國立政治大學112學年度碩士班招生考試
考試科目:基礎數學 系所別:統計學系
Part I: Calculus
解答:State:Let f be continuous on [a,b] and differentiable on (a,b). Then ∃c∈(a,b) such that f′(c)=f(b)−f(a)b−aProve:Let {A(a,f(a))B(b,f(b)), then ¯AB:y=f(b)−f(a)b−a(x−a)+f(a)And let g(x)=f(x)−[f(b)−f(a)b−a(x−a)+f(a)]By Rolle's theorem, ∃c∈(a,b)∋g′(c)=0. That is,g′(x)=f′(x)−f(b)−f(a)b−a⇒g′(c)=f′(c)−f(b)−f(a)b−a=0⇒f′(c)=f(b)−f(a)b−a,Q.E.D解答:Let ϵ>0, we choose δ=1√ϵ. Then, we have ∀x>δ=1√ϵ⇒|1x2−0|=1x2<ϵ(∵x>1√ϵ⇒1x2<ϵ)⇒limx→∞1x2=0,Q.E.D.
解答:(a)∫2x2−x+4x3+4xdx=∫(1x+x−1x2+4)dx=∫1xdx+∫xx2+4dx−14∫1(x/2)2+1dx=lnx+12ln(x2+4)−12tan−1(x2)+C(b)f(x)=1x4⇒f(0)不存在⇒∫3−2f(x)dx不存在(c)∫a0f(x)dx=∫a0f(a−x)dx⇒I=∫π/20sinnxsinnx+cosnxdx=∫π/20sinn(π/2−x)sinn(π/2−x)+cosn(π/2−x)dx=∫π/20cosnxcosnx+sinnxdx⇒I+I=2I=∫π/20sinnx+cosnxsinnx+cosnxdx=∫π/201dx=π2⇒I=π4(d)I=∫∞0e−x2dx⇒I2=(∫∞0e−x2dx)(∫∞0e−y2dy)=∫∞0∫∞0e−(x2+y2)dxdyLet {x=rcosθy=rsinθ, then I2=∫π/20∫∞0re−r2drdθ=∫π/2012dθ=π4⇒I=√π4=√π2(e)I=∫sin5xcos2xdx=∫sinx(1−cos2x)2cos2xdxLet u=cosx, then du=−sinxdx⇒I=∫−(1−u2)2u2du=∫(−u6+2u4−u2)du=−17u7+25u5−13u3+C=−17cos7x+25cos5x−13cos3x+C解答:(tanxy)′=(x+y)′⇒sec2xy⋅(1y−xy′y2)=1+y′⇒1ysec2xy−1=(1+xy2sec2xy)y′⇒dydx=y′=1ysec2xy−11+xy2sec2xy
解答:x2+y2=1⇒{x=cosθy=sinθ⇒z=1−x+y=1−cosθ+sinθ⇒f(x,y,z)=x+2y+3z=cosθ+2sinθ+3−3cosθ+3sinθ=3−2cosθ+5sinθ=3−√29(2√29cosθ−5√29sinθ)=3−√29sin(α−θ)⇒{max(f)=3+√29min(f)=3−√29
Part II: Linear Algebra
解答:(a)λv=Av=A2v=A(Av)=A(λv)=λ(Av)=λ2v⇒λ2v−λv=0⇒λ(λ−1)v=0⇒λ=0,1(b)By (a), the eigenvalues are 0 or 1, then rank(A)=number of 1⇒rank(A)=tr(A)解答:Q is orthogonal ⇒Q−1=QT⇒QQT=I⇒det(QQT)=det(Q)det(QT)=(det(Q))2=1⇒det(Q)=±1,Q.E.D.
解答:N(T)={v∈V∣T(v)=0}(a)T(0)=0⇒0∈N(T)(b)Suppose that {v1∈N(T)v2∈N(T)⇒{T(v1)=0T(v2)=0⇒T(v1+v2)=T(v1)+T(v2)=0⇒v1+v2∈N(T)(c)Suppose that v∈N(T)⇒T(v)=0⇒T(cv)=cT(v)=0⇒cv∈N(T)From (a),(b), and (c), we have N(T) is a subspace of VR(T)={w∈W∣w=T(v)∃v∈V}(a)T(0)=0⇒0∈R(T)(b)Suppose that {w1∈R(T)w2∈R(T)⇒∃v1,v2∈V∋{T(v1)=w1T(v2)=w2⇒w1+w2=T(v1)+T(v2)=T(v1+v2)⇒w1+w2∈R(T)(c)Suppose that w∈R(T)⇒∃v∈V∋T(v)=w⇒T(cv)=cT(v)=cw⇒cw∈R(T)From (a),(b), and (c), we have R(T) is a subspace of W.Q.E.D.
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