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 Sep24 awarded Autobiographer Jul2 awarded Curious May24 awarded Revival Sep30 accepted Every equalizer is monic Sep30 comment How to find the derivative to this equation? The power rule states that the derivative with respect to $x$ of $x^n$ is $nx^{n - 1}$ for all $n \neq 0$. The quotient rule states that the derivative with respect to $x$ of $f(x)/g(x)$ is $(g(x)f'(x) - f(x)g'(x))/g(x)^2$ whenever $g(x) \neq 0$. The original function $h_1(x) = 1/2x^{2/3}$ can be computed using the quotient rule by letting $f(x) = 1$ and $g(x) = 2x^{2/3}$. Since your function can be written as $h_2(x) = x^{-2/3}/2$, it can also be computed using the product rule by letting $n = -2/3$. Can you see how the the power and quotient rules can be used to compute your derivative? Aug19 comment Spivak's Calculus (Chapter 2, Exercise 17) Thank you for responding. I used this method in order to solve the problems in Exercise 6. I attempted to generalize the method to problem 17 by applying the binomial theorem to $(k + 1)^{n + 1}$ in order to obtain an expansion that the method of Exercise 6 requires. What is bothering me is the following: assuming that I didn't make any errors, why did I use the method of Exercise 6 in order to obtain a term of the required form without induction, but Spivak used the method of Exercise 6 to obtain a term of the required for with induction? This leads me to believe that I made an error. Aug19 accepted Spivak's Calculus (Chapter 2, Exercise 17) Aug18 comment Spivak's Calculus (Chapter 2, Exercise 17) I tried using complete induction on $p$. However, I found that expanding $(n + 1)^{p + 1}$ using the binomial theorem generated the sum of the $\alpha_i n^i$ that I needed. It's not clear where I should use induction. Aug18 revised Spivak's Calculus (Chapter 2, Exercise 17) edited body Aug18 asked Spivak's Calculus (Chapter 2, Exercise 17) Aug15 accepted Best Notation for Introducing Finite Numbers of Elements Aug14 revised Best Notation for Introducing Finite Numbers of Elements deleted 12 characters in body Aug14 asked Best Notation for Introducing Finite Numbers of Elements Aug11 revised Why isn't this a regular language? edited body Aug11 comment Difference between “substitution” and “replacement” In general, both replacement and substitution are syntactic operations on strings. If $uvw$ and $x$ are strings, then $uxw$ is the result of replacing $v$ with $x$ in $uvw$. If $u$, $v$ and $w$ are strings, then $u[v \mapsto w]$ is the result of replacing every occurrence of $v$ in $u$ with $w$. The last operation is called substitution, and is defined in terms of replacement. The operation called instantiation is a kind of substitution which eliminates quantifiers by substituting values for free variables. Aug11 revised Why isn't this a regular language? added 59 characters in body Aug11 answered Why isn't this a regular language? Aug11 comment Solving $|a| < |b|$ It's useful to review the definition of the absolute value function: $$\text{abs}(a) = \begin{cases} a & 0 \leq a \\ -a & a < 0 \end{cases}$$ In order to eliminate one absolute value function from an expression, you will need to consider two cases: the case where its argument is non-negative, and the case where its argument is negative. This will give you an expression with two cases with one less absolute value function. Iterating this process will give an expression without the absolute value sign. Aug7 comment Product of Polynomials I had drawn tables of the sums the equality of which I was attempting to prove, but your answer demonstrated how to visualize the indexing that was necessary to complete a proof without induction. I don't think that infinite sums are necessary for the proof, however; I was able to apply your technique to finite sums with additional zero coefficients. Aug7 accepted Product of Polynomials