Johnny5

Suppose i have the following function of the variable z: [math] f(z) = \frac{z^2-1}{z-1} [/math] Is it true or false that f(1) = 2 ?

Years ago' date=' I answered this exact question, by drawing a ray diagram, and then working out the answer mathematically. If you double the distance between you and a tree, the tree shrinks in size by a factor of 2. Let me see if i can find a diagram on google, which leads to the right conclusion. I can't find the answer, but lets see if i can explain. Suppose that you are standing on the ground, and there is a giant california redwood a distance R away from you. So you can draw a triangle, which goes from the base of the tree into your eye, and from the top of the tree into your eye. Now, from a physics standpoint, here is what actually happens. Photons emitted from the surface of the tree go outwards in all directions. That's why everyone can see the tree no matter where they are standing. But the image that your brain ends up processing, is based upon only the actual specific photons that actually enter your eye. These photons travel for all intent and purpose, in a straight line from the atom on the tree which emits them, directly to your eye, and so the sides of the triangle are drawn as straight lines. For the sake of simplicity, suppose that the triangle is a right triangle. The actual height of the tree will be denoted by h, and the base of the triangle will be denoted by R. eye)______________________________| h R Let the angle the hypotenuse, and side R make with your eye be denoted by [math] \theta [/math] Now, suppose you move closer to the tree, to a distance R/2. eye)_______________| h r=R/2 Even though the actual height h of the tree does not change, the tree will now appear larger, you know this is true based upon sensory perception. Your question is merely why. The way that your question gets answered properly, requires that you learn something about lenses, and what they do, because there are lenses in your eye. Let me see if i can find a relevant article on lenses. Here is a site which allows you to play with the positions of an object, and a lens, and then shows you the image height after the rays pass through the lens. This is the kind of thing you need to learn, in order to have the answer. Regards

Ok, I've done Fourier series before, and now that you mention it, I recall that you can get a whole bunch of series in terms of pi. It was in my PDE book, but i dont know where that is at the moment. But i know what you're talking about. But, off the top of my head, I do not remember the entire comprehensive analysis given in that PDE book. I remember off the top of my head some of what has to be done. There are a series of integrals that you have to understand. sin(mx)cos(nx)dx sin(mx)sin(nx)dx cos(mx)cos(nx)dx And as i recall, the answers are either zero or one. And I solved them using integration by parts. Then lets see. The idea behind fourier series goes like this. Suppose that you have some function f(x). Suppose next that it has a Fourier series expansion on some interval L. So we have something like this going on: -L/2 < x < L/2 Possible Fourier sine series: [math] f(x) = \sum_{n=0}^{n=\infty} c_n sin (nx) [/math] Possible Fourier cosine series: [math] f(x) = \sum_{n=0}^{n=\infty} c_n cos (nx) [/math] Now, you have to work out the formulas for the Fourier coefficients, which is what takes the most time. All that was without googling.

Ok pi^3/32, is there an easy way to show this? How exactly does pi enter the answer?

I'm not sure what different methods there are out there to compute the following series: [math] \sum_{n=0}^{n=\infty} \frac{(-1)^n}{(2n+1)^3} [/math]

Surely you're joking. I thought it was quite appropriate to the thread, and I quite enjoyed it myself. Anyways, i showed what I wanted to show, which was a computation of square root two.

You're welcome. To understand more, google for Newton's binomial theorem, power series, convergence of series.

UT1?

When working with series expressions' date=' there is something called the radius of convergence. Consider again the following: [math'] (1+x)^{\alpha}= 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k} [/math] Now, focus on just the series formula: [math] \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k} [/math] Now, look at the function of k, inside the iterated product: [math] f(k) \equiv \frac{\alpha +1-k}{k} [/math] Whether or not any given series converges, can be determined by finding its radius of convergence. Define the radius of convergence as follows: [math] \rho \equiv |\lim_{k \to \infty} f(k)| [/math] Now, in the case of Newtons Binomial expansion we have: [math] \rho \equiv |\lim_{k \to \infty} \frac{\alpha +1-k}{k}| = |-1| = 1 [/math] So, the radius of convergence is one, just as Matt said. Now, it turns out that by taking the limit above, you are actually carrying out the very well known "ratio test" for convergence. Here is a link ratio test. Consider example 14.1.7 in the link above. You are faced with the following series: [math] \sum_{n=1}^{n = \infty} \frac{1}{n^2} [/math] We have: [math] \sum_{n=1}^{n = \infty} \frac{1}{n^2} = 1 + \sum_{n=2}^{n = \infty} \frac{1}{n^2} = 1 + \sum_{n=2}^{n = \infty} \frac{(n-1)!(n-1)!}{n^2(n-1)!(n-1)!}[/math] Now, [math] \sum_{n=2}^{n = \infty} \frac{(n-1)!(n-1)!}{n^2(n-1)!(n-1)!} = \sum_{n=2}^{n = \infty} \frac{(n-1)!(n-1)!}{n! n!}[/math] And [math] \sum_{n=2}^{n = \infty} \frac{(n-1)!(n-1)!}{n! n!} = \sum_{n=2}^{n= \infty} \prod_{k=1}^{k=n} (\frac{k-1}{k})^2 [/math] Thus, we have: [math] \sum_{n=1}^{n = \infty} \frac{1}{n^2} = 1+ \sum_{n=2}^{n= \infty} \prod_{k=1}^{k=n} (\frac{k-1}{k})^2 [/math] Now consider the limit of f(k), as k approaches infinity. [math] \lim_{k \to \infty} (\frac{k-1}{k})^2 = 1 [/math] Now, look at excercise 7.8 here. Let us consider their first example: [math] \sum_{n=1}^{n=\infty} \frac{x^n}{2^n} [/math] For the series above, we have: [math] \sum_{n=1}^{n=\infty} \frac{x^n}{2^n} = \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{1}{2} [/math] Now, investigate the limit as k approaches infinity of f(k). [math] \lim_{k \to \infty} \frac{1}{2} = \frac{1}{2} [/math] The limit above, is one over the radius of convergence, so that: R =2.

I looked back at the site for "coincidence files" and didn't find them. However, I did find this: What are they referring to as "the inertial reference frame"??? As if there is only one. ? And also, what are/where are the coincidence files? Regards

I ran out of time to edit the previous post. The main thing I wanted to show you, was how to use Newton's binomial formula to compute square root of two. In the previous post you will find the following formula: [math] (1+x)^{\alpha}= 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k} [/math] Now, let alpha=1/2, and let x=1 in the formula above to obtain: [math] (1+1)^{\frac{1}{2}}= 1+ \sum_{n=1}^{n=\infty} 1^n \prod_{k=1}^{k=n} \frac{\frac{1}{2} +1-k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1+ \sum_{n=1}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1 + \prod_{k=1}^{k=1} \frac{\frac{3}{2} - k}{k}+ \sum_{n=2}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1 + (\frac{3}{2} - 1) + \sum_{n=2}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.5 + \sum_{n=2}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.5 + \prod_{k=1}^{k=2} \frac{\frac{3}{2} - k}{k} + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.5 + (\frac{3}{2} - 1)( \frac{\frac{3}{2} - 2}{2}) + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.5 - (.5) (\frac{1}{4}) + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] Now, (.5)(1/4)=(.5)(.25)=(.125) from which it follows that: [math] \sqrt{2} = 1.5 - .125 + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.375 + \sum_{n=3}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.375 + \prod_{k=1}^{k=3} \frac{\frac{3}{2} - k}{k}+\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.375 + (-1.25)\prod_{k=3}^{k=3} \frac{\frac{3}{2} - k}{k}+\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.375 + (-1.25) \frac{\frac{3}{2} - 3}{3}+\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.375 + (-1.25)(-.5) +\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] Now, (-1.25)(-.5) = .0625. Hence we have: [math] \sqrt{2} = 1.375 + .0625 +\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4375 +\sum_{n=4}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4375 + \prod_{k=1}^{k=4} \frac{\frac{3}{2} - k}{k}+ \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4375 + (.0625) \prod_{k=4}^{k=4} \frac{\frac{3}{2} - k}{k}+ \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4375 + (.0625)( \frac{\frac{3}{2} - 4}{4}) + \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4375 + (.0625)(-.625) + \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4375 + (-.0390625) + \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.3984375 + \sum_{n=5}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.3984375 + \prod_{k=1}^{k=5} \frac{\frac{3}{2} - k}{k} + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.3984375 + (-.0390625)\prod_{k=5}^{k=5} \frac{\frac{3}{2} - k}{k} + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.3984375 + (-.0390625)( \frac{\frac{3}{2} - 5}{5}) + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.3984375 + (-.0390625)(-.7) + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.3984375 + (.02734375) + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.42578125 + \sum_{n=6}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.42578125 + (-.0205078125) + \sum_{n=7}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4052734375 + \sum_{n=7}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4052734375 +(-.0205078125)(-.7857142857)+ \sum_{n=8}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4052734375 +(.01611328125)+ \sum_{n=8}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.42138671875+ \sum_{n=8}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.42193603515625 +(.01611328125)(-.8125)+ \sum_{n=9}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.42193603515625 -(.01309204101562)+ \sum_{n=9}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.40884399414062 + \sum_{n=9}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] And [math] \sum_{n=9}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} = (-.01309204101562)(-.8333...)+\sum_{n=10}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.40884399414062 + (.01091003417968)+ \sum_{n=10}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] From which it follows that: [math] \sqrt{2} = 1.4197540283203 + \sum_{n=10}^{n=\infty} \prod_{k=1}^{k=n} \frac{\frac{3}{2} - k}{k} [/math] So as you can see, the infinite series really does converge to what your calculator gives you, namely: [math] \sqrt{2} = 1.4142135623731... [/math] Which was the main point of the exercise.

No it is easy, once you understand your first problem. let me finish the computation of square root of two for you. Now' date=' focus on the following quantity: [math'] \prod_{k=0}^{k=1} \frac{\frac{3}{2}-k}{k} [/math] When k=0, we appear to have division by zero error, however, this is avoided because 0! is defined to be equal to 1. This requires some explanation though. Look at 1/n! for a moment. [math] \frac{1}{n!} = \prod_{k=1}^{k=n} \frac{1}{k} [/math] In the case where n=3, we have: [math] \frac{1}{3!} = \prod_{k=1}^{k=3} \frac{1}{k} = (\frac{1}{1}) \prod_{k=2}^{k=3} \frac{1}{k} = (\frac{1}{1}) (\frac{1}{2}) \prod_{k=3}^{k=3} \frac{1}{k} = (\frac{1}{1}) (\frac{1}{2}) (\frac{1}{3}) = \frac{1}{6} [/math] Now, consider again the definition of n! [math] \frac{1}{n!} = \prod_{k=1}^{k=n} \frac{1}{k} [/math] Let n=0. Hence we have: [math] \frac{1}{0!} = \prod_{k=1}^{k=0} \frac{1}{k} [/math] By definition we have: [math] 0! \equiv 1 [/math] Therefore: [math] \frac{1}{1} = 1 = \prod_{k=1}^{k=0} \frac{1}{k} [/math] Multiplication is commutative therefore: [math] \prod_{k=1}^{k=0} \frac{1}{k} = \prod_{k=0}^{k=1} \frac{1}{k}[/math] Now, if we expand the product, we have: [math] \prod_{k=0}^{k=1} \frac{1}{k} = (\frac{1}{0}) (\frac{1}{1}) [/math] By transitivity, it now follows that: [math] 1 = \frac{1}{0} [/math] Which is false. This contradiction was caused by the definition 0!=1. However let us look at our situation... In the case where the upper indice is equal to n, we will want to evaluate the following iterated product: [math] \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} [/math] If we pull out the k from the denominator, we have this: [math] \frac{1}{n!} \prod_{k=0}^{k=n} \frac{3}{2}-k [/math] So here is our series expression for square root of two again: [math] \sqrt{2} = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} [/math] If we pull the k out of the denominator we have: [math] \sqrt{2} = \sum_{n=0}^{n=\infty} \frac{1}{n!}\prod_{k=0}^{k=n} \frac{3}{2}-k [/math] The first term of the series is: [math] \frac{1}{0!}\prod_{k=0}^{k=0} \frac{3}{2}-k = \frac{1}{1} (\frac{3}{2}) = \frac{3}{2} [/math] But this leads to an error in Newton's binomial formula. Let me explain. Here is Newton's binomial formula again: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k} [/math] Consider the case where alpha = 2, and x is an arbitrary real number. By direct substitution, we have: [math] (1+x)^2 = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{2+1-k}{k} [/math] Using only the axioms of the real number system, we can figure out what the left hand side is. [math] (1+x)^2 = (1+x)(1+x) = 1+x+x+xx = 1+2x +x^2 [/math] Now figure out what the RHS must equal. [math] \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{2+1-k}{k} = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{3-k}{k} = \sum_{n=0}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k [/math] The first term of the series is: [math] x^0 \frac{1}{0!} \prod_{k=0}^{k=0} 3-k = x^0 (3-0) = 3x^0 [/math] Using the axioms of algebra, it is easy to prove that if not(x=0) then x^0 = 1. Thus, in the case where x isn't equal to zero, we must have: [math] 1+2x +x^2 = 3 + \sum_{n=1}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k [/math] Now, lets compute the term of the series, when n=1. When n=1, we have: [math] x^1 \frac{1}{1!} \prod_{k=0}^{k=1} 3-k = x (3-0)(3-1) = 6x [/math] Thus, we have: [math] 1+2x +x^2 = 3 + 6x + \sum_{n=2}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k [/math] Now, evaluate the series when n=2. We obtain: [math] x^2 \frac{1}{2!} \prod_{k=0}^{k=2} 3-k = \frac{x^2}{2} (3-0)(3-1)(3-2) = 3x^2 [/math] From which it now follows that: [math] 1+2x +x^2 = 3 + 6x + 3 x^2 + \sum_{n=3}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k [/math] Now, whenever n is greater than or equal to 3, there will be multiplcation by zero since: [math] \prod_{k=3}^{k=3} 3-k = (3-3)=0 [/math] Hence there are only three terms in the expansion formula, because: [math] 0 = \sum_{n=3}^{n=\infty} x^n \frac{1}{n!} \prod_{k=0}^{k=n} 3-k [/math] Thus, we have: [math] 1+2x +x^2 = 3 + 6x + 3 x^2 [/math] Which is false. However, it was almost correct. If the RHS is divided by three instead, we would have had: [math] 1+2x +x^2 = 1 + 2x + x^2 [/math] which is true. Thus, the RHS was off by multiplication of the constant 1/3, but otherwise the series-product expansion formula worked. Now, look again at the formula for (1+x)^alpha: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k} [/math] Suppose instead, that we start off the multiplication from 1, instead of zero, like this: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k} [/math] Let alpha=2, so that we have: [math] (1+x)^2 = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{2+1-k}{k} [/math] We already know that the LHS is equal to 1+2x+x^2, so let us write: [math] 1 + 2x + x^2 = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{2+1-k}{k} [/math] And 2+1=3, so that we have: [math] 1 + 2x + x^2 = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{3-k}{k} [/math] Now, we just have to expand the RHS correctly, so that we obtain equality of the RHS with the LHS. Let us pull out the k from the denominator of the iterated product, so that we have: [math] 1 + 2x + x^2 = \sum_{n=0}^{n=\infty} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k [/math] In the case where x=0, we can readily see that the LHS is equal to 1, from the algebraic field axioms. So in order to have equality, rather than a false statement, the RHS must be equal to 1, in the case where x=0. We know that the series has at most three terms, since when n is greater than or equal to 3, we all higher terms in the series are multiplied by (3-3)=0, due to the iterated product. So without loss of generality, we can write: [math] 1 + 2x + x^2 = \sum_{n=0}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k [/math] The first term of the series is the only term in the series which isnt multiplied by x, and so must be equal to 1, in order for the RHS to equal the LHS in the case where x=0, so we can write: [math] 1 + 2x + x^2 = \frac{0^0}{0! }+ \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k [/math] So that, regardless of the value of x, it needs to be true that: [math] 1 = \frac{0^0}{0!} [/math] Thus, we can write: [math] 1 + 2x + x^2 = 1 + \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k [/math] Now, we just have to expand the remaining terms of the series, to see that the RHS equals the LHS. This time however, the product starts off with k=1, instead of k=0. Thus, we have: [math] \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k = \frac{x^1}{1!} (3-1) + \sum_{n=2}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k [/math] Thus, we have: [math] \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k = 2x + \frac{x^2}{2!} \prod_{k=1}^{k=2} 3-k [/math] Thus, we have: [math] \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k = 2x + \frac{x^2}{2!} (3-1)(3-2) = 2x+ x^2 [/math] And from an earlier line of work we already had: [math] 1 + 2x + x^2 = 1 + \sum_{n=1}^{n=2} \frac{x^n}{n!} \prod_{k=1}^{k=n} 3-k [/math] So that by substition, the remaining two terms of the series on the RHS above, are 2x+x^2, so that we have: [math] 1 + 2x + x^2 = 1 + 2x + x^2 [/math] So it worked, and all is well, provided that we start off the multiplication from k=1, instead of k=0, and that we have: [math] 1 = \frac{0^0}{0!} [/math] Thus, we can write Newton's expansion formula as follows: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k} [/math] Now, because we have started off the multiplication from 1, instead of zero, we are not faced with division by zero error. The only problem is the formula is not as compact as we would like it to be. Before going any further, we can try to use the formula above to compute the square root of two, which is the main point of this post. The secondary point, is to show you how to manipulate the series expressions. Starting off with the formula [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k} [/math] Let x=1, and let alpha = 1/2, and then write this: [math] (1+1)^{\frac{1}{2}}= 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\frac{1}{2} +1-k}{k} [/math] However if we stipulate that: [math] 1 = x^0 \prod_{k=1}^{k=0} \frac{\alpha +1-k}{k} [/math] Then we can write: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k} [/math] And this stipulation had to be made, because of the case where x=0, so that what we really need to stipulate is that: [math] 1 = 0^0 \prod_{k=1}^{k=0} \frac{\alpha +1-k}{k} [/math] However, from the top indice, you can see that we have division by zero error. Let us pull out the k, as n!, so that we have this: [math] 1 = \frac{x^0}{n!} \prod_{k=1}^{k=0} \alpha +1-k [/math] In fact, lets go back and write this for Newton's formula: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \prod_{k=1}^{k=n} \alpha +1-k [/math] Which, provided that not(alpha=-1), can also be written as: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] Now, focus on this part of the formula: [math] \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] If it is permissible for n=0, then we have the zeroth term as: [math] \frac{x^0}{0!} \frac{1}{\alpha+1} \prod_{k=0}^{k=0} \alpha +1-k [/math] Which simplifies slightly to: [math] \frac{x^0}{0!} \frac{1}{\alpha+1} (\alpha +1) = \frac{x^0}{0!} [/math] So lets consider carefully where we are: We are sure that this much is algebraically correct: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] All we want to do is compactify the formula above. Suppose that we stipulate that the n=0 term of the series is equal to 1, then we can write: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] Then, explicitely when n=0 we will have to evaluate: [math] \frac{x^0}{0!} \frac{1}{\alpha+1} \prod_{k=0}^{k=0} \alpha +1-k [/math] And it will have to be the case, that the above formula is equal to 1, so that we must force it to be the case that: [math] 1 = \frac{x^0}{0!} \frac{1}{\alpha+1} \prod_{k=0}^{k=0} \alpha +1-k [/math] The iterated product only has one term, that for which k=0, so that we may write: [math] 1 = \frac{x^0}{0!} \frac{1}{\alpha+1} (\alpha +1-0) [/math] From which it follows that: [math] 1 = \frac{x^0}{0!} [/math] In any case where not(x=0), the numerator of the fraction above is equal to 1, so that in order for Newton's expansion formula to be correct, we must have: [math] 1 = \frac{1}{0!} [/math] From which it follows that we must define 0 factorial to be equal to 1. Let us look at the formula again, and consider the case where x=-1, and alpha =0. [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] So that [math] (1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{(-1)^n}{n!} \frac{1}{1} \prod_{k=0}^{k=n} 1-k [/math] Which can be written as follows: [math] (1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{(-1)^n}{n!} \frac{1}{1} \prod_{k=0}^{k=n} -(k-1) [/math] Which can be written as follows: [math] (1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{(-1)^n}{n!} \frac{1}{1} (-1)^n \prod_{k=0}^{k=n} (k-1) [/math] Which can be written as follows: [math] (1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{(-1)^{2n}}{n!} \frac{1}{1}\prod_{k=0}^{k=n} (k-1) [/math] Which can be written as follows: [math] (1-1)^0 = 0^0 = 1+ \sum_{n=1}^{n=\infty} \frac{1}{n!} \frac{1}{1}\prod_{k=0}^{k=n} (k-1) [/math] From which it follows that: [math] 0^0 = 1+ \frac{1}{1!} \prod_{k=0}^{k=1} (k-1) + \sum_{n=2}^{n=\infty} \frac{1}{n!} \prod_{k=0}^{k=n} (k-1) [/math] Now, when k=1, (k-1)=0, so that we must have: [math] 0^0 = 1 [/math] So look at the formula again: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] Now, look at the definition of n factorial for a moment: [math] n! = \prod_{k=1}^{k=n} k [/math] We can perform an indicial operation and write: [math] n! = \prod_{k=0}^{k=n-1} k+1 [/math] From which it follows that: [math] n! = \frac{1}{n+1}\prod_{k=0}^{k=n} k+1 [/math] Hence, we can write the formula as follows: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{\frac{1}{n+1}\prod_{k=0}^{k=n} k+1} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] Or rather: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{(n+1) x^n}{\prod_{k=0}^{k=n} k+1} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] Or rather: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} (n+1) x^n \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1} [/math] Now, focus on this part of the formula for a moment: [math] \sum_{n=1}^{n=\infty} (n+1) x^n \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1} [/math] If it is permissible for n=0, then the zeroth term of the series above would be given by: [math] (0+1) x^0 \frac{1}{\alpha+1} \prod_{k=0}^{k=0} \frac{\alpha +1-k}{k+1} [/math] Which can be simplified to: [math] x^0 \frac{1}{\alpha+1} \frac{\alpha +1-0}{0+1} [/math] Which can be simplified to: [math] x^0 \frac{\alpha +1}{\alpha +1} [/math] Which can be simplified to: [math] x^0 [/math] So let us begin with the following formula we know to be correct, in any case where not(alpha = -1). [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} (n+1) x^n \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1} [/math] If we stipulate that x^0 = 1, for any x, we can write this instead: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} (n+1) x^n \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1} [/math] As I just showed moments ago. Which for the sake of reading clarity can be written as follows: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \frac{n+1}{\alpha+1} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k+1} [/math] Now, when n=0, the first term of the iterated product is given by: [math] \frac{\alpha +1-0}{0+1} = \alpha +1 = \prod_{k=0}^{k=0} \frac{\alpha +1-k}{k+1} [/math] Hence we can write Newton's binomial formula as follows: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \frac{n+1}{\alpha+1} (\alpha+1)\prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1} [/math] Which simplifies slightly to: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} (n+1) x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1} [/math] Thus, I have now removed any possiblity of division by zero error, in the case where alpha=-1. In other words, using an indicial operation, i took out 1/(1+alpha) and opened up the possibility of division by zero error, but now I've put it back into the iterated product, so that we no longer have to worry about whether or not alpha = -1 in our formula. Thus, the following formula is correct: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} (n+1) x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1} [/math] Which can be written as follows: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} \frac{d}{dx} (x^{n+1}) \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1} [/math] Which can be written as: [math] (1+x)^\alpha = \frac{d}{dx} \sum_{n=0}^{n=\infty} x^{n+1} \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k+1} [/math] Now, lets perform an indicial operation, and write the above formula as follows: [math] (1+x)^\alpha = \frac{d}{dx} \sum_{n=1}^{n=\infty} x^{n} \prod_{k=1}^{k=n-1} \frac{\alpha +1-k}{k+1} [/math] Now, define U(x) as follows: [math] U(x) \equiv \sum_{n=1}^{n=\infty} x^{n} \prod_{k=1}^{k=n-1} \frac{\alpha +1-k}{k+1} [/math] So that we can now write: [math] (1+x)^\alpha = \frac{dU}{dx} [/math] Now, focus on U(x) for a moment... [math] U(x) \equiv \sum_{n=1}^{n=\infty} x^{n} \prod_{k=1}^{k=n-1} \frac{\alpha +1-k}{k+1} [/math] The idea is that if we differentiate U(x), we have to get (1+x)^alpha. Now, go back and look at the following formula: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{1}{\alpha+1} \prod_{k=0}^{k=n} \alpha +1-k [/math] We can rewrite it as follows: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \frac{\alpha + 1}{\alpha+1} \prod_{k=1}^{k=n} \alpha +1-k [/math] Which can be simplified to: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} \frac{x^n}{n!} \prod_{k=1}^{k=n} \alpha +1-k [/math] Which can be simplified to: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{\alpha +1-k}{k} [/math] Here is U(x) again: [math] U(x) \equiv \sum_{n=1}^{n=\infty} x^{n} \prod_{k=1}^{k=n-1} \frac{\alpha +1-k}{k+1} [/math] We have: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \prod_{k=0}^{k=n-1} \frac{\alpha -k}{k+1} [/math] Or equivalently: [math] (1+x)^\alpha = 1+ \sum_{n=1}^{n=\infty} x^n \alpha \prod_{k=1}^{k=n-1} \frac{\alpha -k}{k+1} [/math] Now, let alpha become alpha +1, so that we have: [math] (1+x)^{\alpha +1}= 1+ \sum_{n=1}^{n=\infty} x^n (\alpha +1) \prod_{k=1}^{k=n-1} \frac{\alpha +1 - k}{k+1} [/math] Or equivalently: [math] (1+x)^{\alpha +1}= 1+ (\alpha +1)\sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n-1} \frac{\alpha +1 - k}{k+1} [/math] From which it now follows using the definition of U(x) that: [math] (1+x)^{\alpha +1}= 1+ (\alpha +1) U(x) [/math] Now, we can write the above equation as follows: [math] (1+x)(1+x)^\alpha= 1+ (\alpha +1) U(x) [/math] And recall the earlier formula that: [math] (1+x)^\alpha = \frac{dU}{dx} [/math] Thus, we can write the following differential equation: [math] (1+x) \frac{dU}{dx} = 1+ (\alpha +1) U(x) [/math]

[math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k} [/math] Let x=1, therefore: [math] (1+1)^\alpha = 2^\alpha = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\alpha +1-k}{k} [/math] Let alpha =1/2, therefore: [math] 2^{\frac{1}{2}} = \sqrt{2} = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{1}{2}+1-k}{k} [/math] Therefore: [math] \sqrt{2} = \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} [/math] Obtain the first five terms of the series. [math] \sum_{n=0}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} = 1+\sum_{n=1}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k}[/math] [math] \sum_{n=1}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} = \prod_{k=0}^{k=1} \frac{\frac{3}{2}-k}{k} + \sum_{n=2}^{n=\infty} \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} [/math]

Pretty much i agree with that. Even though no one's perfect, it's fun to try though. Regards. Also, i was just reading this: And also read that a sequence has a limit only if it's a Cauchy sequence. I did a hundred or so proofs using that definition above, but its been quite awhile since i've done one. Actually, maybe far more than that, difficult to say. Might have been 500, but my point is, that it's been awhile.

No i see that, it's in post 23. As I said somewhere, its been over a decade since I took advanced calculus. I did follow your proof. The only thing I had a problem with, was when you wrote >, instead of >, but that is of course just a minor error on your part. Regards PS: I'm glad to see the material again, its been a long time. You did great. Thank you

Very well... Here is the abstract from the link in the first post of this thread:

Well then we can discuss the .pdf, starting with the beginning, and moving fowards, I just read the opening few paragraphs, and they start at SR, in order to introduce tensors. So maybe we should start there. Since I don't already know tensor calculus, you can teach me that. I know we've discussed tensors before, and it will be interesting to see how much i remember. Additionally, since the last time we talked, i discovered that it was Hamilton who introduced the term 'tensor' in his lectures on quaternions, in the 1800's. So how about start off with a bit of tensor analysis?

A couple of reasons. For one, i was hoping to discuss it with someone already familiar with it, and two I had some followup questions.

Let me see if i can reproduce your argument in this post. We want to investigate the behavior of [math] (n!)^{\frac{1}{n}} [/math] As n increases without bound. The above formula actually represents a sequence. The domain of which is the set of natural numbers: [math] \mathbb{N} = \mathcal{f}1,2,3,4... \mathcal{g} [/math] So the first term of the sequence is: [math] (1!)^{\frac{1}{1}} = 1!= 1 [/math] The second term of the sequence is: [math] (2!)^{\frac{1}{2}} = 2^{\frac{1}{2}} = \sqrt{2} =1.414...[/math] The third term of the sequence is: [math] (3!)^{\frac{1}{2}} = 6^{\frac{1}{3}} = 1.817... [/math] The fourth term of the sequence is: [math] (4!)^{\frac{1}{2}} = 24^{\frac{1}{4}} = 2.213... [/math] So as you see, the terms of the sequence are increasing, as n increases. We can represent the sequence as follows: [math] (1,\sqrt{2},6^{\frac{1}{3}}, 24^{\frac{1}{4}},... (n!)^{\frac{1}{n}},...) [/math] So we are out to determine the behavior of the sequence whose nth term is [math] (n!)^{\frac{1}{n}} [/math] as n tends to infinity. In other words, we want to compute the following limit: [math] \lim_{n \to \infty} (n!)^{\frac{1}{n}} [/math] Here is Matt's argument that the sequence diverges. [math] n! = n(n-1)(n-2)...(k+2)(k+1)k(k-1)(k-2)...(4)(3)(2)(1) [/math] Let k denote an arbitrary positive integer. Therefore: [math] n! = n(n-1)...(k+1)(k!) [/math] Or equivalently... [math] n! = (k!)(k+1)(k+2)...(k+(n-k)) [/math] Let me change to product notation. [math] n! \equiv \prod_{j=1}^{j=n} j = (\prod_{j=1}^{j=k} j)(\prod_{j=k+1}^{j=n} j) = k! \prod_{j=k+1}^{j=n} j [/math] Now, [math] \prod_{j=k+1}^{j=n} j = (k+1)(k+2)...n [/math] Also: [math] \prod_{j=k+1}^{j=n} j = \prod_{j=k}^{j=n-1} (j+1) = \prod_{j=k-1}^{j=n-2} (j+2)[/math] So that: [math] \prod_{j=k+1}^{j=n} j = \prod_{j=1}^{j=n-k} (k+j) = (k+1)(k+2)...(k+(n-k)) [/math] So as you can see, the number of terms on the RHS above, is (n-k). That is, the number of terms in the iterated product is 1+(greater indice minus the lesser) i.e 1+(n-k -1). And each of those terms is necessarily greater than k. The point is that: [math] (k+1)(k+2)...(k+(n-k)) > (k)_1(k)_2(k)_3(k)_{n-k} = k^{n-k} [/math] So that: [math] n! = (k)!*(k+1)(k+2)....(k+(n-k)) > k!k^{n-k} [/math] Exactly as Matt has. So it follows that: [math] n! > k!k^{n-k} \geq k^{n-k} [/math] So it follows that: [math] (n!)^{\frac{1}{n}} \geq (k^{n-k})^{\frac{1}{n}} [/math] Therefore: [math] (n!)^{\frac{1}{n}} \geq k^{\frac{n-k}{n}} [/math] k was a given (henced fixed) arbitrary natural number. Taking the limit of both sides we have: [math] \lim_{n \to \infty} (n!)^{\frac{1}{n}} \geq \lim_{n \to \infty} k^{\frac{n-k}{n}} [/math] Now, focus on the RHS. [math] \lim_{n \to \infty} k^{\frac{n-k}{n}} = \lim_{n \to \infty} k^{(1-\frac{k}{n})}} = \lim_{n \to \infty}k(k^{-\frac{k}{n}}) = k\lim_{n \to \infty}(k^{-\frac{k}{n}}) [/math] [math] k \lim_{n \to \infty} k^{-\frac{k}{n}} = k \lim_{n \to \infty} \frac{1}{k^{k/n}} [/math] In the limit as n approaches infinity, the exponent k/n above, approaches zero, so that we have: [math] k \lim_{n \to \infty} \frac{1}{k^{k/n}} = k \frac{1}{k^{0}} = k [/math] Thus, we have found that: [math] \lim_{n \to \infty} (n!)^{\frac{1}{n}} \geq \lim_{n \to \infty} k^{\frac{n-k}{n}} = k [/math] That is: [math] \lim_{n \to \infty} (n!)^{\frac{1}{n}} \geq k [/math] Since k is an arbitrary natural number, it follows that k can be an arbitrarily large natural number, and hence the sequence [math] (n!)^{\frac{1}{n}} [/math] Can be arbitrarily large. Hence it is unbounded from above, so that the sequence diverges. And this was Matt's argument, and it looks good.

Yes, that sounds good, we can pick up from wherever.