结果1 结果2 题目求幂级数的和函数X的2n次方除以2n的阶乘 " /> 求幂级数的和函数X的2n次方除以2n的阶乘相关知识点: 试题来源: 解析 n从0起: ΣX^(2n)/(2n)! =ΣX^(2n)/2^n(n)! =Σ[X^2/2]^n/(n)! =e^(X^2/2) 分析总结。 求幂级数的和函数x的2n次方除以2n的阶乘...
+(2的n次方的阶乘分之x的2n-1次方)? 答案 Private Sub Form_Click()Dim x As Single,s As DoubleDim n As Integers = 0x = Text1.Text ‘文本框里输入x的值n = Text2.Text ’输入n的值For i = 1 To ns = s + x ^ (2 * i - 1) / 2 ^ iNextPrint sEnd Sub请加...相关推荐 1...
(2n)!!=2•4•6•……•(2n-2)•2n=2^n•n!(2n)!=1•2•3•……•n•(n+1)•……•(2n-1)•2n 原函数应该是cosh(x)吧。分母
解答一 举报 Private Sub Form_Click()Dim x As Single,s As DoubleDim n As Integers = 0x = Text1.Text ‘文本框里输入x的值n = Text2.Text ’输入n的值For i = 1 To ns = s + x ^ (2 * i - 1) / 2 ^ iNextPrint sEnd Sub请加... 解析看不懂?免费查看同类题视频解析查看解答 ...
x的2n-1次方的级数求和函数? 当x=1时,x^2n=X^2,而x=0时,x^2n=1,所以当从0开始计数时,应把少的X^2加上,而n=0本身不影响结果,不懂可以在线问~
1-x的2n次方除以1+x的2n次方 值等于-1这到题应属于求极限的题目,即当N趋于无穷大时的值,这种时候往往都是对变量X进行讨论。如:函数f(x)=当n趋向于无穷时,(1-x的2n次... n的阶乘分之2的n次方的极限 具体怎么求? 拆成Ln = (2/1)*(2/2)*(2/3)*(2/4)…*(2/n),这样 Ln 的分母就是n的...
根据题意得(m-n)-(2n-1)=11(m-1)-(4-n)=9解这个方程组得m=13n=1
=Σ[X^2/2]^n/(n)!=e^(X^2/2)结果一 题目 求幂级数的和函数X的2n次方除以2n的阶乘 答案 n从0起:ΣX^(2n)/(2n)!=ΣX^(2n)/2^n(n)!=Σ[X^2/2]^n/(n)!=e^(X^2/2)相关推荐 1求幂级数的和函数X的2n次方除以2n的阶乘 反馈 收藏 ...
Private Sub Form_Click()Dim x As Single, s As Double Dim n As Integer s = 0 x = Text1.Text ‘文本框里输入x的值 n = Text2.Text ’输入n的值 For i = 1 To n s = s + x ^ (2 * i - 1) / 2 ^ i Next Print s End Sub 请加两个文本框测试 ...
所以2^n*n!=1*2*3*...*n*2*2*2*..*2(n个2)=(1*2)*(2*2)*(3*2)*...*(n*2)=2*4*6*8*...*2n所以2n!/(n!*2^n)=1*2*3*4*...*2n/(2*4*6*8*...*2n)=1*3*5*7*...*(2n-1) 解析看不懂?免费查看同类题视频解析查看解答...