Have you ever used Scratch pad like this?

Find the root of following equation.

4y32700(1y)4=04y^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)=4y32700(1y)4f(y)=4y^3-2700(1-y)^4 We have to find root of above function.

2)Find derivative of f(y)f(y). Here,f(y)=12y2+10800(1y)3f^{'}(y)=12y^2+10800(1-y)^3

3)Find yny_n using newton raphson method y0=1y_0=1 yn+1=ynf(yn)f(yn)y_{n+1}=y_n-\frac{f(y_n)}{f^{'}(y_n)} yn+1=y4y32700(1y)412y2+10800(1y)3y_{n+1}=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}

You can use calculator or scratchpad to calculate y1,y2,y3,....y_1,y_2,y_3,.....If this equation has real solution ,these values y1,y2,y3,....y_1,y_2,y_3,.... would converge to root of f(y)f(y).

How to do such iterations smartly?

1) On first line of scratch pad write y=1\color{#69047E}{\text{On first line of scratch pad write y=1}}

2) On second line write\color{#D61F06}{\text{On second line write}}

y=y4y32700(1y)412y2+10800(1y)3y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}

You will see new value of y as result.\color{#20A900}{\text{You will see new value of y as result.}}

3) From line 3, continue pasting\color{#EC7300}{\text{From line 3, continue pasting}}

y=y4y32700(1y)412y2+10800(1y)3y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}

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.\color{#3D99F6}{\text{By using scratchpad, try finding sum of initial 10 fibonacci number.}}

Note by Aamir Faisal Ansari
4 years, 2 months ago

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Yes !i have used it this way.Sometimes in Recursion problems and like that.

Gautam Sharma - 4 years, 2 months ago

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By using scratchpad, try finding sum of initial 10 fibonacci number.

Aamir Faisal Ansari - 4 years, 2 months ago

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