Finding equations of summations from 1 to a non integer

[Gouskin]

That site shows the wrong notation. so, you cannot use it as an example of how it could be right.

Oop! Your computer is correctly interpreting the wrong notation. Looks like you can't use that.

No, you acted stubbornly defiant.

After you proceeded to tell me how wrong the proof to the epitome of my 10th grade year is.

 

[lookagain]

\(\displaystyle \displaystyle\sum_{n = 1}^x \ x^3 \ = \ \int_{.5}^{x + .5}(x^3) dx \ - \ \dfrac{1}{8}\bigg(x^2 + x \bigg)\)
.
\(\displaystyle \displaystyle\sum_{n = 1}^x \ x^2 \ = \ \int_{.5}^{x + .5}(x^2) dx \ - \ \dfrac{1}{12}x \)

 

[Deleted member 4993]

Thank you for the advice. It appears to be very sound. However, I do not know how to go about seeking out a mentor in analytic number theory. But what's fun about mathematics for me is just questioning things. (Call me crazy but I think that 2/+0 = 2 infinity and I think I can prove it, too. Who should I talk to for that?) Right now I am in the 11th grade and I am therefor not familiar with any undergraduate advisers (Unless I'm just stupid and don't know what an undergraduate / undergraduate adviser is). Any tips on that?
Once again, thank you for your advice. I'll definitely consider it.

Be careful playing with \(\displaystyle \infty\) . It is not a "regular" number - and rules of operations with "regular" numbers are not valid. For example:

\(\displaystyle \infty \ + \ n \ = \ \infty\) and


\(\displaystyle \infty \ * \ n \ = \ \infty\)

Somebody will call you crazy if you claim

\(\displaystyle \infty \ * \ 2 \ \ne \ \infty\)

You may (or may not) know that there are three levels of \(\displaystyle \infty\).

You may want to read a wonderful book written by George Gammow - "One, Two, Three, .... infinity"

 

[Gouskin]

Be careful playing with \(\displaystyle \infty\) . It is not a "regular" number - and rules of operations with "regular" numbers are not valid. For example:

\(\displaystyle \infty \ + \ n \ = \ \infty\) and


\(\displaystyle \infty \ * \ n \ = \ \infty\)

Somebody will call you crazy if you claim

\(\displaystyle \infty \ * \ 2 \ \ne \ \infty\)

You may (or may not) know that there are three levels of \(\displaystyle \infty\).

You may want to read a wonderful book written by George Gammow - "One, Two, Three, .... infinity"

The way I'm thinking about it is that, much in the same way that 0 is the start of the number line, infinity is the end of the number line. While it's not technically an finite end, (there's no last number), I consider it to be the end. Sounds weird? If it's discrediting me, then forget about this part as it's more philosophy than math.

Well, I CAN prove that 5*0 does not equal 0 (and thusly prove that 5 * Infinity does not equal infinity). My favorite way of doing this by using factorials:
3! = 3*2! = 3*2*1! = 3*2*1*0! = 3*2*1*0*(-1)!
3*2*1*0 using standard math rules is 0
(-1)! using standard math rules, limits from the right to infinity
3*2*1*0 * (-1)! = 1
0 * infinity = 1
Wait a second. I thought 3! was 6. Not 1. Okay, so (-1)! much equal 6 infinity so that:
0 * 6 * infinity = 6
Now, let's try 4!
4! = 4*3! = 4*3*2*1*0*(-1)!
4*3*2*1*0 using standard math rules is 0
(-1)! using standard math rules, limits from the right to infinity
4*3*2*1*0 * (-1)! = 1
0 * infinity = 1
Wait a second, 4! is 24. So (-1)! must be 24 infinity so that:
0 * 24 * infinity = 24
Whoa, hold on there, 24 does not equal 6. This must mean that the 4*3*2*1*0 = 24*0 and that 3*2*1*0 = 6*0, and thusly, that (-1)! = infinity.
So what this is saying is that 0, when not used as a place holder, has an infinitely small value that is infinitesimally close to the middle of the number line, or middle zero.
(I'm still working on this though)
And what about 3! = 3*2*1*0*-1*-2*-3*(-4)!
3*2*1*0*-1*-2*-3 = -36zero
and (4)! must = -1/6 infinity so that
-36 * 0 * -1/6 * infinity = 6


Now, I know what you're not asking. Is there a formula to calculate these values? Why yes! There is! I have absolutely no idea how I came up with it, but here it is and it works with everything I've tried it on except for log functions.

(when f(x) >= infinity)
in other words, it will allow for you to determine to what power the infinity is, and will return the coefficient of the infinity.

In other words, (-4)!/infinity = lim(x -> -4) 1/(d/dx(1/x!)) = -1/6
In secondary words, I'm crazy, but my way of doing it works.

 

[Gouskin]

\(\displaystyle \displaystyle\sum_{n = 1}^x \ x^3 \ = \ \int_{.5}^{x + .5}(x^3) dx \ - \ \dfrac{1}{8}\bigg(x^2 + x \bigg)\)
.
\(\displaystyle \displaystyle\sum_{n = 1}^x \ x^2 \ = \ \int_{.5}^{x + .5}(x^2) dx \ - \ \dfrac{1}{12}x \)

Thanks for that! However, I don't know if that's easily solvable by hand. I haven't taken a formal calculus class yet.

 

[HallsofIvy]

The way I'm thinking about it is that, much in the same way that 0 is the start of the number line, infinity is the end of the number line.

\
No, 0 is NOT the "start of the number line".

While it's not technically an finite end, (there's no last number), I consider it to be the end. Sounds weird? If it's discrediting me, then forget about this part as it's more philosophy than math.

Well, I CAN prove that 5*0 does not equal 0

I'm sorry to hear that.

(and thusly prove that 5 * Infinity does not equal infinity). My favorite way of doing this by using factorials:
3! = 3*2! = 3*2*1! = 3*2*1*0! = 3*2*1*0*(-1)!

Oh, I see. You do not know what 0! is.

3*2*1*0 using standard math rules is 0
(-1)! using standard math rules, limits from the right to infinity
3*2*1*0 * (-1)! = 1
0 * infinity = 1
Wait a second. I thought 3! was 6. Not 1. Okay, so (-1)! much equal 6 infinity so that:
0 * 6 * infinity = 6
Now, let's try 4!
4! = 4*3! = 4*3*2*1*0*(-1)!
4*3*2*1*0 using standard math rules is 0
(-1)! using standard math rules, limits from the right to infinity
4*3*2*1*0 * (-1)! = 1
0 * infinity = 1
Wait a second, 4! is 24. So (-1)! must be 24 infinity so that:
0 * 24 * infinity = 24
Whoa, hold on there, 24 does not equal 6. This must mean that the 4*3*2*1*0 = 24*0 and that 3*2*1*0 = 6*0, and thusly, that (-1)! = infinity.
So what this is saying is that 0, when not used as a place holder, has an infinitely small value that is infinitesimally close to the middle of the number line, or middle zero.
(I'm still working on this though)
And what about 3! = 3*2*1*0*-1*-2*-3*(-4)!
3*2*1*0*-1*-2*-3 = -36zero
and (4)! must = -1/6 infinity so that
-36 * 0 * -1/6 * infinity = 6


Now, I know what you're not asking. Is there a formula to calculate these values? Why yes! There is! I have absolutely no idea how I came up with it, but here it is and it works with everything I've tried it on except for log functions.

(when f(x) >= infinity)
in other words, it will allow for you to determine to what power the infinity is, and will return the coefficient of the infinity.

In other words, (-4)!/infinity = lim(x -> -4) 1/(d/dx(1/x!)) = -1/6
In secondary words, I'm crazy, but my way of doing it works.

 

[Gouskin]

\
No, 0 is NOT the "start of the number line".
Okay, so the first one is a technicality. I meant start of the positive number line.

I'm sorry to hear that.
That's not really a valid point. That's more of an implied insult.

Oh, I see. You do not know what 0! is.
0! = 1. However this fact doesn't matter because, if you'll look closely, you'll see that 0! was never used in my proof. It was mentioned, but not used in my proof.
The definition of factorial is x! = x*(x-1)*(x-2)*(x-3) ... (x-Z)! Example:
3! = 3*2*1*0! = 3*2*1*0*(-1)! = 6
Not x*(x-(x>Z>0))*0!*(-(0>Z))! Wrong example:
3! = 3*2*1*0!*(-1)!

Please, look:

3! = 3*2*1*0*(-1)!
Using the reflexive property of multiplication:
(3*2*1*0)*(-1)!
Now do the things in parentheses first:
Using current math rules:
(0) * (complex infinity)
Now multiply:
(0) * complex infinity = indeterminate
vs.
Using the math rules I'm developing:
(6*0) * (infinity)
Now multiply:
6*0 * infinity = 6*1 (because 0 and infinity cancel out) = 6

So, normal math rules return 3! = indeterminate
Math rules I'm developing return 3! = 6

 

[JeffM]

Please, look:

3! = 3*2*1*0*(-1)!
Using the reflexive property of multiplication:
(3*2*1*0)*(-1)!
Now do the things in parentheses first:
Using current math rules:
(0) * (complex infinity)
Now multiply:
(0) * complex infinity = indeterminate
vs.
Using the math rules I'm developing:
(6*0) * (infinity)
Now multiply:
6*0 * infinity = 6*1 (because 0 and infinity cancel out) = 6

So, normal math rules return 3! = indeterminate
Math rules I'm developing return 3! = 6

Normal math rules do not return 3! is indeterminate.

The "normal" definition of 3! is 3 * 2 * 1 = 6. Perfectly determinate.

 

[Gouskin]

Normal math rules do not return 3! is indeterminate.

The "normal" definition of 3! is 3 * 2 * 1 = 6. Perfectly determinate.

For 3! when it's written as:
3!, 3*2!, 3*2*1!, 3*2*1*0!
It's determinate.

However check this out:
3! = 3*2*1*0*(-1)!
3! = 6
3*2*1*0*(-1)! = indeterminate
6 does not = indeterminate


This is what I hate about the current definition of zeros and infinities.

a = b
a = c
b = d
c does not = d

 

[Deleted member 4993]

For 3! when it's written as:
3!, 3*2!, 3*2*1!, 3*2*1*0!
It's determinate.

However check this out:
3! = 3*2*1*0*(-1)! ← How is that...n! does not involve multiplying by 0 - neither by (-1)!
3! = 6
3*2*1*0*(-1)! = indeterminate
6 does not = indeterminate


This is what I hate about the current definition of zeros and infinities.

a = b
a = c
b = d
c does not = d

.

 

[Gouskin]

.← How is that...n! does not involve multiplying by 0 - neither by (-1)!

Yes it does. The pattern,
x! = x*(x-1)! = x*(x-1)*(x-2)! = x*(x-1)*(x-2)*(x-3)!
extends forever.
The function for calculating factorials itself literally shows that this pattern extends forever. Go to wolframalpha and type in x! and look at the graph where x < 0:http://www.wolframalpha.com/input/?i=x!,+x+from+-10+to+0
Also try: http://www.wolframalpha.com/input/?i=1.5!+=+1.5*0.5!+=+1.5*0.5*(-0.5)!+=+1.5*0.5*(-0.5)*(-1.5)!
Also try: http://www.wolframalpha.com/input/?i=x!+=+x(x-1)(x-2)(x-3)(x-4)(x-5)! It returns true for all real numbers, including sequences like: 3*2*1*0*(-1)*(-2)!

 

[mmm4444bot]

The definition of factorial is x! = x*(x-1)*(x-2)*(x-3) ... (x-Z)!

You have used a factorial in your definition of x! .

That error is known as circular logic.

What did you have in mind, when you wrote (x-Z)! ?

(When stating a definition, it's important to define all of the symbols.)

Here is a standard definition:

x! = x*(x-1)*(x-2)*(x-3) ... 3*2*1



Here is another definition for x! ; it uses the Gamma Function, and it does not restrict x to a Whole number.


\(\displaystyle x! = \Gamma(x+1)\)


\(\displaystyle \displaystyle \Gamma(x) = \int_{t=0}^{\infty} t^{x-1} \cdot e^{-t} \; dt\)

 

[mmm4444bot]

I left off the parentheses because I think they're ugly.

That's not a valid reason.

If you want to change the Order of Operations, you must type grouping symbols.

 

[Gouskin]

You have used a factorial in your definition of x! .

That error is known as circular logic.
Well, in this case, the actual values can be determined with an initial guess and with this circular logic. I don't remember how to do it right now so I can't give an example, sorry!

What did you have in mind, when you wrote (x-Z)! ?
What I mean is that it can be:
x(x-1)(x-2)(x-3)...(x-998)(x-999)(x-1000)!

As long as it decreases by an integer each time and the last term is a factorial.
Z is all integers, and I believe that it was (Z>0), so all positive integers excluding +0


(When stating a definition, it's important to define all of the symbols.)

Here is a standard definition:

x! = x*(x-1)*(x-2)*(x-3) ... 3*2*1
That's the simplified version. It's expanded version is this:
x! = x*(x-1)*(x-2)*(x-3) ... 3*2*1! and if you define it this way, you aren't stuck with positive integers!


Here is another definition for x! ; it uses the Gamma Function, and it does not restrict x to a Whole number.



\(\displaystyle x! = \Gamma(x+1)\)


\(\displaystyle \displaystyle \Gamma(x) = \int_{t=0}^{\infty} t^{x-1} \cdot e^{-t} \; dt\)

Great response.

I do have a question, though. How would you go about solving the last equation by hand? Or on the computer?

 

[Gouskin]

That's not a valid reason.

If you want to change the Order of Operations, you must type grouping symbols.

What is the answer to this problem?

 

[mmm4444bot]

What I mean is that [the definition for x!] can be:
x(x-1)(x-2)(x-3)...(x-998)(x-999)(x-1000)!

As long as [the factors decrease] by an integer each time and the last term is a factorial.
Z is all integers, and I believe that it was (Z>0), so all positive integers

I'm still not seeing this as a general definition. For example, x-1000 is not a positive Integer, when x is smaller than 1001.

Here is how wolframalpha defines factorials :


The product of the first n positive integers, denoted n!.

and

The factorial n! is defined for a positive integer n as n! = n(n-1)...2·1

 

[Gouskin]

I'm still not seeing this as a general definition. For example, x-1000 is not a positive Integer, when x is smaller than 1001.
As long as it decreases by an integer each time and the last term is a factorial.
It could also be x(x-1)(x-2)(x-3) ... (x-100000000000000000000000000)(x-100000000000000000000000001)(x-100000000000000000000000002)!
This pattern can continue indefinitely.
It doesn't matter if they're not a positive integer because they always work anyway.
For example:
4! = 4*3*2*1*0*-1*-2*-3*-4*(-5)!
= (4*3*2*1*-1*-2*-3*-4) * 0 * (-5)!
= 576 * 0 * lim(x -> -5) 1/(d/dx(1/x!)) * infinity
= 576 * 1/24 * 0 * infinity
= 576/24 * 1
= 24 * 1
= 24


Here is how wolframalpha defines factorials :


The product of the first n positive integers, denoted n!.

and

The factorial n! is defined for a positive integer n as n! = n(n-1)...2·1

x! = Gamma(x+1)
x! = "[As long as the series decreases by 1*TermInSequence each time and the last term is factorialized]"

Yea for different ways to define the same thing!

 

[mmm4444bot]

What is the answer to this problem?

The notation above is unclear. Is the intended meaning 1/1000000^2?

 

[lookagain]

.000001 with a little "2" hanging around

That's exactly what it is. It is meaningless. I already addressed a similar
question posed by the OP.

 

[mmm4444bot]

I already addressed a similar question posed by the OP.

Yes -- and now the OP has posted that contrived example, I supposed in an attempt to demonstrate some validity to their new, experimental notation (one that eliminates "ugly" grouping symbols, requiring their audience to guess).

My response was also an attempt to be clever, as texting 1/1000000^2 requires no parentheses (since the numerator is 1).

The OP is certainly free to blaze their own path. Just as the English steadfastly rejected Leibnitz's notation -- and fell behind the rest of Europe by more than a century regarding scientific advancements, as a result -- before eventually agreeing that clear notation is the better way to go, so, too, may the OP reject established notation for indicating changes to the Order of Operations, before eventually realizing that their stubborness causes more losses than gains.

Also, if the OP's claim is true, it's a real shame that the OP's instructor suffers from the same ignorance because sloppy communication in mathematics -- especially at the introductory level -- needlessly perpetuates problems.