multiplication is point to point and convolustion is point to multi-point ex multiplication-- s[n]=x[n].h[n] s[0]=[x[0].h[0] s[1]=[x[1].h[1] s[2]=[x[2].h[2] . . . .. s[n-1]=[x[n-1].h[n-1] convollustion s[n]=x[n]*h[n] s[0]=[x[0].h[0]+x[0].h[1]+x[0].h[2]+.......+x[0].h[n-1] s[1]=[x[1].h[0]+x[1].h[1]+x[1].h[2]+.......+x[1].h[n-1] s[2]=[x[2].h[2]+x[2].h[1]+x[2].h[2]+.......+x[2].h[n-1] . . . s[n-1]=[x[n-1].h[0]+x[n-1].h[1]+x[n-1].h[2]+.......+x[n-1].h[n-1].
Try 151,600! Permutations & Combinations. P(n,r)=n!(nār)! not P(n,r)=n!/(n!-r!)r! ?
1680 to get this answer you need to get the factorial of 8 and the factorial of 4! like so 8! is n! 8*7*6*5*4*3*2*1=n!_40320 4! is r! 4*3*2*1=r!_24 now divide the two! n!/r!(n-r)!=1680 8!/4!(8-4)!=1680 8!/4!=1680
12%
The formula is (1+r)^n -1. r= rate of return (ie 100%) n=number of compounding periods so (1+1)^15 -1= 32767 or 3276700%
An infinite geometric series can be summed only if the common ratio has an absolute value less than 1. Suppose the sum to n terms is S(n). That is, S(n) = a + ar + ar2 + ... + arn-1 Multipying through by the common ratio, r, gives r*S(n) = ar + ar2 + ar3 + ... + arn Subtracting the second equation from the first, S(n) - r*S(n) = a - arn (1 - r)*S(n) = a*(1 - rn) Dividing by (1 - r), S(n) = (1 - rn)/(1 - r) Now, since |r| < 1, rn tends to 0 as n tends to infinity and so S(n) tends to 1/(1 - r) or, the infinite sum is 1/(1 - r)
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
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr
#include#includevoid main(){int n,a=1,s=1,r;printf("\n enter number of lines ");cin>>n;for(;a=1;b--){printf(" ");}r=pow(s,2);printf("\n");a++;}getch();}// \m/
the square root of -1 is an imaginary number
This browser is totally bloody useless for mathematical display but...The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]Let n -> infinity while np = L, a constant, so that p = L/nthenP(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)Now lim n -> infinity of (1 - L/n)^n = e^(-L)and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.
Void main () { Int n, r, s = 0 ; Printf (" Enter the number") Scanf ("% d", & n); While (n! = 0) { r = n%/10; s = s+r; n = n/10; s = s+r; n = n/10 } Printf ("sun of the dijit of a no. is % d; s)
Sirens is spelled S-I-R-E-N-S.
Square root of -1 is mathematically represented by the IMAGINARY NUMBER i.
An arithmetic series is the sequence of partial sums of an arithmetic sequence. That is, if A = {a, a+d, a+2d, ..., a+(n-1)d, ... } then the terms of the arithmetic series, S(n), are the sums of the first n terms and S(n) = n/2*[2a + (n-1)d]. Arithmetic series can never converge.A geometric series is the sequence of partial sums of a geometric sequence. That is, if G = {a, ar, ar^2, ..., ar^(n-1), ... } then the terms of the geometric series, T(n), are the sums of the first n terms and T(n) = a*(1 - r^n)/(1 - r). If |r| < 1 then T(n) tends to 1/(1 - r) as n tends to infinity.
The r+1 th term is nCr(-x)r where r = 0, 1, 2, ... , n. and where nCr = n!/[r!*(n-r)!]
#include<iostream> using namespace std; int revDigit(int n){ int r,s=0; while(n>0){ r=n%10; n=n/10; s=(s*10)+r; } return s; } main(){ long n,r,r1,s=0,s1=10; cout<<"Enter a number."<<endl; cin>>n; while(n>0){ r=n%2; n=n/2; s=(s*10)+r; } //since we read the binary equivalent in a bottom up manner, we need to reverse it. s1=revDigit(s); cout<<"Binary equivalent:\t"<<s1<<endl; system("pause");//this line is required if you are using dev c++ }