Why do the antiderivatives of y = 1 follow e x ?If you integrate y = 1, you get x. If you integrate that, you get x 2 2 , following the power rule. If you continue this, integrating over and over, the antiderivatives are: 1, x, x 2 2 , x 3 6 , x 4 24 , and so on. It follows the pattern of x n n ! , which happens to be the infinite polynomial of e x . Just by integrating y = 1, you follow the e x p ( ) function. Is there any intuitive way why this is?

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Why do the antiderivatives of   y  =  1 follow       e          x      ?If you integrate   y  =  1, you get x. If you integrate that, you get             x              2              2  , following the power rule. If you continue this, integrating over and over, the antiderivatives are: 1, x,             x              2              2  ,             x              3              6  ,             x              4              24  , and so on. It follows the pattern of             x              n                    n      !      , which happens to be the infinite polynomial of       e          x      . Just by integrating   y  =  1, you follow the   e  x  p  (  ) function. Is there any intuitive way why this is?

concludirgt · Accepted Answer

Step 1As a heuristic explanation, let consider  f  (  x  )  =  1  +      ∫    0    x        I    0    (  t  )  d  t  +      ∫    0    x        I    1    (  t  )  d  t  +      ∫    0    x        I    2    (  t  )  d  t  +  ⋯with       I    0    (  t  )  =  1,       I    1    (  t  )  =      ∫    0    t        I    0    (  u  )  d  u  =  t,       I    2    (  t  )  =      ∫    0    t        I    1    (  u  )  d  u  =            t      2        2  , ...,       I    n    (  t  )  =      ∫    0    t        I          n      −      1        (  u  )  d  u  =            t      n              n      !       such that             d      (      f      (      x      )      )              d      x        =  0  +      I    0    (  x  )  +      I    1    (  x  )  +      I    2    (  x  )  +  …  =  f  (  x  )Step 2That is       f    ′    (  x  )  =  f  (  x  )    f  (  0  )  =  1    ⟹    f  (  x  )  =      e    x

Mackenzie Rios · Accepted Answer

Step 1Here&#039;s a less technical, more intuitive sense: what you&#039;re building up is the series of antiderivatives of any constant C. By doing it an infinite amount of times, and summing all the terms, you&#039;ll get   C      e    x  . Note that 1 is just the C that you won&#039;t notice, because it can be ignored, being the identity for multiplication.If you do this process an infinite number of times, in a &quot;Hilbert hotel&quot; kind of way:- integrating each term and throwing away the potential constant is like &quot;sliding the term to the right&quot;- similarly, deriving each term only &quot;slides it to the left&quot;.Step 2What&#039;s interesting is that, if you run this process to infinity, there&#039;s always one more term to pick from &quot;at the end of infinity&quot; (on the right of your sum). This &quot;infinite source&quot; of terms makes it so that the derivative of this power series (infinite polynomial) is itself.The reason this &quot;resolves to the exponential&quot; specifically, is because the exponential (and more generally, functions of the form       y    0        e          x      −              x        0             are solutions to the equation   f  (  x  )  =      f    ′    (  x  ) for initial conditions   (      x    0    ,  f  (      x    0    )  =      y    0    )

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