Find the root of following equation.
\[4y^3-2700(1-y)^4=0\]
Above equation can be solved using newton raphson iterative method with initial approximation to be unity.
1) First Let f(y)=4y3−2700(1−y)4
We have to find root of above function.
2)Find derivative of f(y).
Here,f′(y)=12y2+10800(1−y)3
3)Find yn using newton raphson method
y0=1
yn+1=yn−f′(yn)f(yn)
yn+1=y−12y2+10800(1−y)34y3−2700(1−y)4
You can use calculator or scratchpad to calculate y1,y2,y3,.....If this equation has real solution ,these values y1,y2,y3,.... would converge to root of f(y).
How to do such iterations smartly?
1) On first line of scratch pad write y=1
2) On second line write
y=y−12y2+10800(1−y)34y3−2700(1−y)4
You will see new value of y as result.
3) From line 3, continue pasting
y=y−12y2+10800(1−y)34y3−2700(1−y)4
until you get a constant value.If your initial approximation is not too away from actual solution,you would get constant value in no more than 10 steps.
By this method you can solve most of the transcendental equation.
By using scratchpad, try finding sum of initial 10 fibonacci number.
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Yes !i have used it this way.Sometimes in Recursion problems and like that.
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By using scratchpad, try finding sum of initial 10 fibonacci number.