Thomas F. X. Smith died on 1996-05-31.
It the the probability that the random variable in question takes any value up to and including the argument. Suppose you have a random variable X and f(x) is the probability that X = x [that is, the rv X takes the value x]. If F(x) denotes the cumulative distribution function of X, then F(x) is the sum of all f(y) where y <= x. Thus, for a fair die, F(1) = f(1) = 1/6 F(2) = f(1) + f(2) = 2/6 F(3) = f(1) + f(2) + f(3) = 3/6 and so on. Note that F(X) = 0 for X < 1, F(a+b) where a is an integer in the interval [1,6] and 0<b<1 is F(a). Thus, for example, F(3.5) = F(3). and F(x) = 1 for x >=6. In the case of continuous probability distributions, the summation is replaced by integration.
In mathematics, a function F(x) is the antidifference of f(x) if F(x+1)-F(x)=f(x).
In mathematics, the curl of a vector is the maximum rotation on a vector field, oriented perpendicularly to the certain plane. The curl of a vector is defined by this form: ∇ x F = [i . . . . j . . . . . k] [∂/∂x ∂/∂y ∂/∂z] [P. . . Q. . . .R. . ] ...given that F = <P,Q,R> or Pi + Qj + Rk Perform the cross-product of the terms to obtain: ∇ x F = (∂R/∂y - ∂Q/∂z)i + (∂P/∂z - ∂R/∂x)j + (∂Q/∂x - ∂P/∂y)k
PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
J. F. X. O'Brien was born in 1828.
F. X. J. D'Ambrosio has written: 'Prepare or deter?'
Judges in the X Factor
J. F. X. Hoery has written: 'Banjire wis surut' 'Wuludomba pancal panggung'
#include<stdio.h> #include<math.h> int main() { float x[10],y[10][10],sum,p,u,temp; int i,n,j,k=0,f,m; float fact(int); printf("\nhow many record you will be enter: "); scanf("%d",&n); for(i=0; i<n; i++) { printf("\n\nenter the value of x%d: ",i); scanf("%f",&x[i]); printf("\n\nenter the value of f(x%d): ",i); scanf("%f",&y[k][i]); } printf("\n\nEnter X for finding f(x): "); scanf("%f",&p); for(i=1;i<n;i++) { k=i; for(j=0;j<n-i;j++) { y[i][j]=(y[i-1][j+1]-y[i-1][j])/(x[k]-x[j]); k++; } } printf("\n_____________________________________________________\n"); printf("\n x(i)\t y(i)\t y1(i) y2(i) y3(i) y4(i)"); printf("\n_____________________________________________________\n"); for(i=0;i<n;i++) { printf("\n %.3f",x[i]); for(j=0;j<n-i;j++) { printf(" "); printf(" %.3f",y[j][i]); } printf("\n"); } i=0; do { if(x[i]<p && p<x[i+1]) k=1; else i++; }while(k != 1); f=i; sum=0; for(i=0;i<n-1;i++) { k=f; temp=1; for(j=0;j<i;j++) { temp = temp * (p - x[k]); k++; } sum = sum + temp*(y[i][f]); } printf("\n\n f(%.2f) = %f ",p,sum); return 0; }
F. X. Martin died in 2000.
#include<stdio.h> #include<conio.h> #include<math.h> void main() { float x[10],y[10],temp=1,f[10],sum,p; int i,n,j,k=0,c; clrscr(); printf("\nhow many record you will be enter: " ); scanf("%d" ,&n); for (i=0; i<n; i++) { printf("\n\nenter the value of x%d: " ,i); scanf("%f" ,&x[i]); printf("\n\nenter the value of f(x%d): " ,i); scanf("%f" ,&y[i]); } printf("\n\nEnter X for finding f(x): " ); scanf("%f" ,&p); for (i=0;i<n;i++) { temp = 1; k = i; for (j=0;j<n;j++) { if (k==j) { continue ; } else { temp = temp * ((p-x[j])/(x[k]-x[j])); } } f[i]=y[i]*temp; } for (i=0;i<n;i++) { sum = sum + f[i]; } printf("\n\n f(%.1f) = %f " ,p,sum); getch(); }
Vegetables that start with the letters f i j u v x and y:fiddlehead fernfenneliceberg lettuceJapanese radishullucovernonia calvoanayamYukon gold potato
John F. X. McGohey died in 1972.
The energy of an electromagnetic wave is given by E = hf, where E is energy, h is Planck's constant, and f is frequency. Rearranging this formula, we get f = E / h. Substituting the values, we get f = 5.0 x 10^20 J / 6.626 x 10^-34 J s ≈ 7.54 x 10^53 Hz.
Thomas F. X. Smith died on 1996-05-31.
The frequency of a photon with an energy of 3.38 x J can be found using the equation E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency. Rearranging the equation to solve for frequency gives f = E/h. Plugging in the values gives f = (3.38 x J) / (6.626 x 10^-34 J·s) = approximately 5.10 x 10^14 Hz.