Confused about following series x(15/27)^x?
919856
It's easier to sum x(15/27)^x (over x) for a general q instead of 15/27, and substitute q = 15/27 only at the end. \nI also change x to n (looks better).\nTwo methods: \n\n1. Method using calculus: \nFor each n, view nq^(n-1) as the derivative, w.r.t. q, of q^n. Note that the (infinite) sum of derivatives = derivative of sum (summation over n, differentiation wrt q). So, instead of summing the derivatives, we'll differentiate the sum (because the sum of q^n is easy). So, sum q^n over n from 0 to infty to get 1/(1-q) -1 (or, which is the same, q/(1-q)), differentiate this wrt q to get 1/(1-q)^2, and multiply by q (because we started from nq^(n-1) instead of nq^n).\n\nAns: q/(1-q)^2.\nFinally, sub q = 15/27 in this.\n\n2. If you want to avoid calculus, view nq^n as the sum of n equal terms (so now you have a double summation), change the order of the double summation, and you'll only have easy sums to do (the geometric series).
1025141
The series is very simple.\n\n0  1  1  2  3  5  8  13  21  34  55  89  144  ... \n\nYou just add the two previous numbers to the next number.\n\nThere are tons of websites related to your question, so you just haven't looked, and so don't know enough to ask a specific question.
700653
Yes you're correct - just plug in -2 for x to get: \n \n20 - -6 + 1 = 27
882163
Huh? The series does not converge to 0, so the sum is divergent:\n\nLim n->inf ( n / (2n+5) ) = Lim n->inf ( 1 / (2+5/n) ) is clearly 1/2\nLim n->inf ln ( n / (2n+5) ) = ln (1/2) = -ln(2) ~= -0.693\n\n=> (Sigma)_n=1_inf ln ( n / (2n+5) ) does not exist\n\n__________________________________________________\nIn response to your mail asking why \nLim n->inf ( n / (2n+5) ) is not indeterminate and therefore would require De L'Hospital's Rule:\n\nYou can extract a common factor n:\nLim n->inf ( n / (2n+5) ) = Lim n->inf ( 1 / (2+5/n) ) = 1/2\n\nDoing what you did neglects that:\nLim n->inf ln ( n / (2n+5) ) != Lim n->inf [ ln (n) - ln (2n+5) ]\n= Inf - Inf = undefined
681334
If there is no remainder, then this is a *whole number without a* decimal.  Or you could say that the decimal portion was zero.\n\ne.g. 27/3 = 9.0\n4/1 = 4.0\n-28/7 = -4.0\n-2/-2 = 1.0\n18/-3 = -6.0\n\netc.
352131
For the second part of your question, you need to differentiate 1/(1+x)^4 for n times and divide by n! and then substitute x=0 to work out the nth coefficient of the power series. The first constant term is found by substituting 0 into 1/(1+x)^4, so it is 1. The second term is found by differentiating 1/(1+x)^4 once and substitute x=0 and then divide by "1!". The third term is produced by differentiate  1/(1+x)^4 twice and substitute x=0 and then divide by "2!" factorial. You can get as many terms as you want, the series looks someting like:\n\n1 -4x + (20/2!)*x^2 -(120/3!)*x^3 .................\n\nSubstitute x= 0.16 to approximate the number, for 7 terms the answer is close to 0.55223. I might have made mistakes in my calc so you might need to calculate these values again.
787465
Write out the exponentials and use the series for e^x. Other than that, do your own homework.
415757
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
892260
#1: Yes, that is correct.\n#2: This series diverges. It is a necessary (but not sufficient) condition for an infinite series to converge that the individual terms approach 0 as n→∞. It is fairly easy to show that the individual terms approach 1 as n→∞.\n#3: First note that ln ((n+1)/n) = ln (n+1) - ln n. So writing out the first few terms of this series:\n\n(ln 2 - ln 1) + (ln 3 - ln 2) + (ln 4 - ln 3) + ...\n\nNote that the second part of each term cancels out the first part of the preceding term. So all that needs to be considered is the first part of the last term and the last part of the first term, because those are the only parts that do not cancel. Formally:\n[n=1, a]∑ln ((n+1)/n) = ln (a+1) - ln 1 = ln (a+1)\n\nThis series does not converge, since [a→∞]lim ln (a+1) = ∞.
818643
You asked about the numerator before, didnt you get an answer?\nWith synthetic substitution, you should find that the roots of the numerator are -1, -1, -2, and 3.\nThis gives (x+1)(x+1)(x+2)(x-3)  see *.\n\nAre you sure you typed the denominator correctly?\nYou typed: \nx^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1\n(by the way, you need to insert spaces into long polynomials like this one or Yahoo does not display them after 30 spaces.  You can place the mouse cursor over them and see the complete polynomial, but many people dont know that, including the previous answerer.)\nThis is not factorable (over the rational numbers)\n\nHowever, if one digit is changed, namely if -30x^3 was written \n -20x^3, then the denominator becomes a very common one (and one that you may notice just by looking at it if you stay at this factoring stuff for a long time):\nnamely (x-1)^6 !!\n\n\n*\n1st root (-1):\n\n-1 || ..1... 1 ... -7 .. -13 ... -6\n...............-1 .... 0 ..... 7 ..... 6\n————————————\n..........1....0.....-7.....-6... || 0\n\nso (x+1) is a root\n\n2nd root (-1):\n\n-1 || ...1....0.....-7.....-6\n................-1 .... 1 .... 6\n——————————\n..........1...-1 ... -6 .. || 0\n\nso (x+1) is a root again.\n\nx^2 - x - 6 remains\n= (x-3)(x+2)\n\nInteresting correlative problem:\nhttp://answers.yahoo.com/question/index;_ylt=Aqz3W3s3NPcz.j__CQYxK0bsy6IX?qid=20061127003230AA2oHuA
535461
To break it up into something easy for a computer to read, break the power down to something divisible by 2 and then split that. Here's an example:\nx^153 = x*x^152\nx*x^152 = x(x^76)^2\nx(x^76)^2 = x((x^38)^2)^2\nx((x^38)^2)^2 = x(((x^19)^2)^2)^2\netc.\nI'm not sure that's what you wanted to know, though...
561879
The series for (1-cos(x))/x^2 is \nsum of (-1)^2n * x^(2n-2) divided by (2n)! (using Taylor series and simple algebra) which is:\n\n(1/2)-(x^2/24)+(x^4/720)-(x^6/40320)+.....\n\nNotice that this series is continuous on the open interval (0,1) which is good for integration.\n\nNow integrate to:\n\n(x/2)-(x^3/72)+(x^5/3600)-(x^7/282240)+......\n\nevaluate at x=1 and x=0 (however notice that the x=0 doesn't commit anything to the evaluation, so only evaluate at x=1)\n\nYou get 1/2-1/72+1/3600-1/282240+...\nNotice that each term contributes less and less to your answer.  Find the term that is less than 10^-5.  \n1/2=0.5\n1/72=0.014\n1/3600=2.778*10^-4 (getting close)\n1/282240=3.543*10^-6 (here it is)\nSince the term above only contributes to the ten-millionths decimal place, you can safely assume you have found an estimate of the original function using a series out to the hundred-thousandths decimal place, or have an error of less than 10^-5.  \n\nEnjoy!
887334
You do this kind of division the same way you do normal long division, with the powers of x analogous to the powers of ten.  But, like normal long division, you have to be careful with the signs!\n\nStart at the left - x-3 goes 2x^3 times into 2x^4, so multiply x-3 by 2x^3, which gives us 2x^4 - 6x^3.  So the first term in our quotient is 2x^3.  We subtract 2x^4 -6x^3 from the dividend, leaving 5x^3+15x^2+6x-18. \n\nWe carry on in the same vein, so we multiply x-3 by 5x^2, the second term in our quotient, getting 5x^3-15x^2, which again we subtract from the dividend.  This leaves us with 6x-18.  Now x-3 goes 6 times into 6x-18, so the third term in our quotient is 6.  6 times x-3 is 6x-18, which we subtract from the dividend, which leaves us with zero, so there is no remainder.  The result of the division is the sum of the three terms we generated, which is 2x^3+5x^2+6.\n\nAs for the factorisation, you will remember that x^2-1 has factors x+1 and x-1, so x^4-1 will have factors x^2-1 aand x^2+1.  Then x^2-1 has factors x-1 and x+1.  So the factorisation of x^4-1 is (x-1)(x+1)(x^2+1).  If you are OK with complex numbers the x^2+1 term may be further factorised to x+i and x-i, so then the factors are (x+1)(x-1)(x+i)(x-i).\n\nYou could have got straight there though by recognising that the fourth powers of +1, -1, +i and -i are all 1!
644170
The answer of "nayanmange" is right but his formula is wrong because it does not predict the sequence correctly. The correct formula is (2n-1)-th and 2n-th term of the series are {(2n-1)^2+1} and {(2n)^2-1} respectively. For n = 3 we find the (2*3)-th i.e. 6th term of the series as {(2*3)^2-1} = 35.
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......
105048
Directly, there is no solution.  If you use a series expansion around x=0, \nsin(x)=sum((-1)^n(x^(2n+1))/(factorial(2n+1)))\nDivide this series by x, then integrate.  After the 3rd term, the error term goes to the ten millionths digit, so I'd say you're pretty safe :)\n\nsin(x)/x = 1 - (x^2)/6 + (x^4)/120 - (x^6)/5040 + .....\nIntegrated is:  x - (x^3)/18 + (x^5)/600 - (x^7)/32580 +\nor \nsum((-1)^n)(x^(2n+1))/(2n+1)*factorial(2n+1))
580109
ans is 70\n(5*2+1) (11*2+2) (24*2+3)\nrule is there r two series 5 11 24 51...\n7 16 34 70\n(7*2+2) (16*2+2) (34*2+2)