We have been told that u ′ ( t ) = u ( t ) and u ( 0 ) = 1 from this information we can conclude that u ( t ) = e x Show that u k = ( 1 + h ) k , k = 0 , 1 , . . . is an approximation for u ( k h ) using Euler's methodMy current thoughts are that h is the interval by which we increment, but I am not sure where to go from there.

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We have been told that       u    ′    (  t  )  =  u  (  t  ) and   u  (  0  )  =  1 from this information we can conclude that   u  (  t  )  =      e    x  Show that       u    k    =  (  1  +  h      )    k    ,  k  =  0  ,  1  ,  .  .  . is an approximation for   u  (  k  h  ) using Euler&#039;s methodMy current thoughts are that h is the interval by which we increment, but I am not sure where to go from there.

skopcze2lelm · Accepted Answer

Call      u    k    =  u  (  k  h  )                  for             k  =  0  ,  1  ,  ⋯Euler&#039;s method can be expressed as      u          k      +      1        =      u    k    +  h      u    k    ′    =      u    k    +  h  (      u    k    )  =  (  1  +  h  )      u    k  You can easily figure out the solution to this expression                              u          1                                    =                    (        1        +        h        )                  u          0                                              u          2                                    =                    (        1        +        h        )                  u          1                =        (        1        +        h                  )          2                          u          0                                              u          3                                    =                    (        1        +        h        )                  u          2                =        (        1        +        h                  )          3                          u          0                                                  ⋮                                            u          k                                    =                    (        1        +        h        )                  u                      k            −            1                          =        (        1        +        h                  )          k                          u          0                    which reduces to       u    k    =  (  1  +  h      )    k   if       u    0    =  u  (  t  =  0  )  =  1

We have been told that u ′ </msup> ( t ) = u ( t )

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