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# Product definition of the exponential function

🔗$$ex=limn→∞​(1+nx​)n (71)$$
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The basic intuition for this is to start from the origin and make small changes to the function based on its known derivative at the origin.
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More precisely, we know that for any base b, exponentiation satisfies:
• .
• .
And we also know that for in particular that we satisfy the exponential function differential equation and so: $$dxdex​(0)=1 (72)$$ One interesting fact is that the only thing we use from the exponential function differential equation is the value around , which is quite little information! This idea is basically what is behind the importance of the ralationship between Lie group-Lie algebra correspondence via the exponential map. In the more general settings of groups and manifolds, restricting ourselves to be near the origin is a huge advantage.
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Now suppose that we want to calculate . The idea is to start from and then then to use the first order of the Taylor series to extend the known value of to .
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E.g., if we split into 2 parts, we know that: $$e1=e1/2e1/2 (73)$$ or in three parts: $$e1=e1/3e1/3e1/3 (74)$$ so we can just use arbitrarily many parts that are arbitrarily close to : $$e1=(e1/n)n (75)$$ and more generally for any we have: $$ex=(ex/n)n (76)$$
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Let's see what happens with the Taylor series. We have near in little-o notation: $$ey=1+y+o(y) (77)$$ Therefore, for , which is near for any fixed : $$ex/n=1+x/n+o(1/n) (78)$$ and therefore: $$ex=(ex/n)n=(1+x/n+o(1/n))n (79)$$ which is basically the formula tha we wanted. We just have to convince ourselves that at , the disappears, i.e.: $$(1+x/n+o(1/n))n=(1+x/n)n (80)$$
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To do that, let's multiply by itself once: $$eyey=(1+y+o(y))(1+y+o(y))=1+2y+o(y) (81)$$ and multiplying a third time: $$eyeyey=(1+2y+o(y))(1+y+o(y))=1+3y+o(y) (82)$$ TODO conclude.
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