I am trying to calculate the global error bound for Euler's method, but I am having trouble. I am given the formula | y ( t i ) − u i | ≤ 1 L ( h M 2 + δ h ) ( e L ( t i − a ) − 1 ) + | δ 0 | e L ( t i − a ) where u i is the Euler approxmation. I am also given M , L , a , δ , δ 0 , h. If I am not mistaken this will give me the error for each step, but how do I find the upper bound for the total error?

Question

I am trying to calculate the global error bound for Euler&#039;s method, but I am having trouble. I am given the formula       |    y  (      t          i        )  −      u          i            |    ≤      1    L    (            h      M        2    +      δ    h    )  (      e          L      (              t                  i                    −      a      )        −  1  )  +      |        δ          0            |        e          L      (              t                  i                    −      a      )       where       u          i       is the Euler approxmation. I am also given   M  ,  L  ,  a  ,  δ  ,      δ          0      , h. If I am not mistaken this will give me the error for each step, but how do I find the upper bound for the total error?

Rebecca Benitez · Accepted Answer

sin          −      1        ⁡      (                  2        x                    1        +                  x          2                      )    =  θ      ⟺      sin  ⁡  (  θ  )  =            2      x              1      +              x        2                ∧    −      π    2    ≤  θ  ≤      π    2  Note that  cos  ⁡  (  θ  )  =      |                  1        −                  x          2                            1        +                  x          2                      |  When       |    x      |    ≤  1  tan  ⁡  (  θ      /    2  )  =            sin      ⁡      (      θ      )              1      +      cos      ⁡      (      θ      )        =                    2        x                    1        +                  x          2                            1      +                        1          −                      x            2                                    1          +                      x            2                                =  xTherefore,  θ  =  2      tan          −      1        ⁡  (  x  )When       |    x      |    &amp;gt;  1  tan  ⁡  (  θ      /    2  )  =            sin      ⁡      (      θ      )              1      +      cos      ⁡      (      θ      )        =                    2        x                    1        +                  x          2                            1      −                        1          −                      x            2                                    1          +                      x            2                                =  1      /    xTherefore,  θ  =  2      tan          −      1        ⁡  (  1      /    x  )Thus,      sin          −      1        ⁡      (                  2        x                    1        +                  x          2                      )    =      {                            2                      tan                          −              1                                ⁡          (          x          )                          if                       |                    x                      |                    ≤          1                                      2                      tan                          −              1                                ⁡          (          1                      /                    x          )                          if                       |                    x                      |                    &amp;gt;          1                        and      tan          −      1        ⁡  (  1      /    x  )  =      {                                        π            2                    −                      tan                          −              1                                ⁡          (          x          )                          if           x          &amp;gt;          0                                      −                      π            2                    −                      tan                          −              1                                ⁡          (          x          )                          if           x          &amp;lt;          0

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