combining rational exponents, but using calculus, @blackpenredpen
combining rational exponents, but using calculus,  @blackpenredpen
Uploaded November 2023 | Updated September 2026, 1 week ago
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The usual exponent rules are pretty easy to see when the exponents are whole numbers. For example, we know e^2 means e*e and e^3 means e*e*e, so e^2*e^3=e*e*e*e*e*e=e^5. But what if we have rational exponents such as e^(1/2)*e^(1/3)? How do we prove the rule of exponent in this case is equal to e^(5/6)?

Proving e^x*e^y=e^(x+y) by power series: youtu.be/r87AfxUwD60
Deriving the power series of e^x centered at 0: youtu.be/1OhMpXFb6yI?si=2u-5z1KINQPLz6sK
Check out the binomial theorem: youtu.be/cvhyJT9c0ac?si=2brCzT8yH5uAslkd

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0:00 We know e^2*e^3=e^(2+3), but what if we have e^(1/2)*e^(1/3)?
1:12 Proving e^(1/2)*e^(1/3) by using power series
3:14 Cauchy product of two infinite series
7:38 Back to the proof
14:42 Check out Brilliant

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combining rational exponents, but using calculus,

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