【知識論】Short Essay #11

2019-12-26 Sin-iu 發表迴響

題目

Given the comparison of the unreflective-creature argument against Feldman’s evidentialism with the clairvoyance argument against Goldman’s reliablism, which theory do you think is more plausible? Why?

申論

假設認知主體 S 在 t 時的持有信念態度(doxastic attitude)D，¹關於 D 是否被 S 在 t 時所證成，需要有一證成項(justifier)，也就是用來證成的東西。Feldman 採取的「證據論證成(EJ)」要求的證成項是「符合 D 的證據」，Goldman 採取的「可靠論證成(RJ)」要求的證成項是「D 的認知形成過程的可靠性」。就證成項的性質而言，EJ 和 RJ 分別屬於證成的「內在論」與「外在論」立場，差異在於：前者要求認知主體透過反思來達取(reflectively access)證成項，²後者則否。整體而言，我覺得內在論與外在論之辯沒有哪一個比較可行，或許應該採取折衷路線。

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知識論

【知識論】Short Essay #10

2019-12-18 Sin-iu 發表迴響

題目

Given Davidson’s account of why our beliefs are mostly right in terms of successful communication, can Davidson’s coherence theory of knowledge be compatible with foundherentism?

申論

我覺得Davidson的融貫理論和Susan Haack的基礎融貫論(foundherentism)不算是相容的，即使就溝通的成功性而言，Davidson的融貫理論確實是實用的。

兩人就證成的機制的解釋，在方法和範疇上均有顯著的差異。Davidson心中的「證成」是由兩個以上「在認知能力和認知脈絡相近」的認知主體透過「溝通」完成的，而所謂溝通包含了表達、詮釋、理解等等程序，他們能夠用語言作為媒介，將對方表達的意義精確的詮釋，也因此，這種意義下的證成非常仰賴語言。我同意Davidson的理論提供一套已臻完備的知識體系，透過兩人的溝通互相檢視彼此的信念，達成融貫的證成序列，也藉此確保真理，但對於認知主體的認知能力和認知脈絡偏差很大時，該理論要如何解釋？也就是說，Davidson的理論似乎無法解釋一個剛出生、還沒有能力與其他人溝通的嬰孩，要如何從這個世界獲取知識；也無法解釋當兩個語言不同的成年人，要如何理解對方的想法，進而共同建立知識；更無法解釋為什麽一個封閉的社群，即便其成員能完美地理解彼此，所有信念也成功地互相支持，但還是存在錯誤的信念（例如地球是平的）。

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微分方程

The Legendre Equation

2019-12-16 Sin-iu 發表迴響

Definition 1

(the Legendre equation)
The Legendre equation of order $\alpha$ is the second-order linear differential equation
$\displaystyle (1-x^2)y'' - 2xy' + \alpha(\alpha+1)y = 0,$
where $\alpha > -1.$

The only singular points of the Legendre equation are at +1 and -1, so it has two linearly independent solutions that can be expressed as power series in powers of $x$ with radius of convergence at least 1.

The substitution $y = \sum c_m x^m$ in the Legendre equation leads to the recurrence relation

$\displaystyle c_{m+2} = -\frac{(\alpha - m)(\alpha + m +1)}{(m+1)(m+2)}c_m$

Proof
Let $\displaystyle y = \sum^\infty_{m=0} c_m x^m$ , so that
$\displaystyle y' = \sum^\infty_{m=1} mc_m x^{m-1}, \quad y'' = \sum^\infty_{m=2} m(m-1)c_m x^{m-2}$ .

$\displaystyle (1-x^2)\sum^\infty_{m=2} m(m-1)c_m x^{m-2} - 2x\sum^\infty_{m=1} mc_m x^{m-1} + \alpha(\alpha+1)\sum^\infty_{m=0} c_m x^m = 0 \\ \sum^\infty_{m=2} m(m-1)c_m x^{m-2}-\sum^\infty_{m=2} m(m-1)c_m x^m \\ \phantom{\sum^\infty_{m=2} m(m-1)} - \sum^\infty_{m=1} 2mc_m x^m + \sum^\infty_{m=0} \alpha(\alpha+1)c_m x^m = 0\\ \sum^\infty_{m=0} (m+2)(m+1)c_{m+2} x^{m}-\sum^\infty_{m=2} m(m-1)c_m x^m \\ \phantom{\sum^\infty_{m=2} m(m-1)} - \sum^\infty_{m=1} 2mc_m x^m + \sum^\infty_{m=0} \alpha(\alpha+1)c_m x^m = 0 \\ \sum^\infty_{m=2}\Big[(m+2)(m+1)c_{m+2}-m(m-1)c_m - 2mc_m + \alpha(\alpha+1)c_m \Big]x^m \\ \phantom{\sum^\infty_{m=2} }+2c_2x^0 + 6c_3x^1 - 2c_1x^1 + \alpha(\alpha+1)c_0 x^0 + \alpha(\alpha+1)c_1 x^1 = 0\\ \sum^\infty_{m=2}\Big\{(m+2)(m+1)c_{m+2}-[\alpha(\alpha+1)-m(m+1)]c_m \Big\}x^m \\ \phantom{\sum^\infty_{m=2} }+[2c_2+\alpha(\alpha+1)c_0]x^0 + [6c_3 - 2c_1+ \alpha(\alpha+1)c_1] x^1 = 0 \\ \sum^\infty_{m=2}\Big[(m+2)(m+1)c_{m+2}-(\alpha-m)(\alpha+m+1)c_m \Big]x^m \\ \phantom{\sum^\infty_{m=2} }+[2c_2+\alpha(\alpha+1)c_0]x^0 + [6c_3 - 2c_1+ \alpha(\alpha+1)c_1] x^1 = 0$

The identity principle requires that

the coefficient of $x^0$ satisfies $\displaystyle 2c_2+\alpha(\alpha+1)c_0 = 0 \Rightarrow c_2 = -\frac{\alpha(\alpha+1)}{2!}c_0$ ;
the coefficient of $x^1$ satisfies $\displaystyle 6c_3 - 2c_1+ \alpha(\alpha+1)c_1 = 0 \Rightarrow c_3 = -\frac{(\alpha-1)(\alpha+2)}{3!}c_1$ ;
the coefficients of $x^m$ for $m = 2,3,\ldots$ satisfy $\displaystyle (m+2)(m+1)c_{m+2}-(\alpha-m)(\alpha+m+1)c_m = 0$ , which yields the recurrence relation