1.f(x)=(x-3)/((x^2)-x-2) f^(100)(0)=??? 這個微分一百次有沒有特別的解法 我一直解不出來 找不到規律
2.Suppose that f is continuous and strictly increasing on [0,1] with f(0)=0 and f(1)=1 If f(x)從0積分到1 =2/3 那麼 f^(-1)(y)(反函數)從0積分到1為多少??
3.lim x→0 (sinx^2-sin^2(x))/(sin^(-1)(y))^4 這題我用羅必達法則解但是怪怪的 請大家幫幫忙 謝謝
微積分問題 請幫幫忙
2011-05-12 8:10 pm
回答 (3)
2011-05-12 10:29 pm
✔ 最佳答案
1.f(x)=(x-3)/[(x-2)(x+1)]=(1/3)[ 4/(x+1)- 1/(x-2)]
=(4/3)* 1/(1+x) + (1/6) * 1/[1- (x/2)]
=(4/3) [1-x+x^2-...+x^100- x^101+...] + (1/6)[ 1+(x/2)+(x/2)^2+...+(x/2)^100+...]
= ...+[(4/3)+(1/6)/2^100] x^100+...
= ...+f^(100)(0)/100! * x^100+ ...
故 f^(100)(0)= 100!*[(4/3)+(1/6)/2^100]
2.
∫[0~1] f^(-1)(y) dy (令 y=f(x))
=∫[0~1] x*f'(x) dx (integration by parts)
= [xf(x)]_[0~1] - ∫[0~1] f(x) dx
= f(1)-0- ∫[0~1] f(x)dx
= 1- (2/3) = 1/3
3.
設題目是 lim(x->0) [sin(x^2)-(sinx)^2]/(arcsin x )^4] (???)
Maclaurin's series: sinx=x- x^3/3!+...
so, sin(x^2)= x^2- x^6/6+... ----(A)
(sinx)^2=(x- x^3/6+...)^2= x^2 - x^4/3+ ... ---(B)
又1/√(1-x)= (1-x)^(-1/2) = 1+(1/2)x- (3/8)x^2+...
so 1/√(1-x^2)= 1+(1/2)x^2 -(3/8)x^4+...
積分得 arcsin(x)= x-(1/6)x^3-(3/40)x^5+...
4次方得 [arcsin(x)]^4= x^4 -(4/6)x^6+... ----(C)
將(A),(B),(C)代入所求limit得
原式= lim(x->0) [(1/3)x^4+...]/[x^4-(4/6)x^6+...] = 1/3
2011-05-14 6:46 am
第三題我有不同解法
只要把題目稍微整理成 幾個式子就OK了
(sinx^2-x^2+x^2-sin^2(x)) / {(sin^(-1)(x))^4 }
=> [ (x^2)( (sinx^2-x^2)/x^6 ) +( (x+sinx) /x )( (x-sinx) /x^3 ) ] / (sin^(-1)(x))^4/x^4)
寫到這就可以得到答案了
已知 lim x→0 (x-sinx)/x^3 = 1/6 lim x→0 arcsinx/x =1
2011-05-13 22:47:17 補充：
小弟我應該沒打錯符號吧 字太多容易打錯 ^^
只要把題目稍微整理成 幾個式子就OK了
(sinx^2-x^2+x^2-sin^2(x)) / {(sin^(-1)(x))^4 }
=> [ (x^2)( (sinx^2-x^2)/x^6 ) +( (x+sinx) /x )( (x-sinx) /x^3 ) ] / (sin^(-1)(x))^4/x^4)
寫到這就可以得到答案了
已知 lim x→0 (x-sinx)/x^3 = 1/6 lim x→0 arcsinx/x =1
2011-05-13 22:47:17 補充：
小弟我應該沒打錯符號吧 字太多容易打錯 ^^
2011-05-13 12:07 am
第三題我忘了利用麥克勞林
高手果真是高手><"
高手果真是高手><"
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