题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Given that when y=arcsin2x
dxdy=2(1−4x2)−21
(a) show that
dx3d3y=(1−4x2)25Ax2+8
where A is a constant to be determined.
(3)
(b) Hence determine the Maclaurin series expansion for arcsin2x in ascending powers of x up to and including the term in x3
(2)
The Maclaurin series expansion for ex is given by
ex=1+x+2x2+…+r!xr+…
(c) Use the Maclaurin series expansion for e3x and the answer to part (b) to show that, for small values of x
e3xarcsin2x≈Cx+Dx2+Ex3
where C, D and E are constants to be determined.
(3)
解答
(a)
解法一
思路
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题目已经给出一阶导,所以从这里开始连续求导即可。第三次求导时要用乘积法则,最后把两项通分到同一个 (1−4x2)5/2 分母。
#### 答题过程
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Given
dxdy=2(1−4x2)−21
Differentiate once:
dx2d2y==2(−21)(1−4x2)−23(−8x)8x(1−4x2)−23
Differentiate again:
dx3d3y====8(1−4x2)−23+8x(−23)(1−4x2)−25(−8x)8(1−4x2)−23+96x2(1−4x2)−25(1−4x2)258(1−4x2)+96x2(1−4x2)2564x2+8
Therefore
A=64
(b)
解法一
思路
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Maclaurin 展开需要 y(0),y′(0),y′′(0),y′′′(0)。由于题目要到 x3,三阶导正好够用。
答题过程
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At x=0,
y(0)=y′(0)=y′′(0)=y′′′(0)=arcsin0=0,2,0,8
Using the Maclaurin expansion,
arcsin2x===y(0)+y′(0)x+2!y′′(0)x2+3!y′′′(0)x3+⋯2x+68x3+⋯2x+34x3+⋯
Therefore, up to and including the term in x3,
arcsin2x≈2x+34x3
(c)
解法一
思路
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相乘后只要保留到 x3。因为 arcsin2x 的最低次项是 x,所以 e3x 只需要展开到 x2 就能覆盖乘积里的 x3。
答题过程
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For small x,
e3x==1+3x+2!(3x)2+⋯1+3x+29x2+⋯
Using part (b),
e3xarcsin2x==≈(1+3x+29x2)(2x+34x3)2x+34x3+6x2+9x3+⋯2x+6x2+331x3
Hence
C=2,D=6,E=331