Debunking Ramanujan's Finite Solutions to Divergent Series

ABSTRACT:

This paper shows how an infinite series yields both an infinite solution and a finite solution and how to determine which one is correct and which one is just plain crazy!

Imagine the following geometric series:

Is s convergent or divergent? Does it blow up to infinity or have a finite value? So far, we can't say. However, according to Ramanujan and his apologists, even if s is divergent, it has a finite value. Let's see if this is really the case. With a little algebra we derive equation 5 below:

What solutions will work for equation 5? Here are two:

One solution is infinity. If k is finite, the other solution is finite. This means, if we want, we can always extract a finite solution from a divergent series. It also means we can always extract an infinite solution from a convergent series! Ramanujan claimed that divergent series have finite solutions. He could have (but didn't) claim that convergent series have infinite solutions. If such claims seem counter-intuitive and just plain crazy, it's because they are. Extracting finite solutions from divergent series (or infinity from convergent series) violates the geometric series rules. Here are the rules:

At 9 we have a system of equations that show that the value of x matters. We don't just get to choose willy-nilly which equation we like best. If we choose the appropriate equation for a convergent series, we get a finite solution as we should. If we choose the appropriate equation for a divergent series, we get infinity as we should when x is greater than or equal to 1. Albeit, our system of equations doesn't cover x when it is equal to or less than -1. Let's start with the case where x equals -1:

Equation 10 doesn't converge, nor does it blow up to infinity. The partial sums oscillate between 1 and 0 forever. Mathematical physicists like to take the average of .5. However, it is obvious there are two solutions: 1 and 0. In the case where x is less than -1, the partial sums oscillate toward infinity and minus infinity. Again, there are two solutions. For negative numbers less than or equal to -1 we need two equations:

The complete system of equations is as follows:

Where x is less than or equal to -1, we can take advantage of a peculiar property of infinity: Infinity minus infinity doesn't have to equal zero, since infinity equals any number plus infinity:

Thus when x is less than -1 we can take the average of the two solutions s1 and s2 and get a finite solution. In fact, since n can be any finite number, there can be an infinite number of finite solutions.

Where x equals -1, there is one finite solution: .5.

Now, given what we know about infinity, finite numbers and the geometric series rules, what kind of solution(s) can we extract from the following series:

This is a tricky one, since there is no x variable to tell us which equation to use. So let's go with the usual method. We first take the average of the "a" series:

Then we do some fancy Ramanujan algebra to get the average of the "b" series:

Now that we have the b-series average, we can find solutions to the c-series:

At 38 and 39 above, we have two solutions: infinity and -1/12. One solution seems correct and the other seems counter-intuitive. Are they both valid? There is a test: simply sum up the series. The fastest way to do that is find the average number in the series, then multiply it by the number of terms. The average number and the number of terms are both infinity:

As expected, c = infinity. For any series, a good rule of thumb is if the series is divergent, a finite solution is invalid and just plain crazy! We can tell if a series equals infinity by looking at the last term. If the last term is infinite, then the series sum is infinite because infinity plus a bunch of other numbers equals infinity. We get a finite solution because the series itself does not know if it is convergent or divergent. Both convergent and divergent series are structured the same way: a sum of numbers--so the algebra is the same for both and yields more than one solution. Thus it is up to us to take a further step and test each solution.

N-body Problem Solutions

ABSTRACT:

This paper reveals a paradox within our current understanding of gravity that makes problems involving more than two bodies seemingly unsolvable. By introducing the alpha factor, the paradox is resolved and the solutions to various n-body problems become clear.

Imagine a skydiver falling to the earth's surface. Normally we can treat this as a 2-body problem--the two bodies being the skydiver and the earth. But suppose we didn't have the benefit of Newton's formula? Suppose we see this as an n-body problem because the earth is made up of n-1 particles, each with an individual mass and distance from the skydiver. The position and momentum of each particle, of course, is uncertain, so let's settle for the expectation values of the particles' positions, i.e., where the particles are most likely to be:

The diagram below is a representation of the distance vectors for each particle to the skydiver (green dot):

To simplify the problem, we determine the average position of the expected value positions of the particles (see blue dots below). Our n-body problem becomes a 5-body problem.

At equation 1 below we sum the individual masses to get the total mass M. At equations 2 through 4 we calculate the resultant vector R which shows us which way the skydiver will move. As it turns out, R is the distance between the skydiver and the earth's center of mass. Via observations we determine the constant G. This leads us to equation 5 which is Newton's formula.

Thus, we have solved an n-body problem along an x-y plane. We could have included the z-axis, but it wasn't necessary. It is often possible to rotate or orient the coordinate system so the problem can be solved in the simpler 2D format.

Note that the direction the skydiver falls is down, or along the y-axis. This is because the x directions cancel each other. The skydiver's movement along x is always zero. This presents a paradox, however! The diagrams below demonstrate that the skydiver will always fall at the same rate no matter how far apart along x the red and blue masses are from each other. The distance to the mass center is always the same.

Here's a 3-body problem: Imagine the red and blue bodies above are an infinite distance apart. The center of mass between them is the same as when the two bodies are both located at the center of mass. In fact, all the earth's particles gravitationally behave as if they are all located at the center of mass, no matter how close or far away they are from the sky diver. Yet, the inverse square law tells us that distance matters. Paradoxically, if distance matters, then Newton's formula, as it is normally applied, wouldn't give us the right answer. It would make more sense to simply sum up each particle's gravitational impact on the skydiver.

Unfortunately, in earth's case, Newton's formula gives us the right answer if we pretend that all the earth's particles are located at the center of mass. It is also unfortunate that it won't enable us to solve the 3-body problem suggested in the above diagrams. Newton's method predicts gravitational acceleration is the same whether the red and blue bodies are close to the green body or far away. Albeit, perhaps there's a way to solve the paradox. Here is my proposed solution:

Equations 6 and 7 above show how the skydiver's position changes along x (delta-x) and along y (delta-y). At the right side of equation 6, dt is the increment of time; X/|X| is the direction along the x-axis; X^2/R^2 is the fraction of acceleration along x; then there's the summation of alpha factors, body masses, and r distances each body is from the skydiver. Ditto for the y-axis at equation 7.

Now, I've introduced something new: the alphas. So what exactly are the alphas and what is their purpose? Examine equations 9 and 10 below:

If each alpha has the value suggested at 10, then the inner product yields the familiar Newton's equation. If each alpha equals one, we get a different outcome:

At equation 11 above we have a solution that makes sense for our paradoxical 3-body problem. So when is it appropriate to use equation 11 instead of equation 10? Both the earth and the 3-body problem appear similar, but there is a difference. This difference becomes more obvious if we think of the earth as accelerating towards the skydiver rather than vice versa. Earth's particles, for the most part, move as one towards the skydiver, so they all move along the same vector instead of individual vectors. If, for example, one or more of the earth's particles are perturbed by an outside force, the remaining particles will also very likely accelerate along the same vector as the perturbed particles. By contrast, if, say, an outside force perturbed the red body in our 3-body problem, the blue body need not move in lockstep with the red body. The two bodies are more independent. The value we assign to each alpha is determined by how independently the bodies move.

Equations 6 and 7 seem fine for falling skydivers, but what about orbiting satellites? Equations 12 through 17 below provide the relevant math for orbiting bodies:

Now, that an n-body solution has been proposed, let's see if we can prove it mathematically. At 18 below we start with what we know: Newton's equaton, then go from there. The final result is equation 24.

Below we use our new formula to solve the seemingly unsolvable 3-body problem. All the alphas are set to 1. The distance between the skydiver and the center of mass is still the same, but the red and blue bodies don't have to move in lockstep, so their distance apart now matters.

Equations 29 and 30 above tell us how much the skydiver moves along x and y during time interval dt. To see some animated simulations of 3-body and 6-body solutions, visit https://garmichaels.pythonanywhere.com/n_body_index/.

References: ChatGPT4 and Wikipedia.com