Next Solution

]]>For the equation.

You can write such a parameterization.

]]>This is one of the 7 things of the pascals triangle which i find very interesting.

]]>Do not worry about it, it is okay. It is enough that you have it. Post if you get stuck on one or find one interesting.

I will not get stuck since i have the solutions book(yes a solutions book is also available there.....). Yeah i will send you the questions which i find difficult and interesting.

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https://brilliant.org/wiki/converse-of-intermediate-value-theorem/

]]>A=2x3x5x7x11x13x17........ up to

m=Any multiple

A-pm= A number factorable by a factor of p

if p has one.To get A-pm, go through the primes, subtracting p as many times as you like.

Example:

p=129

A=2x3x5x7x11

2x3x5x7=210 210-129=81

81x11=891 891-(129x6)=117

A-129m=117

117 will have a factor the same as p if it has any.

117/129=39/43 129 is NOT prime (Common denominator =3)

In this way we can find out if p is composite without ever having to use a number

. Might be useful for computers.]]>Example:

p=130

Rd. Up to nearest prime= 13

Next prime after that = 17

17-1=16

Largest prime gap <130 = 16 (Correct)

This works because the greatest number of composites between two primes occurs when factors are not combined. So what could have been two composites is actually just one, like 15=3x5. To create the greatest possible number of composites I start at 2 not 0. 0 has an infinite number of prime factors, and so the greatest gap between the next repeat will occur after 0. Starting with the smallest composite which is NOT combined factors, I move up. Deleting all numbers factorable by primes less than the square root. The first time I attain TWO primes is when I reach the second prime after the square root. So this -1 is the gap required to create two non-composites with greatest possible occurrence of composites.

Largest prime gap <p =

Rd. Up to second nearest prime -1.]]>The formula is too complicated to remember. It can be remembered much more easily by Implementing Faulhaber's formula. It says,

1ˣ +2ˣ +3ˣ +4ˣ ......nˣ =[1/(1+x)](aB₀nˣ⁺¹+bB₁nˣ+cB₂nˣ⁻¹............yBₓn), where Bₙ is the nth Bernoulli no. and a,b,c,.....y are the consecutive terms of (x+1)th row of Pascal's Triangle.

For e.g.

1¹¹+2¹¹+3¹¹+4¹¹.....7¹¹=(1/12)(1B₀n¹²+12B₁n¹¹+66B₂n¹⁰+220B₃n⁹+495B₄n⁸+792B₅n⁷+924B₆n⁶+792B₇n⁵+495B₈n⁴+220B₉n³+66B₁₀n²+12B₁₁n)

(1+B)ⁿ⁺¹-Bₙ₊₁=0

For e.g.

for,B₁ n=1 so

(1+B)²-B₂=0

⇒1+B₂+2B-B₂=0

⇒1=2B=0

⇒B=-0.5

However in this ( the formula for sums of powers )formula B₂=+0.5]]>

Sorry, haven't been keeping up with the "this is cool" feed. But nice job!

]]>URL:http://www.mathsisfun.com/numbers/infinity.html

]]>I am a guy who came into this forum in 2009. I currently live in the badlands of Florida.

did you become administrator

In March of 1990 at the tender age of 70 I decided to devote most of my remaining brain cells to mathematics. This was primarily the result of a dare by people who believed that I was a loser and could not stick to anything for more than a short time. They were partially right but I could stick to something if I wanted to. So, in that month I decided that I would teach myself mathematics and computers full time. I also made the promise that at the end of 20 years if I could not do any real math, I would quit forever. In April of 2009 ( the 19th year ) I decided it was time to see if I had succeeded.

I noticed the MathsisFun forum and one character in particular, the legendary JaneFairfax. She seemed to be the best problem solver on the forum and naturally I wanted to see how I would do against the best they had.

I did not do anything that I considered amazing for the first couple of weeks until the tongzilla problem appeared. When I solved that one, I knew I had succeeded. I had learned mathematics using my own methods and my own ideas. I had discovered EM. Naturally, my original detractors suddenly forgot their accusations but I did not.

Then I had to calm my mind so that I would not forget that what I had done was not amazing, it was not even difficult. To do that I used the techniques that pappym taught me from your country. They were much harder than mathematics and I am still working on them.

Why am I an administrator? Beats me...only MIF can answer that question. But it has been a privilege to serve here and to know the people that are here. A privilege I do not deserve but have been given to me anyway.

]]>1,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192..........

which is nothing but powers of 2.So,it is actually

2^0,2^0,2^1,2^2,2^3,2^4,2^5,2^6.........

There is also a formula first noticed by Leonhard Euler which proves that the set of primes is endless. The formula is

x²+x+41 is always a prime no. if x is an integer.

This cannot be true: just take x = 41 for instance, which clearly factorises into non-trivial factors. Euler found that this polynomial produces 40 distinct primes for the first 40 values.

In fact, it can be shown that such a polynomial cannot exist.

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