Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1

The Steinitz exchange lemma says that if $w_1,\dots ,w_m$ span a vector space and $v_1,\dots ,v_r$ are linearly independent, then $r\le m$, and after reordering the $w_j$, the family $v_1,\dots ,v_r,w_{r+1},\dots ,w_m$ still spans. Here is a proof. Suppose the first $k-1$ replacements have been made. Express $v_k$ in that spanning family. Some coefficient of a remaining $w_j$ must be nonzero, since otherwise $v_k$ belongs to the span of $v_1,\dots ,v_{k-1}$, contrary to linear independence. Solve for that $w_j$ and replace it by $v_k$; this preserves spanning. If $r>m$, the first $m$ replacements leave a spanning family $v_1,\dots ,v_m$, contradicting linear independence of the first $m+1$ vectors. This proves both assertions.

Now suppose $S$ is linearly independent and spans ${\mathbb{R}}^n$. Express each member of the standard basis using finitely many members of $S$, and let $S_0$ be the finite union of those members. Then $S_0$ spans. No element of $S\setminus S_0$ can exist, since it would be a linear combination of $S_0$ and violate linear independence. Thus $S$ is finite. Applying the Steinitz exchange lemma in both directions to $S$ and the standard basis gives $|S|\le n$ and $n\le |S|$. Hence $\boxed{|S|=n}$.

If $S$ is linearly independent and has $n$ elements, apply the Steinitz exchange lemma with $S$ as the independent family and the standard basis as the spanning family. All $n$ members of the standard basis are replaced, leaving $S$ itself spanning. Thus $\boxed{\mathrm{span}S={\mathbb{R}}^n}$, so $S$ is a basis.

If $S$ spans and has $n$ elements but is linearly dependent, a nontrivial linear combination equal to zero lets us remove one member without changing the span. This produces $n-1$ spanning vectors, whereas the standard basis contains $n$ linearly independent vectors, contradicting the Steinitz exchange lemma. Therefore $\boxed{S\text{ is linearly independent}}$. Together the three deductions establish that any two of the three properties imply the third. For $n=0$, the only independent or zero-element family is the empty basis, and the same conclusions hold.

Relative to the given basis $e_1,e_2$, the proposed vectors are the columns ofThe determinant is nonzero exactly when the columns form a basis. Thus $\boxed{\lambda \ne \pm 1}$. At $\lambda =1$ the vectors coincide, and at $\lambda =-1$ they are negatives of one another.

For complex differentiability at a point, the derivative must have the same value along real and imaginary increments. Along real increments it equals $u_x+i{v}_x$, while along imaginary increments it equals $(u_y+i{v}_y)/i=v_y-i{u}_y$. Equating real and imaginary parts gives the Cauchy-Riemann equationsFor the given real part,Integrating the Cauchy-Riemann equations gives a harmonic conjugateThis open set is open and connected; $F$ is holomorphic there with derivative $-z^{-2}$. The origin must be excluded because the real part is undefined there. On this connected open set, any other harmonic conjugate differs by a real constant: the Cauchy-Riemann equations make both partial derivatives of the difference zero.

In the curvature $-1$ normalization, the hyperbolic triangle area is its angle defect:Join one vertex of a convex geodesic polygon to its nonadjacent vertices. This gives $n-2$ hyperbolic triangles with disjoint interiors. Their angles at every polygon vertex sum to that vertex's interior angle. Adding the hyperbolic triangle areas therefore gives the hyperbolic polygon areaFor a regular hyperbolic polygon with prescribed area, place $n$ equally spaced vertices on a hyperbolic circle of radius $R>0$ and join consecutive vertices by geodesics. Rotations through $2\pi /n$ and reflections in radial lines show that this is a convex regular polygon. The triangle from its centre to a vertex and the midpoint of an adjacent side is right angled, with angles $\pi /n$, $\alpha (R)/2$, and $\pi /2$, and hypotenuse $R$. The angle form of the hyperbolic law of cosines givesConsequently $A(R)=(n-2)\pi -n\alpha (R)$ is continuous and strictly increasing. As $R\downarrow 0$, $\alpha (R)\to \pi -2\pi /n$, so $A(R)\to 0$; as $R\to \infty$, $\alpha (R)\to 0$, so $A(R)\to (n-2)\pi$. The intermediate value theorem proves existence for every permitted $A$. Indeed the required radius isThe argument also proves uniqueness of the radius in this construction; neither endpoint area is attained by a nondegenerate finite-radius polygon.

Take a smooth variation $u_{\epsilon }=u+\epsilon \varphi$ with $\varphi =0$ on the boundary. Assume enough differentiability for the following operations, for example a smooth bounded open set, $u\in C^2$, and a twice continuously differentiable integrand. Differentiation under the integral givesBy integration by parts, the boundary term vanishes and the first variation isStationarity for every such variation, including every compactly supported smooth one, and the fundamental lemma of the calculus of variations give the Euler-Lagrange equationFor the quadratic integrand, $L_u=2k^{2}u$, $L_{u_x}=2u_x$, and $L_{u_y}=2u_y$. Thuswith the prescribed Dirichlet boundary condition. For real $k$, a stationary function is also the unique minimizer, if it exists: for any nonzero admissible $\varphi$, the linear term vanishes and $I[u+\varphi ]-I[u]={\int }_{\mathcal{D}}(|\nabla \varphi {|}^2+k^2{\varphi }^2)>0$. This last positivity uses the zero boundary values; it also holds at $k=0$.

The divergence and scalar vorticity areThus the flow is irrotational but not incompressible on the plane. A velocity potential satisfying $\mathbf{u}=\nabla \varphi$ isIndeed its two partial derivatives are the two velocity components. A usual two-dimensional stream function would satisfy $u_x={\psi }_y$, $u_y=-{\psi }_x$ and hence have identically zero divergence. No such stream function exists on an open region where this divergence is not identically zero; the line $y=0$ is not a two-dimensional flow domain.

Here both the divergence and vorticity vanish:The flow is incompressible and irrotational. With $\mathbf{u}=\nabla \varphi$ and $(u_x,u_y)=({\psi }_y,-{\psi }_x)$,The streamlines are the level sets $x^2-y^2=\text{constant}$. To determine the arrows, put $r=x+y$ and $s=x-y$. A particle satisfies $\dot{r}=-2r$, $\dot{s}=2s$. Thus the separatrix $y=x$ approaches the stagnation point at the origin, whereas $y=-x$ leaves it; the other streamlines are hyperbolas following those asymptotes. The origin itself is a stationary particle.

The Lagrange cardinal polynomials areThus the Lagrange interpolation polynomial isIt has degree at most $n$ and takes the required values. Uniqueness follows because the difference of two such polynomials has $n+1$ distinct roots and degree at most $n$, so is zero. The source's phrase “degree $n$” must allow a smaller degree, for example when $f$ is constant.

Define the divided difference recursively by $f[x_i]=f_i$ andInduction on this recursion, or comparison of the leading coefficient in the Newton interpolation polynomial, givesFor completeness the induction works term by term: an interior term gets the coefficient $(1/(x_i-x_n)-1/(x_i-x_0))/(x_n-x_0)=1/((x_i-x_n)(x_i-x_0))$, multiplying the product over the other interior nodes; the two endpoint terms give the same formula. This is also the leading coefficient of $p$ directly from the cardinal formula, so $p^{(n)}=n!f[x_0,\dots ,x_n]$.

The function $f-p\in C^n[x_0,x_n]$ has $n+1$ distinct zeros. Repeated application of Rolle theorem gives $\xi \in (x_0,x_n)$ with $(f-p{)}^{(n)}(\xi )=0$ when $n\ge 1$. HenceFor $n=0$, this is simply $f[x_0]=f(x_0)$ and we take $\xi =x_0$.

The Rao-Blackwell theorem applies to an estimator $Y$ with finite second moment and a sufficient statistic $T$. The conditional estimator $Y^{\ast }={\mathbb{E}}_{\theta }[Y\mid T]$ can be chosen as a function of the observed $T$ that does not depend on the unknown parameter: this uses the parameter-independent conditional data distribution in the definition of sufficient statistic. The tower property of conditional expectation gives ${\mathbb{E}}_{\theta }{Y}^{\ast }={\mathbb{E}}_{\theta }Y$, so their bias agrees. The law of total variance givesFor any target $a(\theta )$ it follows thatEquality holds exactly when $Y=Y^{\ast }$ almost surely under that parameter value. This proves variance reduction and reduction of mean squared error, whether or not the original estimator is an unbiased estimator. More generally, conditional Jensen inequality proves the corresponding inequality for any convex loss in the estimate, whenever the expectations exist.

Since $\mathbb{P}(X_1=0)=e^{-\lambda }$, the indicator $Y$ is an unbiased estimator of $\theta$. The joint probability mass function isThe Fisher-Neyman factorization theorem therefore makes $T={\sum }_i{X}_i$ a sufficient statistic. More explicitly, conditioning on $T=t$ gives a multinomial distribution with $t$ trials and cell probabilities $1/n$, so $X_1\mid T=t$ has the binomial distribution $\mathrm{Bin}(t,1/n)$. Consequently conditioning by the Rao-Blackwell theorem givesFor $n=1$, interpret the value at $T=0$ as $1$ and at $T>0$ as $0$; thus the same expression uses $0^0=1$ in this probability convention. For $\lambda =0$ all observations and $T$ are zero almost surely. For $n>1$ and $\lambda >0$, the reduction in variance is strict: given $T=1$, the event $X_1=0$ still has a nondegenerate conditional probability. Directly, using the probability generating function of $T\sim \mathrm{Poisson}(n\lambda )$, the new variance is $e^{-2\lambda }(e^{\lambda /n}-1)$.

Introduce nonnegative slack variables $s_1,s_2$ and use the origin as the initial feasible simplex basis:In the simplex method, let $x_2$ enter. The ratio test is $\min (10,15)=10$, so $s_1$ leaves. Solving the first constraint for $x_2$ yields the dictionaryAll nonbasic variables are nonnegative and have negative reduced objective coefficients. Setting them to zero is feasible and optimal:Subtracting $\Delta$ from both right-hand sides changes the same dictionary toIt remains feasible throughout $0\le \Delta \le 10$, including the degenerate endpoint $\Delta =10$, and the reduced coefficients are unchanged. HenceAs an independent optimality certificate, $z\le 2(x_1+x_2+2x_3)\le 2(10-\Delta )$, with equality at the stated point. Negative reduced coefficients also prove uniqueness of the optimal decision vector.

False. A left inverse would force $\alpha$ to be injective: if $\alpha u=0$, then $u=\beta \alpha u=0$. Take the surjective linear map $\alpha :{\mathbb{R}}^2\to \mathbb{R}$ given by $\alpha (x,y)=x$. Since $(0,1)$ belongs to its kernel, every composite $\beta \alpha$ annihilates $(0,1)$ and cannot be the identity map on ${\mathbb{R}}^2$.

True. Choose a basis $v_1,\dots ,v_r$ of $V$. By surjective function, choose $u_i\in U$ with $\alpha u_i=v_i$. Define the linear map $\beta$ by $\beta v_i=u_i$ and extend by linearity. Then $\alpha \beta v_i=v_i$ on every basis vector, and thereforeThis constructs a right inverse and does not require $\alpha$ to be injective. If $V$ is the zero vector space, the unique zero map is the required right inverse.

True. Use the lifted basis $u_1,\dots ,u_r$ above and put $W=\mathrm{span}\{u_1,\dots ,u_r\}$. If $\sum c_i{u}_i=0$, applying $\alpha$ gives $\sum c_i{v}_i=0$, so all $c_i=0$. Thus the $u_i$ form a basis of $W$, and $\alpha {|}_W$ takes this basis to the basis of $V$. It is a linear isomorphism.

In fact the right inverse gives the stronger splittingFor every $u\in U$, write $u=(u-\beta \alpha u)+\beta \alpha u$; the first term is in the kernel and the second in $W$. Their intersection is zero because $\alpha {|}_W$ is injective. This is a direct sum decomposition.

False. Take $U={\mathbb{R}}^2$, $V=\mathbb{R}$, $\alpha (x,y)=x+y$, $X=\mathrm{span}\{(1,0)\}$, and $Y=\mathrm{span}\{(0,1)\}$. Then $U=X\oplus Y$ is a direct sum, but $\alpha (X)=\alpha (Y)=\mathbb{R}$, so their intersection is not zero. Their ordinary sum is $V$, as always follows from surjective function and $U=X+Y$; the failure is precisely the direct sum assertion.

False. Take $\alpha (x,y)=x$, $X=\mathrm{span}\{(1,0)\}$, and $Y=\{0\}$. Then $V=\alpha (X)\oplus \alpha (Y)=\mathbb{R}\oplus \{0\}$, but $X+Y\ne U$ because $(0,1)$ is missing. Thus the premise does not force spanning. It need not force zero intersection either: with the same $\alpha$, take $X=U$ and $Y=\ker \alpha$. Their images still have a direct sum, while $X\cap Y=Y\ne \{0\}$.

The Sylow theorems state that if a finite group has order $|G|=p^{a}m$ with $p$ prime and $p\nmid m$, it has a subgroup of order $p^a$, called a Sylow subgroup. Every p-group subgroup is contained in a Sylow subgroup, and all Sylow subgroups for the same prime are conjugate subgroups. Their number satisfiesMore precisely $n_p=[G:N_G(P)]$ for a Sylow subgroup $P$, where $N_G(P)$ is its normalizer.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

For order $48=2^4\cdot 3$, the Sylow theorems give $n_2\mid 3$ and $n_2\equiv 1(\mathrm{mod}2)$, hence $n_2=1$ or $3$. If $n_2=1$, the Sylow subgroup of order $16$ is a nontrivial proper normal subgroup. If $G$ were simple in the other case, it would be nonabelian (a finite abelian simple group has prime order), and the preceding result would force $48\mid 3!/2=3$, impossible. Thus every group of order 48 is not simple.

An example with no normal Sylow subgroup for $p=2$ isIn $S_4$, $P=⟨(1234),(13)⟩$ is a dihedral group of order $8$: conjugation by $(13)$ inverts $(1234)$. Thus $P\times C_2$ has order $16$ and is a Sylow subgroup of $G$. The only transpositions in $P$ are $(13)$ and $(24)$; conjugating $(13)$ by $(23)$ gives $(12)\notin P$. Therefore $P$ and $P\times C_2$ are not normal. By conjugacy, no other Sylow subgroup for $p=2$ is normal either.

A real function on a metric space is uniformly continuous if for every $\epsilon >0$ there is a $\delta >0$ such that, simultaneously for all $x,y$, $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$. For a continuous function on a finite closed interval $[a,b]$, suppose this fails. Then there are ${\epsilon }_0>0$ and pairs $x_n,y_n$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. By compactness, some subsequence $x_{n_j}$ converges to $x\in [a,b]$; then $y_{n_j}\to x$ too. Continuity contradicts the image separation. This proves the Heine-Cantor theorem in this setting. Boundedness of the interval matters: $x^2$ on $[0,\infty )$ is a counterexample if “closed interval” includes unbounded intervals.

A function is Lipschitz continuous if some finite $K\ge 0$ satisfies $|f(x)-f(y)|\le Kd(x,y)$ for every pair. Choosing $\delta =\epsilon /(K+1)$ proves uniform continuity.

The first assertion is false. Take $f(x)=\sin (x^2)$, which is bounded and continuous, andThen $f(x_n)=1$, $f(y_n)=-1$, while $y_n-x_n=\pi /(y_n+x_n)\to 0$. This violates uniform continuity.

True. If $|f^{\prime }(x)|\le K$ everywhere, the mean value theorem applied between any two distinct points givesThus $f$ is Lipschitz continuous and hence uniformly continuous. No continuity of $f^{\prime }$ is needed: differentiability of $f$ gives continuity on each finite closed interval and is exactly the hypothesis needed for the mean value theorem there.

True. Define $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one. The restriction of $f$ to $[-n,n]$ is uniformly continuous by the Heine-Cantor theorem; its modulus of continuity combined with $|c_n(x)-c_n(y)|\le |x-y|$ proves uniform continuity of $f_n$ on all of $\mathbb{R}$. For each fixed $x$, once $n\ge |x|$ we have $f_n(x)=f(x)$. HenceThis pointwise approximation by uniformly continuous functions even gives eventual equality on every fixed bounded interval. It need not give uniform convergence on the whole line, as the bounded counterexample in part (i) already demonstrates.

If a nonempty subset $S\subset \mathbb{R}$ is not an interval, there are $a,b\in S$ and $c\notin S$ with $a<c<b$. The disjoint nonempty relatively open sets $S\cap (-\infty ,c)$ and $S\cap (c,\infty )$ form a separation, so $S$ is not connected.

Conversely, suppose an interval $I$ had a separation into relatively open nonempty sets $A,B$. Choose $a\in A,b\in B$, interchanging the names if necessary so $a<b$. Put $c=\sup (A\cap [a,b])$. Because $B$ contains a relative neighbourhood of $b$, $c<b$; because $A$ contains a relative neighbourhood of $a$, $c>a$. Also $c\in [a,b]\subset I$. If $c\in A$, relative openness supplies points of $A$ just above $c$, contradicting the supremum. If $c\in B$, a relative neighbourhood of $c$ excludes points of $A$ approaching $c$ from below, again contradicting the supremum. Thus every interval is connected, including a singleton and unbounded or open intervals.

TakeIt is closed, because its only accumulation point is $0$ and that point is included, and bounded. The Heine-Borel theorem makes it compact. The connected components of its complement are preciselyEach is an interval and therefore connected; none can be enlarged inside the complement across one of the intervening points of $K$.

Let $a,b\in {\mathbb{R}}^2\setminus T$. If $a=b$, use the constant path. Otherwise consider the countably many lines through $a$ and a point of $T$, and through $b$ and a point of $T$. Choose a line $L$ whose direction is not parallel to any of these lines; such a direction exists because there are uncountably many directions but only a countable set of forbidden ones. Every forbidden line meets $L$ in at most one point. Choose $c\in L$ outside those countably many intersections and outside $\{a,b\}$.

Neither segment $[a,c]$ nor $[c,b]$ meets $T$: a forbidden point on either segment would put $c$ on the corresponding forbidden line. The piecewise linear pathlies in the complement. This proves that the plane minus a countable set is path connected.

If a homeomorphism $h:{\mathbb{R}}^2\to \mathbb{R}$ existed, its restriction after removing a point $a$ would give a homeomorphism between ${\mathbb{R}}^2\setminus \{a\}$ and $\mathbb{R}\setminus \{h(a)\}$. The first is a path-connected space and therefore connected; the second consists of two separated intervals. Since homeomorphisms preserve connectedness, $\boxed{{\mathbb{R}}^2\ncong \mathbb{R}}$.

Fix $\mathrm{Im}\lambda >-c$. Truncate $C_1$ at $\pm R-ic$ and close it with the upper semicircle centred at $-ic$, oriented counterclockwise. For large $R$ the only pole of the integrand inside is $z=\lambda$. Uniformly on the arc, $|z-\lambda |$ is comparable to $R$ and $f(z)=o(R^{-1})$, so the arc integral is $o(R^{-1})$ and tends to zero. The line integral is absolutely convergent at infinity for the same decay reason. The Cauchy integral formula givesThe right-hand side is holomorphic on this open half-plane by the hypothesis, proving the required holomorphic function. The contour never passes through $\lambda$ because its imaginary part is strictly above the line.

Write $r_{+}={\mathrm{Res}}_{z={\alpha }_{+}}g(z)$. For $\mathrm{Im}\lambda >-c$ and $\lambda \ne {\alpha }_{+}$, close $C_1$ upwards as before. The residue theorem includes the poles at $\lambda$ and ${\alpha }_{+}$, and the assumed decay in the upper region makes the arc contribution vanish. ThusIt is essential to include the apparently exceptional point $\lambda ={\alpha }_{+}$. The Laurent series there is $g(z)=r_{+}/(z-{\alpha }_{+})+g_0(z)$ with $g_0$ holomorphic. At that point the integrand is $r_{+}/(z-{\alpha }_{+}{)}^2+g_0(z)/(z-{\alpha }_{+})$, whose residue is $g_0({\alpha }_{+})$. HenceThe displayed difference is interpreted by its removable singularity. It therefore defines a holomorphic function at every point of the required half-plane, including ${\alpha }_{+}$.

For $\mathrm{Im}\lambda <c$, close $C_2$ upwards. The pole at $\lambda$ is outside this contour, while ${\alpha }_{+}$ is inside; the pole ${\alpha }_{-}$ is also outside. The residue theorem and the minus sign in the definition of $J$ giveThis rational function is holomorphic throughout $\mathrm{Im}\lambda <c$, since its only pole lies above $c$. In particular the formula includes $\lambda ={\alpha }_{-}$: the original line-integrand has no singularity on $C_2$ there, and no evaluation of $g({\alpha }_{-})$ is needed. Closing downwards would require a decay hypothesis in the lower half-plane that has not been given; closing upwards avoids that unjustified step.

In the open strip $-c<\mathrm{Im}\lambda <c$, neither pole of $g$ is present. The two evaluations therefore giveBoth summands are holomorphic on this strip. This is a half-plane decomposition by Cauchy integrals; the pole subtraction in $H$ is what makes its extension holomorphic even at the upper pole of $g$.

Use the Fourier transform convention $\tilde{h}(k)={\int }_{\mathbb{R}}{e}^{-ikx}h(x)dx$, with inverse $(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{ikx}\tilde{h}(k)dk$. Splitting at the origin, with $a>0$ to ensure convergence, givesThus

The original PDF has forcing $e^{-|x|}$; the supplied TeX loses the absolute-value bars. Set $q=|\omega |$. First assume $q>0$ and $q\ne 1$. The Fourier transform of a derivative, obtained by integration by parts with the boundary terms zero, is $\widetilde{u^{\prime \prime }}=-k^2\tilde{u}$. Applying it and the preceding transform givesUsing the inverse Fourier transform pairs from part (i),Both decay conditions hold. The first derivatives of the two exponential terms cancel at zero, so $u$ is continuously differentiable there; the second derivatives have matching limits and the differential equation holds at zero as well as on both half-lines. For uniqueness, a difference solves the homogeneous equation. Decay gives $A{e}^{-qx}$ on the positive half-line and $B{e}^{qx}$ on the negative half-line. Continuity gives $A=B$ and derivative continuity gives $-qA=qB$, forcing both to be zero.

The PDF's condition $\omega \ne \pm 1$ also allows $\omega =0$, but no decaying solution exists when $\omega =0$. Indeed on $x>0$, $-u^{\prime \prime }=e^{-x}$ and decay forces $u=-e^{-x}$; on $x<0$, decay forces $u=-e^x$. Their one-sided derivatives at zero are $1$ and $-1$, which cannot match. The missing condition $\omega \ne 0$ is therefore a genuine source omission, not a transcription repair. The transform expression at $q=0$ has a nonintegrable $k^{-2}$ singularity, consistently with this obstruction.

For smooth rapidly decaying initial data, apply the Fourier transform in $x$ and differentiate under the integral. The Fourier transform of a derivative turns the wave equation intoSolving this constant-coefficient ordinary differential equation givesAt $k=0$ the quotient means its continuous limit $t$, so $\tilde{v}(0,t)=\tilde{f}(0)+t\tilde{g}(0)$.