Видео

25

Sir

carry on sir...

Thanks

Ektu kosto kore likhe janas kemon laglo toder ... Karon, for me 1 comment=>>♾️ motivation

Kire kemon lagche toder....? Ei video ta dekhe concept ta clear kor tarpor exam purpose e video anbo

বুঝতে পারছিস তো সবাই ? Janas ektu..

খুব সুন্দর 👍

Kemon lagche toder jodi somoy thake janas...😊

Dedication ❤❤❤

Toder kemon laglo janas..

Thanks ❤❤

আপনার বাসা কোথায়?

ভিডিও তে একটু গুণ টুন ভুল আছে তোমরা একটু চেক করে নেবে

Are bhai 11025 hobe 225×7=1475

It feels so satisfying when a teacher teach something with great enthusiasm. I wish i had a physic teacher like you. Make sure to never satisfied with your teaching effort. You need be to tanacious in order create exceptional contents.

Sir apni math ta valo bojate pareni 😥

Dada wali comedy Akon Tomar dekhe parinaam

Sir খুব ভালো বুঝতে পারলাম

খুব ভালো বুঝতে পারলাম

এই অধ্যায় খুব ভালো বুঝতে পারলাম আপনি খুব ভালো পড়ান

1575hobe

🎉🎉

Thank you

👍🏻

Thnx sir

🎉

😂😂😂😂😂

❤❤❤❤❤❤

❤❤❤❤❤❤

Thank you sir

Nice 🎉

স্যার মাধ্যমিকের কিছু করুন কারণ সামনে মাধ্যমিক পরীক্ষা 🙏☺️

Superb content 🔥🔥🔥

Amra to Jani v=v vag p kintu akhane askoyar also Kotha theka

Nice❤❤

🔥

🎉🎉Op video 🎉🎉❤

🎉🎉Op video ❤❤🎉

Thank you 💚

Keep it up bro ...💙

👍

Thank you sir ato sundor kore kew bojayni

ভাই একটা ফ্রিজ তো আটঘন্টা চলে না 24 ঘন্টা চলে । সে ক্ষেত্রে আপনার হিসাব অনুযায়ী একটা ফ্রিজের বিল 15/ 1600 টাকা চলে আসবে। এটা কি করে সঠিক হতে পারে

THE RK 4 METHOD #include<stdio.h> #include<math.h> int main() { long int n,i; double f(double,double),x[100],y[100],k[6],sum,w,h; printf("enter x_0 : "); scanf("%lf", &x[0]); printf(" enter y_0 : "); scanf("%lf", &y[0]); printf(" enter x : "); scanf("%lf", & w); printf(" enter h : "); scanf("%lf", &h); n=(w-x[0])/h ; for(i=1;i<=n;i++) { k[1]=h*f(x[i-1],y[i-1]); k[2]=h*f(x[i-1]+(h/2),y[i-1]+(k[1]/2)); k[3]=h*f(x[i-1]+(h/2),y[i-1]+(k[2]/2)); k[4]=h*f(x[i-1]+h,y[i-1]+k[3]); sum=(k[1] + 2*k[2] +2*k[3] +k[4])/6; y[i]=y[i-1]+sum; x[i]=x[i-1]+h; } printf(" rk_4 method gives "); printf(" y(%lf) = %0.9lf",x[n],y[n]); printf(" correct to 9 decimal places ."); return 0; } double f(double a,double b) { return ((a*a) + (b*b/4)); }

Modified Euler method #include<stdio.h> #include<math.h> int main() { long int c,z,j,i; double x[50],y[50],h,m[50],m1,m2,a; double s[100],f(double,double),w; printf("enter x_0 : "); scanf("%lf",& x[0]); printf(" enter h : "); scanf("%lf",& h); printf(" enter x : "); scanf("%lf",& a); printf(" enter y_0 : "); scanf("%lf",& y[0]); z=(a-x[0])/h; s[0]=y[0]; for(i=1;i<=z;i++) { x[i]= x[0]+ i*h; m[i]=f(x[i-1],y[i-1]); c=0; do { m1=f(x[i],s[c]); m2=(m[i]+m1)/2; s[c+1]=y[i-1]+m2*h; w=s[c]-s[c+1]; w=fabs(w); c=c+1; } while(w>0.0000000001); y[i]=s[c]; } printf(" modified euler method gives"); printf(" y ( %lf ) = %0.9lf", x[i-1],y[i-1]); printf(" correct to 9 decimal places ."); return 0; } double f(double a,double b) { return (((2- b*b)/(5*a))); }

Euler method #include<stdio.h> #include<math.h> int main() { int i; double x[100],k,y[100],f(double,double),n,w,h; printf("enter X_0 : "); scanf("%lf", & x[0]); printf(" enter Y_0 : "); scanf("%lf", & y[0]); printf(" enter step length : "); scanf("%lf", & h); printf(" input the value of x : "); scanf("%lf", & k); n=(k-x[0])/h; for(i=1;i<=n;i++) { x[i]=x[i-1]+h; } for(i=1;i<=n;i++) { y[i]=y[i-1]+ h*f(x[i-1],y[i-1]); } printf(" euler method gives"); printf(" y( %lf ) = %0.9lf",x[i-1],y[i-1]); printf(" correct to 9 decimal places . "); return 0; } double f(double u ,double z ) { return (z-u)/(z+u); }

THE PARABOLIC CURVE FITTING #include<stdio.h> #include<math.h> int main() { double a0,a1,a2,D,Da0,Da1,Da2,y,x; double sx=0,sx2=0,sx3=0,sx4=0,sxy=0,sx2y=0,sy=0; int i,n; printf("enter no of points : "); scanf("%d",&n); for(i=0;i<n;i++) { printf(" enter x(%d) y(%d) : ",i,i); scanf("%lf %lf",&x,&y); sx+=x; sy+=y; sx2+=x*x; sx3+=x*x*x; sx4+=x*x*x*x; sxy+=x*y; sx2y+=x*x*y; } D=n*(sx2*sx4-sx3*sx3)-sx*(sx*sx4-sx2*sx3)+sx2*(sx*sx3-sx2*sx2); Da0=sy*(sx2*sx4-sx3*sx3)-sx*(sxy*sx4-sx2y*sx3)+sx2*(sxy*sx3-sx2*sx2y); Da1=n*(sxy*sx4-sx3*sx2y)-sy*(sx*sx4-sx2*sx3)+sx2*(sx*sx2y-sx2*sxy); Da2=n*(sx2*sx2y-sx3*sxy)-sx*(sx*sx2y-sx2*sxy)+sy*(sx*sx3-sx2*sx2); a0=Da0/D; a1=Da1/D; a2=Da2/D; printf(" equation of parabola is "); printf(" ( %0.6lf ) x*x + ( %0.6lf ) x + %0.6lf ",a2,a1,a0); printf(" Correct to 6 decimal places ."); return 0; }