I've just discoveblack that I can use Euler's method to estimating the slope of y = ( x + c ) 2 with perfect accuracy, regardless of what Δ x is used. What causes this?Does it happen with other functions, what must they look like for it to happen?I am using this method to estimate the slope, m m ≈ ( x + Δ x + c ) 2 − ( x − Δ x + c ) 2 2 Δ x Doesn't matter what Δ x is, the slope is always correct ...I would expect this only to work for tiny Δ x, but works with large ones too.

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I&#039;ve just discoveblack that I can use Euler&#039;s method to estimating the slope of   y  =  (  x  +  c      )    2   with perfect accuracy, regardless of what   Δ  x is used. What causes this?Does it happen with other functions, what must they look like for it to happen?I am using this method to estimate the slope,   m  m  ≈            (      x      +      Δ      x      +      c              )        2            −      (      x      −      Δ      x      +      c              )        2                    2      Δ      x      Doesn&#039;t matter what   Δ  x is, the slope is always correct ...I would expect this only to work for tiny   Δ  x, but works with large ones too.

Abbigail Vaughn · Accepted Answer

In the quadratic function   y  =      x    2  , you can very easily show that the secant line between points   (  a  ,      a    2    ) and   (  b  ,      b    2    ) where   b  ≠  a is parallel to the tangent at the midpoint   (      1    2    (  a  +  b  )  ,      1    4    (  a  +  b      )    2    ).The slope of the secant is                     b        2            −              a        2                    b      −      a        =  b  +  a. Slope of the tangent at the midpoint (by calculus) is   (  2  )  (      1    2    (  a  +  b  )  )  =  a  +  b, so they&#039;re equal and the secant is parallel to the tangent.That&#039;s the reason for your observation. Adding a constant, or using a constant multiplier, etc. won&#039;t change the result so it applies for any function of the form   y  =  k  (  x  +  c      )    2    +  m.

Taniya Burns · Accepted Answer

Thanks a lot

I need to solve the following Cauchy problem using Runge Kutta method (do 2 iterations). y′(t)=x(t)y(t)+x^2(t) x′(t)=y^2(t) y(0)=x(0)=1

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