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While this may convince students, you haven't actually prove that any exponential function has a slope. The usual presentation doesn't even demonstrate that suc

by btilly 2y ago

While this may convince students, you haven't actually prove that any exponential function has a slope. The usual presentation doesn't even demonstrate that such functions are defined at irrational numbers. That said, it is worthwhile to go through the algebra exercise to convince yourself that, for large n, (1+x/n)^n expands out to approximately 1 + x + x^2/2 + x^3/6 + ... Hint. The x^k terms come out to (x/n)^k (n choose k). This will turn out to be x^k/k! + O(x^k/n). As n goes to infinity, the error term drops out, and we're just left with the series that we want.

ogogmad 2y ago
> for large n, (1+x/n)^n expands out to approximately 1 + x + x^2/2 + x^3/6 + ... The rigorous version of this argument uses the Dominated Convergence Theorem in the special case of infinite series.
- btilly 2y ago
  There are several ways to make this rigorous. An explicit epsilon-delta style proof is not that hard to produce, it's just a little messy. What you have to do is, for a given x and 0<ε, pick N large enough that you can bound the tail after x^N/N! onwards with a geometric series adding up to at most ε/2. Now pick n large enough enough that the sum of errors in the terms up to N is also bounded by ε/2. From that n on, (1+x/n)^n is within ε of the power series for e^x.