Gate 2016 ce paperi mathematicsq1
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gate 2016 engineering mathematics ce/paper i/q1--the newton raphson method
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Gate 2016 ce paperi mathematicsq1
- 3. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx
- 4. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that x0 = 0.333
- 5. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that x0 = 0.333 So f(x0) = -0.069
- 6. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that x0 = 0.333,f(x0) =-0.069 The Newton-Raphson formula is as follows x1 = x0 – f’(x0)/f(x0)
- 7. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that f(x0) = 0.333,f(x0)=--0.069 The Newton-Raphson formula is as follows x1 = x0 – f’(x0)/f(x0) We have f’(x0)=2.55
- 8. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that f(x0) = 0.333,f(x0)=--0.069 The Newton-Raphson formula is as follows x1 = x0 – f’(x0)/f(x0) We have f’(x0)=2.55 So x1= 0.333+ 0.069/2.55=0.360