拉普拉斯反变换 [p*(e^-p)]/1+（p^2） 同时另外一个问题 ：Show that,for a > 0,L[x^a] = (a/p)L[x^(a−1)] (1)This was used in class,in conjunction with the result L(1) = 1/p,to show thatL[x^n] = /p^(n+1)where n is a non-negative int
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/28 09:05:40
拉普拉斯反变换 [p*(e^-p)]/1+（p^2） 同时另外一个问题 ：Show that,for a > 0,L[x^a] = (a/p)L[x^(a−1)] (1)This was used in class,in conjunction with the result L(1) = 1/p,to show thatL[x^n] = /p^(n+1)where n is a non-negative int
拉普拉斯反变换 [p*(e^-p)]/1+（p^2）
同时另外一个问题 ：
Show that,for a > 0,
L[x^a] = (a/p)L[x^(a−1)] (1)
This was used in class,in conjunction with the result L(1) = 1/p,to show that
L[x^n] = /p^(n+1)
where n is a non-negative integer.It was also shown in class that
L[x^−1/2] =根号π除以p^1/2 .
Use this result and equation
(1) to evaluate L[x^n−(1/2)],where n is a non-negative integer.Suggesta value for (n − 1/2
拉普拉斯……
拉普拉斯反变换 [p*(e^-p)]/1+（p^2） 同时另外一个问题 ：Show that,for a > 0,L[x^a] = (a/p)L[x^(a−1)] (1)This was used in class,in conjunction with the result L(1) = 1/p,to show thatL[x^n] = /p^(n+1)where n is a non-negative int
我只随便说下玩…太高级了,是在什么阶段学的啊
是这样?[p*(e^-p)]/[1+（p^2）]
似乎好像仿佛可以直接用黎曼-梅林反演吧,负无穷到正无穷积分,恰好可以用留数定理,还是单级点 呃…行吗
英文的：（或许想的肤浅了 但也莫喷= =）
他给的第一个式子a取n+1/2 左[]=(n+1/2)/p乘右边[] ,右边[]就是要算的
左右的都直接带第2行公式,得
(n+1/2)!/p^(n+3/2)=(n-1/2)!/p^(n+1/2) * [(n+1/2)/p]
就可以得(n-1/2)!*(n+1/2)=(n+1/2)!
看来非整数阶乘也是相差1的
然后他第三个式子可知 (-1/2)!=√π,(n-1/2)!就是出来了
√π开始乘数字
。。。。好专业啊