Find a closed form of expression for generating function of the sequence G, where G(k+3) =G(k+2) +G(k+1) +G(k) for k>=0, given G(0) =G(1) =G(2) =1?
I am not sure if this is a homework question or something like that. If it were a homework question for a tenth grader or similar, then it is sort of a toughie. Nevertheless, I don't want to give the exact answer.My suggestion would be to first understand a simpler but more well-known sequence.G(k+2) = G (k+1) + G(k)How do you get closed form function for this? This is the popular fibbonacci sequence. If you can understand this, you will easily be able to go to the next step.Now the above equation looks "like a differential equation", and we approach it like one. We expect the solution to be exponential wrt "k", we don't know what the actual closed-form will look like; in other words, we don't know what the base will be, we also don't know what the multiplier will be. Therefore we start with:[math]G(k)=Ar^k[/math](In above equation, we don't know what A and r would be)If we substitute, we would get:[math]Ar^k(r^2-r-1)=0[/math]Hence, "r" has two values, the zeros of "[math]r^2-r-1[/math]". Lets call them r1 and r2. You can very easily calculate them.Therefore, the generic solution will be:[math]G(k)=A1\times r1^k + A2\times r2^k[/math]Now, you can solve for A1 and A2 using the "Seed" conditions.Let us extend try and extend this to "third-order" Fibonacci:G(k+3) = G (k+2) + G(k+1) + G(k)We now start with an arbitrary solution:[math]G(k)=Ar^k[/math]If we substitute, we would get:[math]Ar^k(r^3-r^2-r-1)=0[/math]Now, this will have three solutions (ugly ones and not very easily calculable, but still solutions). Lets call them r1, r2, and r3.Therefore, the generic solution will be:[math]G(k)=A1\times r1^k + A2\times r2^k + A3\times r3^k[/math]Now, you can solve for A1, A2, and A3 using the "Seed" conditions.Edit: For all linear recurrence dependencies of any order, the above is the procedure. The general linear recurrence of d-order is of the form:[math]G_k = c_1G_{k-1} + c_2G_{k-2}+\cdots+c_dG_{k-d}[/math]
Generating functions?
Find the exponential generating function for the number of alphanumeric strings of length n formed from the 26 uppercase letters of the English alphabet and 10 decimal digits if • each vowel must appear at least one time; • the letter T must appear at least three times; • the letter Z may appear at most three times; • each even digit must appear an even number of times; and • each odd digit must appear an odd number of times. I'm having trouble setting up the exponential functions, like for the first bullet, would it be (xe^x)^5? I'm really confused on these
Find the generating function for the sequence (an)=(a0,a1,a2,a3....)?
First, we write a generating function factor for each point possibility: **Note that the exponents reflect the possibilities with the points in each factor. Safety: x^0 + x^2 + x^4 + ... = 1/(1 - x^2) Field Goal: x^0 + x^3 + x^6 + ... = 1/(1 - x^3) TD, no conversion: x^0 + x^6 + x^12 + ... = 1/(1 - x^6) TD with conversion: x^0 + x^7 + x^14 + ... = 1/(1 - x^6). Hence, the generating function for the NFL scoring is their product: g(x) = 1/[(1 - x^2)(1 - x^3)(1 - x^6)(1 - x^7)]. [The coefficient to x^n in g(x) equals a(n).] I hope this helps!
How do you create a generating function?
According to me the numerical estimation of the producing capacity at a specific estimation of the free variable is of no interest, thus creating capacities can appear to be fairly unintuitive at first. In any case, the vital point is that the producing capacity all in all encodes the greater part of the data in the likelihood conveyance in an exceptionally valuable manner. Creating capacities are vital and significant devices in likelihood, as they are in different zones of arithmetic, from combinatorics to differential conditions.I know more about sparenparts here all types of industrial products and goods are available.
How can I visualize the complex exponential function?
"Visualizing" any complex-valued function of a complex variable is a challenge, regardless of the function, because it is essentially a map from R^2 to R^2, and thus would require 4 dimensions to visualize both the real and imaginary (or magnitude and angle) parts simultaneously. Thus the best we dimensionally-challenged humans can do is "visualize" (i.e., graph) either "full" images of one dimensional subsets of the domain space (e.g., the images of straight lines or segments, and other curves), or one "component" of the image of a two dimensional subset of the domain space (eg, the real OR imaginary part, or the magnitude OR the angle, of the image of an area, eg, a disc or a square, in the domain space). (I'm neglecting the possibility of using time, i.e., "making a movie," as the axis for one of the image components, because one can't really represent that component truly continuously, though one could approx. that as precisely as one's resources and patience might allow, and because it still would not gives us a "sense of the whole" as we feel we get when we draw a "complete" graph of a map representable in R^2 or R^3.)That said, the common "first things" on which to observe the "action" of any complex map are the horizontal and vertical lines, respectively y and x = constant, starting with the coordinate axes, y=0 and x=0. Start with these and then move on to things like the lines y=mx, m not equal to 0 (why?) and origin-centered circles, and then on to things like y=mx+b, b not equal to 0 (why?) and circles centered at other than the origin. Once you get comfortable with those, you can (try to) graph the real and imaginary parts and the magnitude and angles of the images of an arbitrarily-positioned rectangle and circle. The point is, unlike maps from R->R^1 or 2 or from R^2->R^1, where you have some chance of drawing a reasonably "global" representation of the map, the best one can do to develop a "visual intuition" for the behavior of a complex-valued function of a complex variable is build up a "repertoire" of "sub-images" one knows for a given map, exponential function included. (I hope this doesn't discourage you: I can't overstate the importance of doing precisely what I have described for the exponential function--it is the most important function in complex variables, and it is extremely useful to develop this "visual intuition" for it.)
For each of these generating functions, provide a closed formula for the sequence it determines.?
All of these functions ARE sequences... I decide to determine the series for you a) Not possible series since you just get the remainder with no series... b) Same here... c) Σ(n = 0,∞) (5x)ⁿ d) x³ Σ(n = 0,∞) (-3x)ⁿ = Σ(n = 0,∞) (-1)ⁿ(3)ⁿx^(n + 3) e) You just get... x² + 3x + 7 + [Σ(n = 0,∞) x^(2n)] where x² + 3x + 7 is your remainder, so you don't get the full series. f) Same here.. You get the incomplete series... You work out other problems... Note that these functions have similar sequence... g) Hint: Use differentiation and you get somewhat the series like Σ(n = 0,∞) xⁿ. Then, differentiate the function to get the answer... Good luck!
Generating Functions - Math?
kinda stuck on this exponential generating function: Find the exponential generating function for the number of alphanumeric strings of length n formed from the 26 uppercase letters of the english alphabet and 10 decimal digits if, 1 each vowel must appear at least one time; 2 the letter T must appear at least three times; 3 the letter Z may appear at most three times; 4 each even digit must appear an even number of times; and 5 each odd digit must apear an odd number of times. So what I sort of have is, for 1, (x + x^2/2! + x^3/3! + ...)^5 saying each vowel should appear at least once. = xe^5x ?? 2, (x^3/3! + x^4/4! + x^5/5! + ...) i think this would be x^3e^x since we shift everything over by 3 3, (1 + x + x^2/2! + x^3/3!) I don't know how to write this exponential function out for this one 4, (1 + x^2/2! + x^4/4! + x^6/6! + ...) = (e^x + e^-x)/2 5, (x + x^3/3! + x^5/5! + ...) = (e^x - e^-x)/2 Then where do I go from here? I know this isn't an easy problem but if you help, I will give you best answer :)
The ways of selecting 4 letters from word examination?
E^1 X^1 A^2 M^1 I^2 N^2 T^1 O^1 (11 letters total) Depending on whether the question requires the count of (i) order-independent or (ii) order-dependent selections. We have to evaluate the appropriate coefficients of (i) an ordinary generation function and (ii) an exponential generating function. (i) [x^4](1+x)^5(1+x+x^2)^3 (ii) [x^4/4!](1+x)^5(1+x+x^2/2!)^3 where [x^4] is the "evaluate coefficient x^4" operator and similarly [x^4/4!] says "evaluate the coefficient of x^4/4!". In case (i) there is a binomial (1+x) associated to each singleton letter E, X, M, T and O and a trinomial (1+x+x^2) associated to each doubleton letter A, I and N. These are then multiplied. In case (ii) there is an exponential binomial (1+x/1!) associated to each singleton letter E, X, M, T and O and an exponential trinomial (1+x/1!+x^2/2!) associated to each doubleton letter A, I and N. These are also multiplied. It is straightforward to multiply manually but this is merely an exercise in tedium so we can instead opt for a computer algebra system like wolfram alpha or sage. Using the following inputs in sage we get (i) Input: expand( (1+x)^5*(1+x+x^2)^3).coefficient( x^4) Output: 136 <---- (ii) Input: 24*expand( (1+x)^5*(1+x+x^2/2)^3).coefficient( x^4) Output: 2454 <---- Our answers :) --------------------------- Additional: The full expansions for (i) order-independent and (ii) order dependent selections of various sizes are given by (i) (1+x)^5(1+x+x^2)^3 = x^11 + 8*x^10 + 31*x^9 + 77*x^8 + 136*x^7 + 179*x^6 + 179*x^5 + 136*x^4 + 77*x^3 + 31*x^2 + 8*x + 1 (ii) (1+x)^5*(1+x+x^2/2)^3 = 1/8*x^11 + 11/8*x^10 + 29/4*x^9 + 24*x^8 + 441/8*x^7 + 735/8*x^6 + 113*x^5 + 409/4*x^4 + 133/2*x^3 + 59/2*x^2 + 8*x + 1 for a k-selection in (i) simply evaluate [x^k] and for (ii) evaluate [x^k/k!]. For example a 3-selection is (i) 77 (ii) 3!*133/2 = 399
How can I find the number of permutations and combinations with the letters of the word "EXAMINATION", taking 4 at a time?
Thanks for A2APlease go through this thread: How many permutations of 4 letters can be made out of the letters of the word 'examination'?