Which one of these functions (linear, quadratic, rational, exponential) is related to the arithmetic sequence?
372518
The linear functions are related to the arithmetic sequences.  Because geometric sequences go by intervals of a number my mulitplying or dividing.  Arithmetic sequences go up or down by intervals of a number by addition or subtraction.  Linear functions prove arithmetic sequences right.  Exponential functions prove geometric sequences.\n\n1,3,5,7,9,11     Arithmetic added 2 each time\n\n1,2,4,8,16,32    Geometric  multiplied by 2
243446
It's called the Fibonacci sequence, but there's actually a couple more terms at the front: 0,1,1,2,3,5,8,13...  You take the last 2 terms and add them together to get the next term.  So 8+13 = 21; 21+13 = 34, etc.\n\nHere is a good start if you're interested in learning more about it:\n\nhttp://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/
981709
To extend the above two discussions and bring the matter to a conclusion, more or less.\n\nLet us define a sequence by T0 = 1, Tn = x^Tn-1.  Call the limit of this sequence as n -> infinity  as f(x).\n\nFor any given y = f(x), we see that:\n\n     lim Tn = lim Tn-1 = y\n>>\n    lim Tn = lim(x^Tn-1)=x^limTn-1\n>>\n     y = x^y\n>> \n    x = y ^ (1/y)\n\nFor y >= 1:\nThis can easily be seen to satisfy the  special cases of y = 1, 2 and 4.\n\nx, as a function of y has a maximum at y = e.~ 2.718, which leads to x~1.447.\n\nFor values of  x > e^(1/e),  f(x) does not exist; and the sequence T cannot have a limit > e.\n\nFor y < 1, we can similarly see that limits < 1/e are not possible.  The smallest x for which the limit exists is (1/e)^e ~ 0.066, giving a limit 1/e ~ 0.368
893957
well darling ;\nimagine this like that\n\ny^2 + ( a+b)y + ab \nyou need to find " a+b " & " ab" \nin our function \na+b = 16 & ab = 64 \n( y + a)( y + b) \n( y + 8 ) ( y + 8)\n{ ' 8 + 8 = 16 '   &  ' 8 * 8 = 64 ' }\nso\n( y + 8)^2\n\nGood Luck sweetheart.
944543
Okay . let see \n\nPart 1 \nVertical asymptotes correspond to the zeroes of the denominator of a rational function \nx^2+2x+1= 0 \n(x + 1)(x + 1)= 0\nx = -1 \nV.A = -1 \n\nPart 2 \nhorizontal asymptotes ; limit of f(x) as x → ∞\nlim (2x - 4) / (x^2+2x+1)\nx → ∞\nlim 2x / x^2   { 1 < 2 } \nx → ∞ \n= 0\nhorizontal asymptotes is " 0 "\n\nGoodLuck
1023953
If f is a linear function where f(x)=y, then it's of the form y = mx + b.\n\nYou know\n8 = 4m + b\n4 = 1m + b\nSubtract the second equation from the first.\n4 = 3m\n4/3 = m\nSubstitute this back into the second equation:\n4 = 4/3 + b\n8/3 = b\nSo y = 4/3x + 8/3.  Since for f(12) x=12,\ny = 4/3 (12) + 8/3\n   = 48/3 + 8/3\n   = 56/3, or 18 2/3
484068
Example of Fibonnaci: 1,1,2,3,5,8,13,21\nEach number in the sequence is the sum of the 2 numbers before. I.E. 1+1=2; 1+2=3; 2+3=5; 3+5=8, etc.\n\nHope this helps.
752791
you mean f(x)=3x+2, for example?\n\nreplace f(x) with Y, graph and solve as a normal equation. Linear functions have infinite whole solutions, depending on what you want Y to be.\n\nOf course, I could be minunderstanding you completely. If so, sorry.
1080867
This is the Gaussian binomial coefficient [n,1] for q=2. \n\nGreetings from The On-Line Encyclopedia of Integer Sequences!
844451
to the right, i.e. x-> infinity,  3x^2 - 5x -7^-x is more like 3x^2,\nnow , to the left, ie x-> -infinity, 3x^2-5x-7^{-x} is more like the exponential -7^{-x} \nis this what you mean, may be if you provided an example, i could help you better.\n'
787465
Write out the exponentials and use the series for e^x. Other than that, do your own homework.
631314
so you've basically asked every possible series formula question here...but i'll still help you!  the formula for the sum of an arithmetic sequence is:\n\nS(n)=(n/2)[a(1)+a(n)]\n\nthe sume of first 20 terms is then:\n\nS(20)=(20/2)[a(1)+a(20)]\n\nso you need to know the first term, a(1), and the 20th term, or a(20).  if the first term is 5 and the 20th term is 45, then:\n\nS(20)=(10)(5+45)\nS(20)=500
190708
you've asked a lot of questions regarding rational and irrational. Here's a site that may be of help:\n\nhttp://www.purplemath.com/modules/numtypes.htm
203992
a) [1,2] is a sub set of domain of g so for every x in [1,2]\nwe have a unique value of g.\n\nb) if the sequence converge, in the limit Xn+1 must be equal to X, so if the limit is L we have:\n\nL=(3L^2+3)^1/4\n\nL^4=3L^2+3\n\nL^4-3L^2-3=0\n\ndelta=9+12=21\n\nL^2=(3+√21)/2  (L^2>0)\n\nL = √[(3+√21)/2]  (L>0 because X0>0)\n\n4<√21<5  ---> 7<3+√21<8\n\n---> 3<(3+√21)/2<4  ---> 1<L<2\n\nc) we have (Xn+1)^4=3Xn^2+3 let Yn be Xn^2\n\n---> (Yn+1)^2=3Yn+3 ---> Yn
832512
either square root or erauqs
778985
The rational zeros will be of the form p/q where p is a factor of 3 (the lowest order term) and q is a factor of 2 (the coefficient of the highest order x term, in this case 2x^3).  That means the possible answers are +/- 3/1, 3/2, 1/1 and 1/2.  Just plug those in to see which works.  (1, -1, 1.5)
925404
OK, let's see.\n\nIt's an arithmetic sequence, so the difference between consecutive terms is a constant.  The first term is 1, so we can express each term as:\n\nx(i) = 1 + k*(i-1), i = 1, 2, 3, ....\n\nNext the problem says that the 2nd, 10th, and 34th terms are the first three terms in a geometric sequence.  That means that the ratio of the 10th term to the 2nd term must be the same as the ratio of the 34th term to the 10th, because when you divide any two terms of a geometric sequence, you get the same constant.\n\ni(2) = 1 + 1*k\ni(10) = 1 + 9*k\ni(34) = 1 + 33*k\n\nSo we said i(10)/i(2) must equal i(34)/i(10):\n\n1+9k    1+33k\n-------- = --------\n1+1k      1+9k\n\nCross multiplying: (1+9k)^2 = (1+1k)(1+33k).\n\nThen 1 + 18k + 81k^2 = 1 + 34k + 33k^2.\n\nPutting it all on the left side, 48k^2 - 16k = 0, or 16k(3k - 1) = 0.\n\nSo either k = 0 (not interesting) or k = 1/3.\n\nTherefore the arithmetic sequence must be:\n\n1, 1 1/3, 1 2/3, 2, 2 1/3, 2 2/3, 3, 3 1/3, ....\n\nThe second term is 1 1/3.  The 10th term is 4 (1 + 3 * 1/3).  The 34th term is 12 (1 + 33 * 1/3).  So the geometric progression would be:\n\n4/3, 4, 12, 48, 192, .....