國立臺北大學112學年度碩士班一般入學考試
系(所)組別:統計學系
科 目:基礎數學
一、(50%)CALCULUS
解答:L=(1x)x⇒lnL=xln1x=ln1x1xlimx→0+ln1x1x=limx→0+(ln1x)′(1x)′=limx→0+−1x−1x2=0⇒limx→0+L=e0=1解答:f(x)=∫x2x√1+t3dt⇒f′(x)=√1+x6⋅(2x)−√1+x3⋅1=2x√1+x6−√1+x3
解答:(a)x=sinu⇒{√1−x2=cosxdx=cosudu⇒∫0−11+x√1−x2dx=∫0−π/21+sinucosu⋅cosudu=∫0−π/2(1+sinu)du=[u−cosu]|0−π/2=π2−1(b)f(x)=1+x√1−x2⇒f′(x)=1√1−x2+x(1+x)(1−x2)3/2⇒{f(0)=1f′(0)=1⇒∫0−1(f(0)+f′(0)x)dx=∫0−1(1+x)dx=[x+12x2]|0−1=12
解答:L(σ)=n∏i=11√2πσe−12(xi−μσ)2=(2πσ2)−n/2e−12∑(xi−μσ)2⇒lnL(σ)=−n2ln(2π)−n2lnσ2−12σ2n∑i=1(xi−μ)2⇒∂∂σ2lnL(σ)=−n2σ2+12σ4n∑i=1(xi−μ)2=12σ2[1σ2n∑i=1(xi−μ)2−n]=0⇒σ2=∑ni=1(xi−μ)2n⇒σ=√∑ni=1(xi−μ)2n
二、(50%)
解答:(b){C−1B−1BC=C−1C=IBCC−1B−1=BB−1=I⇒C−1B−1=(BC)−1,Q.E.D.
解答:(a)det(A−λI)=|0−λ2−123−λ−2−1−20−λ|=−λ3+3λ2+9λ+5⇒the characteristic polynomial of A:−λ3+3λ2+9λ+5det(A−λI)=0⇒−(λ+1)2(λ−5)=0⇒eighenvalues: λ=−1,5(b)λ1=−1⇒(A−λ1I)v=0⇒[12−124−2−1−21][x1x2x3]=0⇒x1+2x2=x3⇒v=x2(−210)+x3(101)⇒E−1(A)={s(−210)+t(101)∣s,t∈R}λ2=5⇒(A−λ2I)v=0⇒[−52−12−2−2−1−2−5][x1x2x3]=0⇒{x1+x3=0x2+2x3=0⇒v=x3(−1−21)⇒E5(A)={(−1−21)∣k∈R}(c)A=[02−123−2−1−20]=[−21−110−2011][−1000−10005][−21−110−2011]−1[−21110−2011]By Gram-Schmidt process→[−2√51√30−1√61√52√30−2√605√301√6]⇒A=[−2√51√30−1√61√52√30−2√605√301√6][−1000−10005][−2√51√30−1√61√52√30−2√605√301√6]T⇒P=[−2√51√30−1√61√52√30−2√605√301√6],D=[−1000−10005]
解答:(b){C−1B−1BC=C−1C=IBCC−1B−1=BB−1=I⇒C−1B−1=(BC)−1,Q.E.D.
解答:(a)det(A−λI)=|0−λ2−123−λ−2−1−20−λ|=−λ3+3λ2+9λ+5⇒the characteristic polynomial of A:−λ3+3λ2+9λ+5det(A−λI)=0⇒−(λ+1)2(λ−5)=0⇒eighenvalues: λ=−1,5(b)λ1=−1⇒(A−λ1I)v=0⇒[12−124−2−1−21][x1x2x3]=0⇒x1+2x2=x3⇒v=x2(−210)+x3(101)⇒E−1(A)={s(−210)+t(101)∣s,t∈R}λ2=5⇒(A−λ2I)v=0⇒[−52−12−2−2−1−2−5][x1x2x3]=0⇒{x1+x3=0x2+2x3=0⇒v=x3(−1−21)⇒E5(A)={(−1−21)∣k∈R}(c)A=[02−123−2−1−20]=[−21−110−2011][−1000−10005][−21−110−2011]−1[−21110−2011]By Gram-Schmidt process→[−2√51√30−1√61√52√30−2√605√301√6]⇒A=[−2√51√30−1√61√52√30−2√605√301√6][−1000−10005][−2√51√30−1√61√52√30−2√605√301√6]T⇒P=[−2√51√30−1√61√52√30−2√605√301√6],D=[−1000−10005]
解答:B is a basis of V⇒→w=n∑i=1ci→vi,∀→w∈V⇒T(→w)=T(n∑i=1ci→vi)=n∑i=1ciT(→vi)⇒T(→w)∈Span{T(→vi)}ni=1⇒Range(T)⊆Span{T(→vi)}ni=1→w′∈Span{T(→vi)}ni=1⇒→w′=n∑i=1ciT(→vi)=T(n∑i=1ci→vi)⇒→w′∈Range(T)⇒Span{T(→vi)}ni=1⊆Range(T)⇒T(B) spans the range of T,Q.E.D.
解答:(a){xTAx=⟨x,Ax⟩=⟨Ax,x⟩xTAx=xT(−AT)x=−(xTAT)x=−(Ax)Tx=−⟨Ax,x⟩⇒⟨Ax,x⟩=−⟨Ax,x⟩⇒⟨Ax,x⟩=0⇒xTAx=0,Q.E.D.(b)If I+A is not invertible, we can find nonzero vector x, such that (I+A)x=0⇒x+Ax=0⇒Ax=−x⇒xTAx=⟨Ax,x⟩=⟨−x,x⟩=0(by(a))⇒x=0, a contradiction.⇒I+A is invertible,Q.E.D.
解答:(a){xTAx=⟨x,Ax⟩=⟨Ax,x⟩xTAx=xT(−AT)x=−(xTAT)x=−(Ax)Tx=−⟨Ax,x⟩⇒⟨Ax,x⟩=−⟨Ax,x⟩⇒⟨Ax,x⟩=0⇒xTAx=0,Q.E.D.(b)If I+A is not invertible, we can find nonzero vector x, such that (I+A)x=0⇒x+Ax=0⇒Ax=−x⇒xTAx=⟨Ax,x⟩=⟨−x,x⟩=0(by(a))⇒x=0, a contradiction.⇒I+A is invertible,Q.E.D.
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