sqrt(n) = S*r2sqrt(n)/S = r2 sqrt(sqrt(n)/S) = ror4th root of n/sqrt(S) = r.If your equation was sqrt(n) = S*r-2sqrt(n) = S*(1/r2)sqrt(n)/S = 1/r2S/sqrt(n) = r2 sqrt(S/sqrt[n]) = rorsqrt(S)/4th root of n = r
S = {5n + 2 | n = 1, 2, ... , 10}
438904 s = 4.38904×10^5 s
well, ace of 1 (or the first number in the equation) =1, d=1 (or what it's added by each time), which shows that it's arithmetic), and n=1000 (the number you're trying to get)the equation is s of n=n/2 (ace of 1 + ace of n) s of n is the sum of the numbersso, s of n=500(1+1000)s of n=500500so 500,500 is the sum of the numbers from 1 to 1000.you can find more athttp://www.youtube.com/watch?v=VgVJrSJxkDk&feature=youtube_gdatahttp://www.youtube.com/watch?v=U_8GRLJplZg&feature=youtube_gdata
Assuming you mean the first n counting numbers then: let S{n} be the sum; then: S{n} = 1 + 2 + ... + (n-1) + n As addition is commutative, the sum can be reversed to give: S{n} = n + (n-1) + ... + 2 + 1 Now add the two versions together (term by term), giving: S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1) → 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1) As there were originally n terms, this is (n+1) added n times, giving: 2S{n} = n(n+1) → S{n} = ½n(n+1) The sum of the first n counting numbers is ½n(n+1).
Sn = n2 S12 = 122 = 144.
A zeta function is the function of the complex variable s which analytically continues the sum of the series 1/n^3 for all values from n=1 to infinity, which converges when the real part of s is greater than 1.
Given a set S and the set of positive integers Nwhere n Є N, any function from N into S is called a sequence, notated as u. If u Є S, then u is usually written as us.See related links for more information.
Isaac N. Youngs has written: 'A short abridgment of the rules of music' -- subject(s): Hymns, Music theory, Musical notation, Shakers
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The scientific notation for 142,000 s is: 1.42 × 105s
Possible. int cmp; cmp= strcmp (p, q); if (cmp<0) printf ("%s < %s\n", p, q); else if (cmp>0) printf ("%s > %s\n", p, q); else printf ("%s == %s\n", p, q);
A sigmoid is a mathematical function that is shown on a curve. It essentially shows data on a graph or chart but only ends up showing that data in the shape of an S.
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It is a function which returns itself.Whenever function is called from itself then there is a chance that the function will enter in to an infinite loop.To prevent the program from entering in to a infinite loop.To prevent a program from entering in to an infinite loop in such a manner,every recursive function must have terminating condition that decide whether to call the function once more or not.This terminating conditions is also known as stopping rule in recursive function.Using Recursion to Print the Fibonacci Series #include#includevoid main(){int c=1,f=0,s=1;int i;printf("%d%20d\n",f,f);printf("%d%20d\n",s,s);for (i=2; i
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