...所以 f'(x) > 0 for all x 表示函數 f 是嚴格遞增，因此 f 是一對一 ( one -to- one ) 的 另外 lim_{x->∞} f(x) = ∞ 且 lim_{x->-∞} f(x...
...and eigenfunction pairs. You did the first case only. (i) (E)--> y(x)=A*cos(kx... general solution y(x)=A*x+B. Applying (BC)'s we find A=0 while B...
...positive coefficients dk. Besides, as proved already, at least one cj (negative) exist. ---(2). Next, applying vn inner product to vn - summation {cj*vj} = summation {dk*vk}, which is equivalent to...
... which are differentiable over their domains you may apply the following “theorem ” to ensure that f is strictly ...
... integral equation ( 積分是∫0t ) of the first kind (沒有g(t)項)with a convolution kernel( kernel 函數...要確定N(0)=0 [t=0代入]，否則此題無意義。 2. Apply Laplace變換 to N(t)=g(t)*F(t...
...coordinate transform T from (x,y) to (u,v) is applied . Any Calculus text book will have this topic...a Calculus text is above average. e.g. the one written by J.Stewart would be suffice...
...df(z) =∫_[0,x]f(t)dt+∫_[0,x]zdf(z) (∵ f is one to one ,f(z)=f(0) f(x)=f(z) =>z=0,z=x ) =∫_[0,x]f(t)dt+∫_[0,x]z...
