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How would I solve for d in the equation ( e times d ) mod n equals 1?
ed mod n = 1 ed = kn + 1 for integer k so d = (kn + 1)/e for integer k.
Gaussian elimination in c?
#include<stdio.h> #include<stdlib.h> #include<math.h> #include<conio.h> void main(void) { int K, P, C, J; double A[100][101]; int N; int Row[100]; double X[100]; double SUM, M; int T; do { printf("Please enter number of equations [Not more than %d]\n",100); scanf("%d", &N); } while( N > 100); printf("You say there are %d equations.\n", N); printf("From AX = B enter elements of [A,B] row by row:\n"); for (K = 1; K <= N; K++) { for (J = 1; J <= N+1; J++) { printf(" For row %d enter element %d please :\n", K, J); scanf("%lf", &A[K-1][J-1]); } } for (J = 1; J<= N; J++) Row[J-1] = J - 1; for (P = 1; P <= N - 1; P++) { for (K = P + 1; K <= N; K++) { if ( fabs(A[Row[K-1]][P-1]) > fabs(A[Row[P-1]][P-1]) ) { T = Row[P-1]; Row[P-1] = Row[K-1]; Row[K-1] = T; } } if (A[Row[P-1]][P-1] 0) { printf("The matrix is SINGULAR !\n"); printf("Cannot use algorithm --- exit\n"); exit(1); } X[N-1] = A[Row[N-1]][N] / A[Row[N-1]][N-1]; for (K = N - 1; K >= 1; K--) { SUM = 0; for (C = K + 1; C <= N; C++) { SUM += A[Row[K-1]][C-1] * X[C-1]; } X[K-1] = ( A[Row[K-1]][N] - SUM) / A[Row[K-1]][K-1]; } for( K = 1; K <= N; K++) printf("X[%d] = %lf\n", K, X[K-1]); getch(); }
What letter logically follows T Q N K H?
T,Q,N,K,H,E next number
Prove that 2 power n lessthan equal to n factorial lessthan equal to n power n?
Assume 2^k < k! for all n > k here n > 2, then 2^n = 2^(n - 1)*2 < (n-1)! * n = n! Done. Connie and John
1 plus 3 plus 5 plus 7n how will you do this program?
Print "Type the upper limit (n) ?" Input n K = -1 WHILE K < = n K = K + 2 Sum = Sum + K WEND Print "The sum of all odd numbers up to "; n; "is "; Sum
How can you prove e-infinity equals 0?
I think you mean e to the (- infinity) power. The proof would be a limit proof. The limit (as n-->infinity) of [( en) ] = 0 You should have some other limits in class that you have proven. Show that your limit is less than one of those given for all values of n then you have your proof. For instance, if you already know that lim (as n-->infinity) of [(1/n) ] = 0 then for n = 1, 1/e1 < 1/1 true for n = 2, 1/e2 < 1/2 true Then assume that it is true for n = k so for n = K, 1/eK < 1/K assumed true therefore: eK > K multiply both by e e(k+1) > ke but we know ke > k+1 because e>1 SO: e(k+1) > ke > k+1 now take the reciprocal (reverses the inequalities) 1/e(k+1) < 1/ke < 1/ (k+1) by transitive prop of inequalities eliminate the middle term so that 1/e(k+1) < 1/ (k+1) this proves the case for n=K+1 and therefore will be true for all values of n since k was never a specified value. And if: 1/e(k+1) < 1/ (k+1) by one of the properties of limits, since the lim of 1/n is zero, then the lim of 1/en is also zero when n --> infinity.
How do you spell nicker?
If you mean knicker, like trousers, k-n-i-c-k-e-r
How would I solve for d in the equation ( e times d ) mod n equals 1?
ed mod n = 1 ed = kn + 1 for integer k so d = (kn + 1)/e for integer k.
Can a sequence of numbers be both geometric and arithmetic?
Yes, it can both arithmetic and geometric.The formula for an arithmetic sequence is: a(n)=a(1)+d(n-1)The formula for a geometric sequence is: a(n)=a(1)*r^(n-1)Now, when d is zero and r is one, a sequence is both geometric and arithmetic. This is because it becomes a(n)=a(1)1 =a(1). Note that a(n) is often written anIt can easily observed that this makes the sequence a constant.Example:a(1)=a(2)=(i) for i= 3,4,5...if a(1)=3 then for a geometric sequence a(n)=3+0(n-1)=3,3,3,3,3,3,3and the geometric sequence a(n)=3r0 =3 also so the sequence is 3,3,3,3...In fact, we could do this for any constant sequence such as 1,1,1,1,1,1,1...or e,e,e,e,e,e,e,e...In general, let k be a constant, the sequence an =a1 (r)1 (n-1)(0) with a1 =kis the constant sequence k, k, k,... and is both geometric and arithmetic.
Sum of 1 plus 2 plus 3 plus 4 plus .. plus n?
n(n+1)/2 You can see this from the following: Let x=1+2+3+...+n This is the same as x=n+(n-1)+...+1 x=1+2+3+...+n x=n+(n-1)+...+1 If you add the corresponding terms on the right-hand side of the two equations together, they each equal n+1 (e.g., 1+n=n+1, 2+n-1=n+1, ..., n+1=n+1). There are n such terms. So adding the each of the left-hand sides and right-hand sides of the two equations, we get: x+x=(n+1)+(n+1)+...+(n+1) [with n (n+1) terms on the right-hand side 2x=n*(n+1) x=n*(n+1)/2 A more formal proof by induction is also possible: (1) The formula works for n=1 because 1=1*2/2. (2) Assume that it works for an integer k. (3) Now show that given the assumption that it works for k, it must also work for k+1. By assmuption, 1+2+3+...+k=k(k+1)/2. Adding k+1 to each side, we get: 1+2+3+...+k+(k=1)=k(k+1)/2+(k+1)=k(k+1)/2+2(k+1)/2=(k(k+1)+2(k+1))/2=((k+2)(k+1))/2=(((k+1)+1)(k+1))/2=((k+1)((k+1)+1)/2
How do you derive the mean and variance for the Rayleigh distribution?
Use the following integral to find the first and second moments. The mean is, of course, the first moment and the variance is the second moment minus the first moment squared. :<math>\int_{0}^{\infty} x^{n} e^{-ax^2}\,\mathrm{d}x = \begin{cases} \frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n>-1,a>0) \\ \frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, k \;\text{integer}, a>0) \\ \frac{k!}{2a^{k+1}} & (n=2k+1,k \;\text{integer}, a>0) \end{cases} </math> (!! is the [[Double factorial]])
Gaussian elimination in c?
#include<stdio.h> #include<stdlib.h> #include<math.h> #include<conio.h> void main(void) { int K, P, C, J; double A[100][101]; int N; int Row[100]; double X[100]; double SUM, M; int T; do { printf("Please enter number of equations [Not more than %d]\n",100); scanf("%d", &N); } while( N > 100); printf("You say there are %d equations.\n", N); printf("From AX = B enter elements of [A,B] row by row:\n"); for (K = 1; K <= N; K++) { for (J = 1; J <= N+1; J++) { printf(" For row %d enter element %d please :\n", K, J); scanf("%lf", &A[K-1][J-1]); } } for (J = 1; J<= N; J++) Row[J-1] = J - 1; for (P = 1; P <= N - 1; P++) { for (K = P + 1; K <= N; K++) { if ( fabs(A[Row[K-1]][P-1]) > fabs(A[Row[P-1]][P-1]) ) { T = Row[P-1]; Row[P-1] = Row[K-1]; Row[K-1] = T; } } if (A[Row[P-1]][P-1] 0) { printf("The matrix is SINGULAR !\n"); printf("Cannot use algorithm --- exit\n"); exit(1); } X[N-1] = A[Row[N-1]][N] / A[Row[N-1]][N-1]; for (K = N - 1; K >= 1; K--) { SUM = 0; for (C = K + 1; C <= N; C++) { SUM += A[Row[K-1]][C-1] * X[C-1]; } X[K-1] = ( A[Row[K-1]][N] - SUM) / A[Row[K-1]][K-1]; } for( K = 1; K <= N; K++) printf("X[%d] = %lf\n", K, X[K-1]); getch(); }
A geometric sequence is defined recursively by an equals 12an-1 The first term of the sequence is 0.001 Which of the following is the explicit formula for the nth term of the sequence?
n=1 a(1) = 0.001 n=2 a(2) = 0.012 n=3 a(3) = 0.144 n=4 a(4) = 1.728 n=5 a(5) = 20.736 .... n=k a(k) = (12^(k-1))/1000 let n = k+1 a(k+1) = 12^(k)/1000 12(ak) = a(k+1) 12(12^k-1)/1000 = 12^k/1000 the 12 gets absorbed here. 12^(k-1+1)/1000 = 12^k/1000 Valid for k and k+1 therefore our equation A(n) = 12^(n-1)/1000, n(greater than or equal to) 1
Why is the sum of the reciprocals of all of the divisors of a perfect number equal to 2?
Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)which implies that 1/f(r) = f(k+1-r)/NThen 1/f(1) + 1/(f(2) + ... + 1/f(k)= f(k)/N + f(k-1)/N + ... + f(1)/N= [f(k) + f(k-1) + ... + f(1)] / N= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N
What are some eight letter words with 2nd letter E and 4th letter N and 6th letter E and 8th letter K?
According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 1 words with the pattern -E-N-E-K. That is, eight letter words with 2nd letter E and 4th letter N and 6th letter E and 8th letter K. In alphabetical order, they are: penneeck
C program for optimal merge pattern?
#include<iostream.h> #include<conio.h> void main() { clrscr(); int i,k,a[10],c[10],n,l; cout<<"Enter the no. of elements\t"; cin>>n; cout<<"\nEnter the sorted elments for optimal merge pattern"; for(i=0;i<n;i++) { cout<<"\t"; cin>>a[i]; } i=0;k=0; c[k]=a[i]+a[i+1]; i=2; while(i<n) { k++; if((c[k-1]+a[i])<=(a[i]+a[i+1])) { c[k]=c[k-1]+a[i]; } else { c[k]=a[i]+a[i+1]; i=i+2; while(i<n) { k++; if((c[k-1]+a[i])<=(c[k-2]+a[i])) { c[k]=c[k-1]+a[i]; } else { c[k]=c[k-2]+a[i]; }i++; } }i++; } k++; c[k]=c[k-1]+c[k-2]; cout<<"\n\nThe optimal sum are as follows......\n\n"; for(k=0;k<n-1;k++) { cout<<c[k]<<"\t"; } l=0; for(k=0;k<n-1;k++) { l=l+c[k]; } cout<<"\n\n The external path length is ......"<<l; getch(); }
Words can you make with the following letters n a t i o n a l s c i e n c e w e e k n a t i o n a l s c i e n c e w e e k?
national science week nation weak weaker known to a sick i
