose the row Rp = 9 are shown below.

9 -> 9^2: 81 80 77 72 65 56 45 32

As you can see, the numbers 9 and 9^2 are also included as numbers in the set.

I'll explain more later, unfortunately the library is closing. Please do take a look at what you see so far, and let me know if you find any flaws in the concept. Your help is most appreciated!

Thanks,

Edwin

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