Why is f ( x + h ) = f ( x ) + h f ′ ( x ) + O ( h 2 ) ?

Question

Why is   f  (  x  +  h  )  =  f  (  x  )  +  h      f    ′    (  x  )  +  O  (      h    2    ) ?

Serenity Mason · Accepted Answer

Taylor&#039;s Theorem with Peano&#039;s Form of Remainder: Let f be a function such that       f          (      n      )        (  a  ) exists. Then we have  f  (  a  +  h  )  =  f  (  a  )  +  h      f    ′    (  a  )  +            h              2                    2      !            f    ″    (  a  )  +  ⋯  +            h              n                    n      !            f          (      n      )        (  a  )  +  o  (      h          n        )For n=1  (which is also the first result in your question with x in place of a) this is an immediate result of definition of derivative       f    ′    (  a  ) and hence it does not require any proof. The proof for   n  &amp;gt;  1 follows by the application of L&#039;Hospital&#039;s Rule applied (n-1) times.Let&#039;s now see what happens when we put n=2 in Taylor&#039;s theorem above. We get  f  (  a  +  h  )  −  f  (  a  )  =  h      f    ′    (  a  )  +      h          2            f    ″    (  a  )      /    2  +  o  (      h          2        )which implies that  f  (  a  +  h  )  −  f  (  a  )  =  h      f    ′    (  a  )  +  O  (      h          2        )Thus for you second result we need the existence of f′′ at the point under consideration. You can see that we don&#039;t need the continuity of f′′ at the point under consideration.

Why is f(x+h)=f(x)+hf'(x)+O(h^2) ?

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