This question was asked in CSIR. please help me to find out correct choiceLet y ( t ) satisfy the differential equation y ′ = λ y ; y ( 0 ) = 1. Then the backward Euler method for n &gt; 1 and h &gt; 0 y n − y n − 1 h = λ y n ; y ( 0 ) = 1yields1. a first order approximation to e λ n h 2. a polynomial approximation to e λ n h 3. a rational function approximation to e λ n h 4. a Chebyshev polynomial approximation to e λ n h

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This question was asked in CSIR. please help me to find out correct choiceLet   y  (  t  ) satisfy the differential equation      y    ′    =  λ  y  ;  y  (  0  )  =  1. Then the backward Euler method for   n  &amp;gt;  1 and   h  &amp;gt;  0                    y        n            −              y                  n          −          1                      h    =  λ      y    n    ;    y  (  0  )  =  1yields1. a first order approximation to       e          λ      n      h      2. a polynomial approximation to       e          λ      n      h      3. a rational function approximation to       e          λ      n      h      4. a Chebyshev polynomial approximation to       e          λ      n      h

Leslie Rollins · Accepted Answer

The correct answer is (1).Let us prove it:An the solution is to apply Taylor series. Let       y    n   be the approximation point or the center of series, because it is occurs twice per numerical scheme.  y  (  x  )  =  y  (  c  )  +      1          1      !        (  x  −  c  )      y    ′    (  c  )  +  .  .  .  +      1          n      !        (  x  −  c      )    n        y          (      n      )        (      x    0    )  +  .  .  .Let   c  :=      x          n        ;    x  :=      x          n      −      1        ;        y    n    :=  y  (      x    n    )  ;        y          n      −      1        :=  y  (      x          n      −      1        )  ;    h  :=      x    n    −      x          n      −      1            y          n      −      1        =      y          n        −  h      y    ′    (      x          n        )  +            h      2        2        y    ″    (      x          n        )  −            h      3        6        y    ‴    (      x          n        )  +  .  .  .                    y                  n                    −              y                  n          −          1                      h    =      y    ′    (      x          n        )  −      h    2        y    ″    (      x          n        )  +            h      2        6        y    ‴    (      x          n        )  −  .  .  .We may show that residual depends on step h linearly:   0  =                    y                  n                    −              y                  n          −          1                      h    −      y    n    =                              (                      y            ′                    (                      x                          n                                )          −          λ                      y            n                    )                ⏟                    =      0        −      h    2        y    ″    (      x          n        )  +            h      2        6        y    ‴    (      x          n        )  −  .  .  .  =  O  (  h  )So it is first order approximation.

Addison Trujillo · Accepted Answer

Since the explicit solution of that recursion is      y    n    =            (        1  −                    λ            n      h        n                      )                    −      n      you get for   n  h  =  t fixed a well-known approximation of the exponential       e                  λ            t      . Using the logarithm series you can find the approximation order.2. and 3. are ambiguous since they do not specify the variable of the function, i.e., what is considered fixed in those expressions.

This question was asked in CSIR. please help me to find out correct choice Let y ( t

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