339 reputation
19
bio website
location Olympia, WA
age 26
visits member for 3 years, 11 months
seen Nov 12 '13 at 3:08

New to programming: interested in functional languages. Philosophy major.


Sep
24
awarded  Autobiographer
Jul
2
awarded  Curious
May
24
awarded  Revival
Sep
30
accepted Every equalizer is monic
Sep
30
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?
Aug
19
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.
Aug
19
accepted Spivak's Calculus (Chapter 2, Exercise 17)
Aug
18
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.
Aug
18
revised Spivak's Calculus (Chapter 2, Exercise 17)
edited body
Aug
18
asked Spivak's Calculus (Chapter 2, Exercise 17)
Aug
15
accepted Best Notation for Introducing Finite Numbers of Elements
Aug
14
revised Best Notation for Introducing Finite Numbers of Elements
deleted 12 characters in body
Aug
14
asked Best Notation for Introducing Finite Numbers of Elements
Aug
11
revised Why isn't this a regular language?
edited body
Aug
11
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.
Aug
11
revised Why isn't this a regular language?
added 59 characters in body
Aug
11
answered Why isn't this a regular language?
Aug
11
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.
Aug
7
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.
Aug
7
accepted Product of Polynomials