Johnny5

Here is wolfram on partial sums Partial sums You can use the concept to move from Riemann sums, to the definition of a definite integral, if memory serves me. So, let me see if I can answer your question the way you want it answered... Using partial sums. Suppose you are given the following sequence: [math] a(k) [/math] The first term of the sequence is a(1), the second term of the sequence is a(2), the third term of the sequence is a(3), the nth term of the sequence is a(n), and so on... So lets take a randomly chosen specific example. Suppose that: [math] a(k) = 3k^2 -5k+2 [/math] We can begin to generate the terms of the sequence ourselves. a(1) = 3(1)(1)-5(1)+2=3-5+2=-2+2=0 a(2) = 3(2)(2)-5(2)+2=12-10+2=2+2=4 a(3) = 3(3)(3)-5(3)+2=27-15+2=14 a(4) = 3(4)(4)-5(4)+2=48-20+2=30 and so on So, we can represent the sequence as: (0,4,14,30,...) And we can use the method of "differencing" to go from the terms of the sequence back to the polynomial generating function, but I will save that for some other time. Definition Of Partial Sum Given a sequence a(k) at random, the sum SN of the first N terms of the sequence, beginning at a(1), and terminating at a(N), is given by: [math] S_N \equiv \sum_{k=1}^{k=N} a(k) [/math] So let us define e^x, as follows: [math] e^x \equiv \sum_{n=0}^{n=\infty} \frac{x^n}{n!} [/math] and investigate the sequence of partial sums, given that a(k) = x^k/k!. The first term in the sequence of partial sums is S1. The second term in the sequence of partial sums is S2. The third term in the sequence of partial sums is S3. The fourth term in the sequence of partial sums is S4, and so on. [math] a(k) = \frac{x^k}{k!} [/math] a(1) = x a(2) = x^2/2! a(3) = x^3/3! a(4) = x^4/4! and so on S1= a(1)=x S2= a(1)+a(2)=x+x^2/2! S3= a(1)+a(2)+a(3) = x+x^2/2!+x^3/3! S4= a(1)+a(2)+a(3)+a(4) = x+x^2/2!+x^3/3! +x^4/4! and so on The sequence of partial sums is given by: (S1,S2,S3,...Sn,...) Now, you want to go from considering the sequence of partial sums, to first proving that e^x is differentiable, and secondly prove that e^x is equal to its own derivative. In order to prove that a function f(x) is differentiable at a point Z, you first have to check to see that f(Z) is defined. If it is, you then have to check to make sure that the limit as x approaches Z from the right, is equal to the limit as X approaches Z from the left. Let me check to make sure that's right. Here is something on Banach Spaces

Are you trying to imply that using a quadratic in the formulas, leads to something strange about time, i.e. it is two dimensional, when your intuition says it's one dimensional? What are you trying to imply with the above argument?

I am looking for a quick proof that [math] 0^0=1 [/math] some kind of argument, doesnt have to be fancy Thank you e.g. let x = 0^0 therefore ln x = 0 ln 0 It's provable from the field axioms that 0*y=0, for any number y. Hence ln x = 0 therefore x=1 QED I am looking for other proofs. PS: And i know that lim x-->0+ of x = -infinity [math] \lim_{x \to 0^+} ln x = - \infty [/math] That's why I want a different proof, because the above isn't one.

Don't you mean that the celestial meridian is your line of latitude projected onto the stars? You said longitude. Regards

Since his latex isn't working, i will fill in the missing step [math] e^x \equiv \sum_{n=0}^{n=\infty} \frac{x^n}{n!} =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+... [/math] Now, differentiate everything with respect to x, like so: [math] \frac{d(e^x)}{dx} \equiv \frac{d}{dx} \sum_{n=0}^{n=\infty} \frac{x^n}{n!} =\frac{d}{dx}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+...) [/math] Which leads to this: [math] \sum_{n=0}^{n=\infty}\frac{d}{dx} ( \frac{x^n}{n!} ) =1 +\frac{2x}{2!}+\frac{3x^2}{3!}+...+\frac{nx^{n-1}}{n!}+... [/math] Which leads to this: [math] \sum_{n=0}^{n=\infty} n\frac{x^{n-1}}{n!} = 1 +\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^{n-1}}{(n-1)!}+... [/math] Which leads to: [math] \sum_{n=0}^{n=\infty} \frac{x^{n-1}}{(n-1)!} = 1+x+\frac{x^2}{2!}+...+\frac{x^{n-1}}{(n-1)!}+... [/math] As you can see, the infinite series on the RHS is the original series, with the arbitrary term written as x^(n-1)/(n-1)! rather than x^n/n!... but the series is the same. Kind regards

It's going ok, I run through things at home at night. I have been looking for the book I first learned it from, all over the house in fact, and can't find it. I could finish that book in 2 days. Right now, all I have is Schaum's outline, but it will have to do if I can't find my old text. Well no that's not true, I have a few books, but I know which one would go the quickest. In all honesty, I have been learning quite a few new things about astronomy, which are fascinating. But this weekend I will have some time, so ask me the same question monday. Regards

... your line of longitude projected above you... ...the "celestial meridian" is just your line of longitude projected above you... your line of longitude, is a meridian around the earth. Suppose you are in Greenwich England. Your latitude is 51 degrees north, and your longitude is zero, because you are on the earth's prime meridian. Prime meridian is a great circle that wraps around the the earth. It passes through both earth's north pole, and earth's south pole. Now I want to project this great circle onto the celestial sphere? I think so... You said that there are three points which are on the "celestial meridian" The north pole of earth... projected up of course onto the sphere, and the south pole of earth down, and to fix it in place, the point directly above your head... which is called the zenith if I'm not mistaken. Did I get all that correctly? Thanks

I am not sure I understand your question. I think you are just asking how to integrate x times e raised to the 2x power. If that is all, there are several ways you can do this, I am not sure if it matters to you how you solve the problem. The first thing that should jump out at you, is that you can solve it using integration by parts. Another method, would be to convert the integrand to a series, then integrate term by term, and then convert the final answer back into elementary functions. I can do it both ways for you. [math] \int x e^{2x} dx [/math] Here is the integration by parts formula: [math] \int u dv = uv - \int v du [/math] In your specific problem, let u=x, and let v = 1/2 e^(2x) From which it follows that: du=dx and dv = e^(2x)dx From which it follows that: udv = x e^(2x)dx From which it follows that: [math] \int u dv = \int x e^{2x} dx [/math] So by the transitive property of equality: [math] \int x e^{2x} dx = uv - \int v du [/math] From which it follows that: [math] \int x e^{2x} dx = \frac{xe^{2x}}{2} - \frac{1}{2} \int e^{2x} dx [/math] And the integral of e^(2x)dx is just 1/2 e^(2x). Therefore: [math] \int x e^{2x} dx = \frac{xe^{2x}}{2} - \frac{1}{4} e^{2x} + C[/math] Where C is just an arbitrary constant of integration, and can be absorbed into the other constant. And I promised you another method. This method uses the fact that: [math] e^x = \sum_{n=0}^{n=\infty} \frac{x^n}{n!} [/math] For e^(2x), we have: [math] e^{2x} = \sum_{n=0}^{n=\infty} \frac{(2x)^n}{n!} [/math] Multiply both sides of the previous equation by x, to obtain: [math] xe^{2x} = x\sum_{n=0}^{n=\infty} \frac{(2x)^n}{n!} [/math] Summation is over n, and n is not dependent upon x, and therefore can pass through the summation sign. Therefore: [math] xe^{2x} = \sum_{n=0}^{n=\infty} x \frac{(2x)^n}{n!} [/math] Multiply both sides of the equation above by dx, to obtain: [math] xe^{2x} dx = \sum_{n=0}^{n=\infty} x \frac{(2x)^n}{n!} dx [/math] Now, integrate both sides of the previous equation, to obtain: [math] \int xe^{2x} dx = \int \sum_{n=0}^{n=\infty} x \frac{(2x)^n}{n!} dx [/math] Now, you can integrate term by term. The RHS is a series, so you have something like this: integral of [a+b+c+d+e...] and an integral of a sum, is the sum of the integrals, in other words: [math] \int (a+b+c) = \int a + \int b + \int c [/math] in words"integral of a sum is the sum of the integrals." This can be proven, but that is a different excercise. back to the problem... So we can pass the integral sign through the summation sign to obtain: [math] \int xe^{2x} dx = \sum_{n=0}^{n=\infty} \int x \frac{(2x)^n}{n!} dx [/math] Now, you see the (2x)^n quantity, and if the exterior was also (2x) then you could get (2x)^(n+1), so multiply by 2/2 to get: [math] \int xe^{2x} dx = \frac{1}{2} \sum_{n=0}^{n=\infty} \int 2x \frac{(2x)^n}{n!} dx [/math] Therefore: [math] \int xe^{2x} dx = \frac{1}{2} \sum_{n=0}^{n=\infty} \int \frac{(2x)^{n+1} }{n!} dx [/math] Therefore: [math] \int xe^{2x} dx = \frac{1}{2} \sum_{n=0}^{n=\infty} \frac{1}{n!} \int (2x)^{n+1} dx [/math] And the integral that must now be performed is trivial. For example suppose that n=3, then the integral would be: [math] \int (2x)^4 dx [/math] Let U=2x, therefore dU=2dx, therefore 1/2dU=dx, now substitute, to obtain: [math] \frac{1}{2} \int U^4 dU [/math] Which is just 1/2( U^5/5 ) So we can make that substitution for any n, therefore: Let U=2x [math] \int xe^{2x} dx = \frac{1}{2} \sum_{n=0}^{n=\infty} \frac{1}{n!} \int U^{n+1} \frac{1}{2} dU[/math] Therefore: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{1}{n!} \int U^{n+1} dU[/math] Therefore: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{1}{n!} [ \frac{U^{n+2}}{n+2} + C] [/math] Replacing U by 2x, we obtain: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{1}{n!} [ \frac{(2x)^{n+2}}{n+2} + C] [/math] Whence it follows that: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(2x)^{n+2}}{(n+2) n!} + C \sum_{n=0}^{n=\infty} \frac{1}{n!}} [/math] Where I let the arbitrary constant c absorb the constant 1/4. Now, the integral is totally over. We can leave the answer as is, or we can attempt to write it in terms of elementary functions. Although integration by parts was quicker, the method I've just shown is more powerful. Let me attempt to write the infinite sums in terms of elementary functions. [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(2x)^{n+2}}{(n+2) n!} + Ce} [/math] Where I have made use of the fact that: [math] e = \sum_{n=0}^{n=\infty} \frac{1}{n!} = 2.71828...[/math] Which is just a pure number, and can be absorbed in to the arbitrary constant of integration. So... [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(2x)^{n+2}}{(n+2) n!} + C} [/math] Now, multiply both the numerator and denominator by (n+1), like so: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(n+1)(2x)^{n+2}}{(n+2)(n+1) n!} + C} [/math] and (n+2)(n+1)n! is just n plus two factorial, that is: (n+2)(n+1)n! = (n+2)! Therefore: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=0}^{n=\infty} \frac{(n+1)(2x)^{n+2}}{(n+2)!} + C} [/math] From which it follows that: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(n-1)(2x)^n}{n!} + C} [/math] From which it follows that: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{n(2x)^n}{n!} - \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} + C} [/math] From which it follows that: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{(n-1)!} - \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} + C} [/math] Where I simply converted n! into n(n-1)!, and then cancelled out n, from the numerator and denominator. It now follows that: [math] \int xe^{2x} dx = \frac{1}{4} \sum_{n=1}^{n=\infty} \frac{(2x)^{n+1}}{n!} - \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} + C} [/math] It follows that: [math] \int xe^{2x} dx = \frac{1}{4} 2x \sum_{n=1}^{n=\infty} \frac{(2x)^n}{n!} - \frac{1}{4} \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} + C} [/math] We can now make use of the fact that: [math] e^{2x} = \sum_{n=0}^{n=\infty} \frac{(2x)^n}{n!} = 1+2x+ \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} [/math] And the previous fact leads to: [math] e^{2x} -1 - 2x = \sum_{n=2}^{n=\infty} \frac{(2x)^n}{n!} [/math] Which we can now use above, to obtain: [math] \int xe^{2x} dx = \frac{1}{4} 2x \sum_{n=1}^{n=\infty} \frac{(2x)^n}{n!} - \frac{1}{4} (e^{2x} -1 - 2x) + C} [/math] And since: [math] e^{2x} -1 = \sum_{n=1}^{n=\infty} \frac{(2x)^n}{n!} [/math] It follows that: [math] \int xe^{2x} dx = \frac{1}{4} 2x (e^{2x} -1) - \frac{1}{4} (e^{2x} -1 - 2x) + C} [/math] From which it follows that: [math] \int xe^{2x} dx = \frac{x}{2} e^{2x} - \frac{x}{2} -\frac{e^{2x}}{4} +\frac{1}{4} +\frac{x}{2} + C} [/math] From which it now follows that: [math] \int xe^{2x} dx = \frac{x}{2} e^{2x} -\frac{e^{2x}}{4} + C} [/math] Which is the exact same answer we got using integration by parts. QED Regards PS: By the way, the series solution was more for me to run through some things for myself, I recommend integration by parts for your original problem.

Dr Swanson, I think that's incorrect. I think the north south line intersects the umm celestial meridian, which is a great circle on the celestial sphere. But would different earth observers necessarily have different celestial meridians? Thank you

I am just beginning to try and understand things like solar day, sidereal day, apsides, aphelion, perihelion, vernal equinox, autumn equinox, winter solstice, summer solstice, celestial sphere, and a host of other terms associated with basic astronomy. In particular, right now I am trying to understand "celestial meridian." They spoke about zenith, and nadir. But none of the sites I went to was clear on these terms. Suppose that you are located somewhere on the equator. Send a line through your location and the center of the earth. Is it necessarily true that where the line meets the celestial sphere above you, "zenith" and where the line below you "nadir" meets the "celestial sphere", that those two points must lie on the "celestial meridian"? So that someone standing somewhere else on earth, would have a different celestial meridian? Did I understand that right? Thank you

You are confused for no reason at all, because it is not necessary to have a name for these things, as Tom Mattson said. Mathematicians deal with symbolic logic, and there are just so many ways of vary (if x then y), using 'not', all of which can be understood using a truth table. But I understand why you want a name to refer to the variations. The ones I am familiar with are: inverse, converse, contrapositive. Original statement: If A then B Converse: If B then A Contrapositive: If not B then not A Inverse: If not A then not B The contrapositive is redundant, because it can be proven to be logically equivalent to the original statement. But this happens often in logic, that is no reason not to know how to manipulate logical expressions. Regards

I'm not sure what you are asking. solar day... hmmm sidereal day... Apparently not the same thing. The amount of time it takes the earth to spin 360 degress in the rest frame of the stars, is called a sidereal day. i got it. Suppose you use the sun, to measure the value of a day, instead of a star in the constellation Orion, as I first suggested. Then you are not using a star in the celestial sphere, to define "day" you are using our star(the sun) which is much much closer to us than any star in the celestial sphere. But the sun is not really "fixed." It too, is orbiting the center of mass of our solar system. Kepler approximated things, by treating planetary motion in a frame in which the sun was at rest. So naturally, a solar day will not be equivalent to a sidereal day. Thank you Dr. There is still an unanswered question, "Approximately how far does the earth rotate in a solar day?" I am reading this article here. That animation was exactly the answer. In order to understand the answer given in the animation, it is necessary to understand "celestial meridian," so I am now reading this (Celestial sphere).

Dr Swanson, can you explain stellar aberration to me, I just read about it somewhere else, and I didn't know what it is. If it's really complicated, don't bother. Kind regards PS: Actually, I'll do the work on my own, so nevermind. Just ignore the question.

What should I do here? I want to end up understanding things the best way possible. Right now, I just want to understand what NASA is doing, and that will probably take 2 hours. I liked what Dr Cooper said. Also, I see 'sidereal' time coming in now, and I never did understand that.

Walking away from something doesn't make it smaller. Walking away from something makes it appear to your eye to be smaller. This is because what you see are photons, which left the object. You don't see the object, you see photons. You can draw a picture to help you understand, a ray diagram. You draw lines into your eyeballs.

Definition: Ephemeris is a set of parameters used by a global navigation satellite receiver to predict the location of a satellite and its clock behavior. Each satellite contains and transmits ephemeris data about its own orbit and clock. Ephemeris data is more accurate than the almanac data but is applicable over a short time frame from four to six hours. Ephemeris data is transmitted by the satellite every 30 seconds. I need to make sure I understand this properly. I went to the column, and selected two numbers, the first labeled delta, and the second labeled delta dot. Here they are: Delta = 1.0062406299 Delta dot = 0.78301 So I am thinking that 'dot' is a derivative with respect to time, so that if delta represents distance, then delta-dot is a speed. Now, the two locations I entered were: 1. New York 2. Sun So I presume that delta gives me the distance from New York to the center of mass of the sun. So now, I want to understand this distance in meters, but the explanation of the table says that delta is in astronomical units AU. Here is what link to nasa has: 1 AU = 149,597,870.691 kilometers and 1 kilometer equals 1000 meters. Now, the ephermeris generator gave the following number for the distance between New York and the Sun, at time 0:00:00 today... Delta = 1.0062406299 And here is what the explanation as to the meaning of "Delta" is: delta deldot = Target apparent range ("delta") and range-rate ("delta-dot") relative to observer. Units: AU and KM/S So the units of delta are astronomical units, which are units of distance, and the units of delta-dot are kilometers per second, which are units of speed. Very good. Ok so... I see that I can compute the distance from New York to the center of mass of the sun, in meters, at 0:00:00 this morning. Let me try and figure it out... [math] 1.0062406299 AU = (1.0062406299 AU) \frac{149,597,870.691 Km}{AU} [/math] [math] (1.0062406299 AU) \frac{149,597,870.691 Km}{AU} = 150531455.635811 Km [/math] But I wanted the distance in meters... [math] 150531455.636 Km \frac{1000 meters}{Km} = 150,531,455,636 \ \ \text{meters} [/math] so 150 billion meters or 150 million kilometers I wonder what it is in parsecs. Here is the conversion formula, to go from astronomical units AU, to parsecs: 206265 AU = 3.26 light year Here is the definition of PARSEC Let me see if I can make sense out of this. What the hell is the definition of AU again, average distance from earth to sun i think, but best to check... Here is the exact definition of an Astronomical Unit (AU) straight from NASA: And NASA has is at: 365.2569 days So we have three competing values. I think I'll go with NASA' date=' they had more decimal places. Here is Dr Alan Cooper talking about NASA's value: What does the AU mean? he makes the following important comments on the Gaussian constant: [math] \beta \alpha \rho \iota [/math] Pronounced "var-ee" in Greek, "barry" in English. Greek word for "heavy." So i guess i was right, "barycenter of solar system" means center of mass of the solar system. International Celestial Reference System Professionally speaking, this is the best reference frame to understand the motion of the planets, and sun in, relative to the "fixed stars" But still the rest frame of the sun is a good approximation to the ICRS. Right now I am trying to understand this "Gaussian constant." --------------------------------------------------------------- Ok, looks like I'm finally making sense out of things here. This site really helped a lot: Definition of sidereal year Also useful at that site are: 1) Gaussian year 2) Astronomical unit So this is really important: Sidereal year Screw it, I'm just going to let NASA teach me. NASA on our solar system

That's not exactly correct bascule, there was some confusion about this at a site I was just reading. Let me see if I can find it again. Here is one of the links I was reading: Nasa on Solar Polarity Reversal The date of the article is february 15th, 2001. In the first and second paragraphs, you can clearly see that they say, "the magnetic field of the sun can flip." Meaning that north and south poles switched positions. A little further down, the article says that pole reversal will occur again, in 2012, and that this happens every 11 years, just as Dr Swanson said. But a little further down, I got confused about something they wrote let me see if i can find it... Nevermind, it was another article I was reading, and I can't find that one.

The calculations look good, but what do they have to do with whether or not I should write a more general Lorentz force formula using a contribution from magnetic monopoles, and then trying to connect two inverse square formulas on one side of an equation, to Newtonian gravity on the other? And hopefully explain precession in the process, and have results which would have predicted the Zeeman effect??????? I was just planning to do what is mathematically necessary to explain gravitation as an electromagnetic phenomenon.

Well I was planning to use that idea to connect electromagnetism to gravity Dr Swanson.

Here is the data from the table in the other thread: 2005 Phenomenon----- Date--hour:minute Perihelion-------- Jan 2, 01:00 Vernal Equinox--- Mar 20, 12:33 Autmn Equinox--- Sept 22, 22:23 Aphelion--------- July 5, 05:00 Summer Solstice- June 21, 06:46 Winter Solstice-- Dec 21, 18:35 Vernal equinox Ok, let me sort of fumble towards the answer. Nice picture Analemma Newton/Kepler mathematical connection PS: I don't mind doing the calculation at all Dr swanson, it's good practice, plus it forces me to improve the simple solar system model I started off with, but what I really was looking for, was a site which has the current values continuously computed. Something professional which I can rely on, because i am going to need it again, and I cannot stop and do the calculation every time I need it. Regards