Consider the following problem y ′ = ln ⁡ ln ⁡ ( 4 + y 2 ) , x ∈ [ 0 , 1 ] , y ( 0 ) = 1We can formulate the problem the approximate the solution y n + 1 = y n + h ln ⁡ ln ⁡ ( 4 + y n 2 ) , n = 0 , 1 , . . . , N − 1 , y 0 = 0with mesh points x n = n h. I am tasked with finding the truncation error. Now considering a Taylor series we find that | T n | ≤ h 2 ! | ( ln ⁡ ln ⁡ ( 4 + ξ 2 ) ) ′ | ξ ∈ ( x n , x n + 1 )Now ln ⁡ ln ⁡ ( 4 + y 2 ) ′ = d d y ln ⁡ ln ⁡ ( 4 + y 2 ) = 2 y ln ⁡ ln ⁡ ( 4 + y 2 ) ln ⁡ ( 4 + y 2 ) ( 4 + y 2 ) ≤ 2 y 4 + y 2 ≤ 1 / 2 ∴ | T n | ≤ h 4 Is this the correct way to go about obtaining this?

Question

Consider the following problem      y    ′    =  ln  ⁡  ln  ⁡  (  4  +      y    2    )  ,    x  ∈  [  0  ,  1  ]  ,  y  (  0  )  =  1We can formulate the problem the approximate the solution      y          n      +      1        =      y    n    +  h  ln  ⁡  ln  ⁡  (  4  +      y    n    2    )  ,    n  =  0  ,  1  ,  .  .  .  ,  N  −  1  ,        y    0    =  0with mesh points       x    n    =  n  h. I am tasked with finding the truncation error. Now considering a Taylor series we find that      |        T    n        |    ≤      h          2      !            |    (  ln  ⁡  ln  ⁡  (  4  +      ξ    2    )      )    ′        |      ξ  ∈  (      x    n    ,      x          n      +      1        )Now  ln  ⁡  ln  ⁡  (  4  +      y    2        )    ′    =      d          d      y        ln  ⁡  ln  ⁡  (  4  +      y    2    )  =            2      y      ln      ⁡      ln      ⁡      (      4      +              y        2            )              ln      ⁡      (      4      +              y        2            )      (      4      +              y        2            )        ≤            2      y              4      +              y        2              ≤  1      /    2  ∴      |        T    n        |    ≤      h    4  Is this the correct way to go about obtaining this?

bequejatz8d · Accepted Answer

I believe that the only mistake is in your application of the chain rule. I find it helpful to break the function into its composite parts and apply the chain rule one step at a time.Let   f  ,  g  ,  h be the functions given by                    f        (        y        )                            =        4        +                  y          2                                    g        (        y        )                            =        ln        ⁡        (        y        )                            h        (        y        )                            =        g        (        y        )        .            Then  k  (  y  )  =  h  (  g  (  f  (  y  )  )  =  ln  ⁡  (  ln  ⁡  (  4  +      y    2    )is the function of interest and   k is defined and differentiable for all   y. Moreover,      k    ′    (  y  )  =      h    ′    (  g  (  f  (  y  )  )      g    ′    (  f  (  y  )  )      f    ′    (  y  )  =      1          ln      ⁡      (      4      +              y        2            )            1          4      +              y        2              2  y  .The derivative       k    ′   can be bounded as follows      |        k    ′    (  y  )      |    ≤      1          ln      ⁡      (      4      )                  2              |            y              |                    4      +              y        2              ,because ln is monotone increasing. You have already applied the helpful inequality      |    a  b      |    ≤      1    2    (      a    2    +      b    2    )  ,which in our case, where   a  =  2 and   b  =      |    y      |  , allows for the estimate      |        k    ′    (  y  )      |    ≤      1          2      ln      ⁡      (      4      )        .Breaking complicated functions into composite parts is especially useful when programming computers. One line per component produces a program which is easy for a human being to debug/verify.

Consider the following problem y′=lnln(4+y^2),x in [0,1],y(0)=1

Answered question

Answer & Explanation

New Questions in Integral Calculus