Adding Past Infinity (WARNING: Math Ahead)

Minute Physics provides an energetic and entertaining view of old and new problems in physics -- all in a minute!

In this episode we take a break from physics and do a little fuzzy math. But not really: this is actually relevant to physics! Come back and I'll explain later. Created by Henry Reich

[primaryforest]

9 years later and he still hasn't said what this has to do with physics

[trevormutter]

"today a physicist does math"
don't physicists do math, like everyday? Lol

[ibooai.]

Hey, wait a minute! You can't do this! It is an infinite series. Since the terms do not go to zero, it is a divergent series, meaning the sum of the terms is not a fixed value, it is infinity. Then you subtract another infinite series, which is also divergent. That is infinity - infinity, which is indeterminate, not 0, or 1, or -1, or anything like that. It is like dividing by zero.

[juliep.]

if you move the positive line (upper) to the left, you get a positive outcome, if you move the negative one to the left, you get a negative outcome. so basically you can get ANY result ranging from negative to positive infinity just by randomly choosing where you begin subtracting

[UnforsakenXII]

Therefore the square root of infinity is imaginary. :^ )

[SenseiFilth]

finally a way to measure you moms weight

[KebunH]

You could also change (2-1) for (4-3) and suddenly infinity equals -3. I think the mistake/cheat here is the fact that you are assuming that the second string of numbers (the -1 part) is always 1 step ahead, otherwise the difference between the two (the 2 and the -1 strings of numbers) wouldve been roughly the size the last added part of the 2 string, or every addition in the -1 string added together, which in the case of infinity, is an infinitely big number

[XXXXL]

Adding positives never becomes negative unless youre in Java

[codediporpal]

See what he did there at 0:24?

= (2 + 4 + 8 + ...) - (1 + 2 + 4 + 8 + 16 + ...) 
= -1 + (2 + 4 + 8 + ...) - (2 + 4 + 8 + ...)

He brought out the -1 as if the associative property applies.  Does it really?  If so I'll show you that 1=0, 2=1, and that cat poop is candy.

S = 1 + 1 + 1 + ....

So,

0 = S - S
  = (1 + 1 + 1 + ....) - (1 + 1 + 1 +
  = 1 + (1 + 1 + 1 + ....) - (1 + 1 + 1 + ....) 
  = 1

Therefore 0 = 1

0+1 = 1+1
1 = 2

Cat poop = candy left as exercise to the reader.

[christianmorales]

Thanks for putting the math warning. Saved my life

[MaskedMammal]

I don't think this really works. Because if you limit how many numbers you use...
(2-1) * (1+2+4) = (2+4+8) - (1+2+4) = 7
(2-1) * (1+2+4+8) = (2+4+8+16) - (1+2+4+8) = 15
(2-1) * (1+2+4+8+16) = (2+4+8+16+32) - (1+2+4+8+16) = 31

Even though we're going toward infinity, you simply cannot disregard that the second parenthases will always be one power of two (minus 1) behind the progression of the first. No matter what point you choose to stop at to examine the results, you will always find that your result is a power of two minus 1.
Not coincidentally, if you just look at the sequence before you multiplied (2-1) into it, you get the exact same result. 

Hopefully that makes sense. I just don't see how we can come to the conclusion that just because we're extending both sides into infinity, we can ignore the fact that one side is always one power of two ahead of the other. I don't think going to infinity means that that no longer applies.

[JosephHarner]

Except that you didn't align your terms properly, and yes, it does matter. If you take a limit of the terms as they are added to the sequence, instead of deciding to unbalance the equation, you get lim (x->inf) of 2x-1x, which approaches infinity, not -1.

[deathwilldie]

In a way, yes this is correct, but its similar to the proof that states the sum of all integers is -1/12 in the way that you are "shifting" the infinite series which is a mathematical fallacy. I know that physicists do this all the time to get the answer they need for an equation, but it really is a bad habit.

[PrimusProductions]

When restricting "The sum of positive terms is always positive" rule to only a finite number of terms, the rest of Algebra still works. The infinite sum satisfies all properties of -1

Let s=1+2+4+8+16+...

If s=-1 then s+1=0
Addition is commutative so 1+s=s+1
1+s=0
1+1+2+4+8+16+...=0
2+2+4+8+16+...=0
4+4+8+16+...=0
8+8+16+...=0
16+16+...=0
32+...=0
Although the smallest term gets larger and larger, the number of terms decreases. Doing this infinitely will give you the empty sum which is 0

If s=-1 then s+2=1
s=-1
1+2+4+8+16+...=-1
Add 2 to both sides but on the left hand side insert the term in between the 1 and the 2
1+2+2+4+8+16+...=-1+2
1+4+4+8+16+...=1
1+8+8+16+...=1
1+16+16+...=1
1+32+...=1
Everything on the left hand side past the 1 eventually disappears.

The video is easier to understand if you know binary:






\therefore

[XouZ]

This rule only apply on numbers, you can't treat infinity as a number since it isn't a number.

[stevexysabc]

INFINITY: NO CAN STOP ME AHAHAHA

minutephysics: I am about to end this man whole career

[karlhansson]

That statement is *ACTUALLY NOT TRUE*...

Since in math, you'll write following:
(2-1)*(1+2+4+8+16+...+n) and n=∞

This will give us following:


then as he explanied. 2-2=0 4-4=0 etc.

-then we have: 2n-1-n where n=∞-

-2n-n=n => n-1 where n=∞-

-The real answer is ∞-1 since it is wrong to not include ∞ when we are counting with infinity-

*CORRECTION*

now since we were adding numbers untill we get to "n" (infinity number), the sum of all numbers before it will be n-1, so the real answer is actually 2n-1 since n+n-1=2n-1

So my mistake was by subtracting n from 2n since n has allready been canceled out by 2*(n/2)=n

  -2+4+8+-...+-n-+2n
- 1--2-4-8--...-n/2-n-

and ofcourse n/2 was canceled out by n/2 in the upper section...

[phoenixdown]

Argh. This breaks some very basic theorems about convergent and divergent series. Specifically (Theorem 3.24 in Rudin's Principles of Mathematical Analysis, 3rd edition), a series of nonnegative terms converges if and only if its partial sums forms a bounded sequence. The partial sums are not bounded, so the series cannot converge.

At an even more basic level, a necessary (but not sufficient) condition for a series to converge is for the absolute value of the terms in the sequence to converge to 0. That is, if {a_n} is a sequence, then if sum_1^infty a_n converges, then for any epsilon > 0, there exists an N such that |a_n| <= epsilon for every n >= N. By the contrapositive statement, if there is an epsilon > 0 such that for any N, there exists an n >= N with |a_n| > epsilon, then the series sum_1^infty a_n diverges. By considering epsilon = 1/2, we find that |a_n| > epsilon for every n and every N. Thus the series diverges and thus has no finite limit. It follows that cancelling terms and reaching a finite solution does not make sense, as those only make sense if the series converges.

[sergechamps]

The sum of positive terms all of which are higher than 1/12 to inifinite equals -1/12.
{1+2+3+4+5+6+....} = -1/12 according to Ramanujan summation of divergent series.

[pavelZhd]

You know...
As a percon with decent Math education, I can tell this stuff is.... somewhat correts. But not 100% correct.

Ususally the summ of 1+2+4+8+... is defined by Limit of insinite summ of N steps while N is headed to infinity.
In that case tou can not easily multiply insides and outsides of a Limit and live with it.
So mathematically correct those calculation will result in "Limit of ((2 times N) -1) while N heads to infinity" So it's basically (2 times infinity) -1. Which is also called infinity, but... that exactly shows that there are many different infinities there.

Anyway, thank you for a brainWhak that made me refresh my knowlage. ^_^