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Computing derivatives with fractional exponents. Ask Question Asked 7 years, 1 month ago.
Let. () = − +. We want to make this look nicer. The fractional exponents are unpleasant. We can get rid of them all by multiplying through by x1 2 x 1 2. But then to keep f(x) f (x) unchanged, we will need to divide by x1 2 x 1 2. Now we carry out the strategy: () = x (x − x + x−) = x − x +. x.
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In the first convention, 'continuity' is important. If two exponents are 'near' each other, then they should produce 'nearby' values when used to exponentiate. However, despite the fact $2/3$, $3/5$, and $\pi/5$ are all similarish in size, $(-5)^{2/3}$ and $(-5)^{3/5}$ are widely separated by the fact one 'should' be positive and the other ...
I am trying to simplify the following polynomial with fractional exponents. I have $3x^\\frac{5}{3}-\\frac{5x^\\frac{2}{3}}{3}-\\frac{4x^\\frac{-1}{3}}{3}$. How do I ...
To simplify a fraction with powers in the numerator and denominator a possible method is to factor each power base into prime factors. With practice it can be done directly if the bases are small numbers. 2 2 and 3 3 are prime numbers. So we need only to factor 6 = 2 ⋅ 3 6 = 2 ⋅ 3: 2200132003 62002 = 2200132003 (2 ⋅ 3)2002 = 2200132003 ...
$\begingroup$ You know that this extension makes you cross the boundary between algebra (without topology) to analysis (with topology creeping into the scene) just because binomial theorem with, for example, exponent $1/3$ means expanding $(1+x)^{1/3}=1+(1/3)x+...$ into a series, and there are convergence issues for the proof (radius of convergence= ?).
For this equation to logically hold, the exponents must be equal, and so we can say that. x1 =xab 1 = ab x 1 = x a b 1 = a b. By the Multiplicative Inverse Property (see section on Reintroducing Arithmetic), we know that if ab = 1 a b = 1 then a a and b b must be multiplicative inverses, and so Here is a general proof for all rational numbers ...
The exponents follow the rule. (ab)c = abc. meaning that the power of a power is the power to the product of the exponents. Then assume a rational power p / q. We write. ap / q = b and raise both members to the qth: a (p / q) q = ap = bq. So b is the number such that when raised to the qth power yields ap. In other words, it is the qth root of ap.
Explanation on how this solves for the real part of a complex fraction. 0. De Moivre and trignometry ...