Using the index of a sequence as the domain and the value of the sequence as the range is a seq. a function?
482286
Yes, every sequence is a function, but defined only at positive integers.
663163
go to run-regedit\nand then open HKEY_CLASSES_ROOT&#92;CLSID&#92;{450D8... key\nthen see the SortOrderIndex and make its value 54.\n\n:-)
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
1159041
If you are thinking of f : ? -> { 0, 1, 2, 3, ... } (i.e., the domain is mapped into the set of non-negative integers) then clearly the domain "?" contains [at least] all finite bit strings.  If you keep the same restriction on the range, you can even extend the domain to include those infinite bit strings (however you wish to define that) which contain only finitely many ones.  If you extend the range to also include infinite cardinals and/or ordinals then you can include even more infinite bit strings into the domain.
109954
get "x" by it's selef
978426
for example:\n\n-1 < 2\nthen\n1 > -2\n\nanother:\n x < -5\nthen\n-x > 5\n\n-2 x > 9\nx< -9/2\n\ns
881644
Something like this:\n\nint **myArray;\nint s1,s2,i;\n\nprintf("Input row size: ");\nscanf("%d",&s1);\nprintf("Input column size: ");\nscanf("%d",&s2);\n\nmyArray[0] = malloc(s1*s2*sizeof(int));\nif (myArray[0] != NULL ) {\n  for (i = 0;i<s1,i++) {\n         myArray[i] = (i * s2) + myArray[0];\n  }\n}\n\nmyArray can then be used as normal\n\nmyArray[2][4] = 20;
19284
It's a notational convenience.\n\n1) It helps avoid the problem of running out of unique symbols when working with a large, possibly infinate, set of variables.\n\n2) It provides a way to keep track of ordering in a set of variables when that ordering makes a difference.\n\n3) It makes it possible to express the values of a sequence of variables when their value depends on their position in the sequence. An example would be the Fibonacci sequence where each number in the sequence is the sum of the two previous numbers (the first two being defined as 1). It can be expressed as: \n\na(sub)1 = 1\na(sub)2 = 1\na(sub)n = a(sub)(n-1) + a(sub)(n-2)
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
572733
Please check out this site it gives averything you want on VBA.\n\nwww.functionx.com
394920
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
151678
[Domain], Kingdom, Phylum, Class, Order, Genus, Species.  Can be remembered easily using the following mnemonic:\nKing Philip Came Over For Good Spaghetti.\n[Domain is sometimes used to classify eukaryotes and prokaryotes.]\nHope this helps!
3995
Yes: there is 1 range value for each domain value\n\nThe domain values are the first numbers in each pair.  For each of them, -2, -1, 0, and 3, there is one range value.
952602
This is what always helps me.\n\nA function is a machine.\n\nThe domain is everything you can put into the machine so that it doesn't break.\n\nThe range is everything you can get out of the machine.\n\nIn the case of your function, there's only one possible value of x that will "break" the function, in the sense that you'd get an infinite answer.  That value is x=0, because then you'd have 9/0 + 7, and 9/0 is infinity.\n\nSo the range is {x | x is not equal to 0}.\n\nAssuming that you meant 9/x + 7, rather than 9/(x+7).  If you meant the latter, x cannot be equal to -7.
865710
This function I've wrote takes the right cell as argument, and, if  that cell is not empty, returns it's value , otherwise the value of the cell above the cell where the function resides: \n\nFunction MyFunction(rightcell As Range)\n   Application.Volatile True\n   MyFunction = IIf (rightcell (1, 1).Value <> "", rightcell (1, 1).Value, rightcell(1, 1).Offset(-1, -1).Value)\nEnd Function\n\nTo make the function available proceed as following:\n-Press Alt+F11- this will bring up the VBA editor\n-navigate to Insert > Module to open a blank module screen\n-paste the above code there\n-close the VBA editor\n\nAs in your example,in  B3 cell, you will call te function like this:\n =MyFunction(C3)
516203
Here's the sequence and a site you can load it up for mapping.
145915
Stick the sequences in a computer and tell it to look for stretches of a certain length where a certain percent of the nucleotides match.  How close a match and how long a sequence depends on exactly what you're looking for.