Doubt in Eulers method with smaller step size

I have a doubt in applying eulers method with smaller step size. While I got all the problems with step size =1 correct, I am unable to get any in the other category correct. Isnt eulers method this?

${ y }_{ n+1 }={ y }_{ n }+\quad (step\quad size\quad *\quad dy/dx)$ ?

Im trying probs like dy/dx= x^3 , f(1)=5, step size = 1/3, find f(12)- f(7). I even wrote a program to do this but still im getting the wrong answer. Can someone please help?

#Calculus #Euler #Problem #Trouble #Smallstepsize

Easy Math Editor

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Comments

class euler { public static void main(String args[]) { float i,fn=5,d,step,f7; for(i=1;i<=12;i=i+1/3) { d=iii; step=1/3;

fn=fn+step*d; if(i==7) { f7=fn; }

} System.out.println(fn - f7); } } ye try kr .. P.S: n compiler here :D

Mayank Uniyal - 6 years, 2 months ago

while(x<=limit) { diff=xxx; printf("%f \t %f \t %f \n",x,y,diff); y=y+ (step*diff); x=x+step;

Guess ive used the same thing. Check the screenshot, its from the results of the same code.

Romil Punetha - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$