Starting from L ( 1 ) = 1 s , use the basic properties of Laplace transform to show that L ( t n ) = n ! s n + 1 for every positive integer n.

Name: Starting from L ( 1 ) = 1 s , use the basic properties of Laplace transform to show that L ( t n ) = n ! s n + 1 for every positive integer n.
Uploaded: 2022-02-23T17:22:57+00:00
Description: Starting from L ( 1 ) = 1 s , use the basic properties of Laplace transform to show that L ( t n ) = n ! s n + 1 for every positive integer n.

Question

Starting from       L    (    1    )    =          1      s      , use the basic properties of Laplace transform to show that       L    (          t      n        )    =                  n        !                    s                  n          +          1                     for every positive integer n.

Jolie Padilla · Accepted Answer

Apply induction on n and use       L    (    1    )    =                  1        s             in the base case. We need to show that      L    (          t      n        )    =                            n          !                          s                      n            +            1                              for every positive integer n.Base case: When   n  =  0, we have       t    0    =  1. Therefore                              L                {                  t          0                }                            =                  L                {        1        }        =                              1            s                          =                                            0              !                                      s                              0                +                1                                                        Induction Hypothesis: Let   n  ∈      N   with   n  ≥  1. Then      L    {      t    n    }  =                    n        !                    s                  n          +          1                    Induction Step: We need to show that      L    {      t          n      +      1        }  =                    (        n        +        1        )        !                    s                  n          +          2                    starting from the LHS      L    {      t          n      +      1        }  =      ∫    0    ∞        t          n      +      1            e          −      s      t           d  twe now use integration by parts, which says that  ∫  f      g    ′    =  f  g  −  ∫      f    ′    g     d  tlet   f  =      t          n      +      1       and       g    ′    =      e          −      s      t      . Then,       f    ′    =  (  n  +  1  )      t    n   and   g  =  −            1      s            e          −      s      t       so that  ∫      t          n      +      1            e          −      s      t           d  t  =  −                    t                  n          +          1                    s            e          −      s      t        +                    n        +        1            s        ∫      t    n        e          −      s      t           d  tevaluating t=0 and   t  →  ∞, we get                              L                {                  t                      n            +            1                          }                            =                                          −                                          1                s                                                    t                              n                +                1                                                    e                              −                s                t                                      |                    0          ∞                +                                            n              +              1                        s                                    L                {                  t          n                }                                          =                                          −                                                                                s                                          −                      1                                                                            t                                          n                      +                      1                                                                                        e                                      s                    t                                                                        |                    0          ∞                +                                            n              +              1                        s                                    L                {                  t          n                }                                          =        0        −        0        +                                            n              +              1                        s                                    L                {                  t          n                }                                          =                                            n              +              1                        s                          ×                                            n              !                                      s                              n                +                1                                                                                      =                                            (              n              +              1              )              !                                      s                              n                +                2                                                        Hence, by the principal of mathematical induction we see that       L    (          t      n        )    =                            n          !                          s                      n            +            1                               is true for every positive integer n.

obojeneqk · Accepted Answer

Apply integration by parts.      L    (      t    n    )  =      ∫    0    ∞        t    n        e          −      s      t        d  t  =  (  −  1      /    s  )      e          −      s      t            t    n              |        0    ∞    +  (  1      /    s  )      ∫    0    ∞    n      t          n      −      1            e          −      s      t        d  t  =  (  n      /    s  )      L    (      t          n      −      1        )Thus      L    (  t  )  =  (  1      /    s  )      L    (  1  )  =  1      /              s      2      and so forth.

Starting from L(1)=1/s, use the basic properties of Laplace transform to show that L(t^n)=(n!)/(s^(n+1)) for every positive integer n.

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