Solution:

Given:

\sum_{n=1}^{\infty}\frac{(-1)^n}{n-\log n}

We must show that the series converges, but not absolutely.

Step 1: Convergence by Leibniz’s alternating-series test

Put b_n=\frac{1}{n-\log n}. For n\ge1, b_n>0. Also,

\lim_{n\to\infty}b_n = \lim_{n\to\infty}\frac{1/n}{1-(\log n)/n} =0,

because (\log n)/n\to0.

To check monotonicity, consider g(x)=x-\log x. Then

g'(x)=1-\frac1x=\frac{x-1}{x}>0\qquad(x>1).

Thus n-\log n is increasing for n\ge2, and therefore b_n is decreasing. Hence b_n\downarrow0. By Leibniz’s test,

\sum_{n=1}^{\infty}\frac{(-1)^n}{n-\log n}

converges.

Step 2: Failure of absolute convergence

The absolute-value series is

\sum_{n=1}^{\infty}\frac{1}{n-\log n}.

For n>1, \log n>0, so

n-\log n<n \quad\Longrightarrow\quad \frac1{n-\log n}>\frac1n.

Since the harmonic series \sum 1/n diverges, the comparison test gives

\sum_{n=1}^{\infty}\frac{1}{n-\log n}=\infty.

Final Answer:

\boxed{\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n-\log n} \text{ is conditionally convergent.}}

Solution:

Let \phi:S_3\to\mathbb Z_3 be a group homomorphism.

Relevant result: First Isomorphism Theorem

For a homomorphism \phi:G\to H, \ker\phi is normal in G, \operatorname{Im}\phi is a subgroup of H, and

G/\ker\phi\cong\operatorname{Im}\phi.

The only subgroups of \mathbb Z_3 are \{0\} and \mathbb Z_3.

Case 1: If \operatorname{Im}\phi=\{0\}, then

\phi(\sigma)=0\qquad(\sigma\in S_3),

which is the trivial homomorphism.

Case 2: Suppose \operatorname{Im}\phi=\mathbb Z_3. Then \phi is onto. Since |S_3|=6 and |\mathbb Z_3|=3,

\frac{|S_3|}{|\ker\phi|} = |\operatorname{Im}\phi| =3 \quad\Longrightarrow\quad |\ker\phi|=2.

But \ker\phi must be normal in S_3. The order-two subgroups are

\langle(12)\rangle,\qquad \langle(13)\rangle,\qquad \langle(23)\rangle,

and none is normal. For example,

(123)(12)(123)^{-1}=(23),

so conjugation takes \langle(12)\rangle to a different subgroup. Hence no normal subgroup of order 2 exists, so a surjective homomorphism is impossible.

Final Answer:

\boxed{\phi(\sigma)=0\text{ for every }\sigma\in S_3}

is the only homomorphism S_3\to\mathbb Z_3.

Solution:

The equation is

xp+yq=-xy,

where p=z_x and q=z_y. This is Lagrange’s equation Pp+Qq=R with

P=x,\qquad Q=y,\qquad R=-xy.

Auxiliary equations

\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{-xy}.

From the first two ratios,

\frac{dx}{x}=\frac{dy}{y} \quad\Longrightarrow\quad \log|x|-\log|y|=C \quad\Longrightarrow\quad \frac{y}{x}=C_1.

Thus one first integral is u=y/x.

Along a characteristic let y=kx, where k is constant. Using

\frac{dx}{x}=\frac{dz}{-xy},

we get

\frac{dz}{dx} = \frac{-xy}{x} =-y=-kx.

Integrating,

z=-\frac{kx^2}{2}+C_2 =-\frac{xy}{2}+C_2.

Hence the second first integral is

v=z+\frac{xy}{2}.

The general solution is an arbitrary relation F(u,v)=0:

F\left(\frac{y}{x},z+\frac{xy}{2}\right)=0.

Equivalently,

z=-\frac{xy}{2}+\phi\left(\frac{y}{x}\right).

Verification

Let u=y/x. Then

u_x=-\frac{y}{x^2},\qquad u_y=\frac1x.

Therefore

z_x=-\frac y2-\frac{y}{x^2}\phi'(u), \qquad z_y=-\frac x2+\frac1x\phi'(u).

Thus

xz_x+yz_y = -\frac{xy}{2}-\frac{y}{x}\phi'(u) -\frac{xy}{2}+\frac{y}{x}\phi'(u) =-xy.

Final Answer:

\boxed{z=-\frac{xy}{2}+\phi\left(\frac{y}{x}\right)}

on a region where x\ne0.

Solution:

For y'=f(x,y), Euler’s method is

y_{n+1}=y_n+h f(x_n,y_n).

Here f(x,y)=y and h=0.1, hence

y_{n+1}=y_n+0.1y_n=1.1y_n.

Starting with x_0=0,\ y_0=1:

y_1=1+0.1(1)=1.1,\qquad x_1=0.1. y_2=1.1+0.1(1.1)=1.21,\qquad x_2=0.2. y_3=1.21+0.1(1.21)=1.331,\qquad x_3=0.3. y_4=1.331+0.1(1.331)=1.4641,\qquad x_4=0.4. y_5=1.4641+0.1(1.4641)=1.61051,\qquad x_5=0.5.

Final Answer:

\boxed{y(0.5)\approx1.61051\approx1.611.}

Solution:

Theorem: Every subgroup of a cyclic group is cyclic.

Proof

Let G=\langle a\rangle be cyclic and let H be a subgroup of G.

If H=\{e\}, then H=\langle e\rangle, so H is cyclic.

Now suppose H\ne\{e\}. Since every element of G is a power of a, there is a positive integer n such that a^n\in H. Hence the set

S=\{n\in\mathbb N:a^n\in H\}

is nonempty. By the well-ordering principle, let m be its least element. Then a^m\in H, so

\langle a^m\rangle\subseteq H.

Take any h\in H. Since G=\langle a\rangle, write

h=a^r,\qquad r\in\mathbb Z.

By the division algorithm,

r=qm+s,\qquad 0\le s<m.

Therefore

a^s=a^{r-qm}=a^r(a^m)^{-q}.

Both a^r and (a^m)^{-q} belong to H, hence a^s\in H. If s>0, this contradicts the minimality of m. Thus s=0.

Hence r=qm and

h=a^r=(a^m)^q\in\langle a^m\rangle.

So H\subseteq\langle a^m\rangle. Combining both inclusions,

H=\langle a^m\rangle.

Final Answer:

\boxed{\text{Every subgroup of a cyclic group is cyclic.}}

Solution:

Definition: The order of an element a in a group G is the least positive integer n such that

a^n=e,

if such an integer exists. Otherwise the element has infinite order.

Let |G| be even. Pair each element g with its inverse g^{-1}. Whenever

g\ne g^{-1},

the two elements form a distinct pair \{g,g^{-1}\}. Thus the number of non-self-inverse elements is even.

An element is self-inverse exactly when

g=g^{-1} \quad\Longleftrightarrow\quad g^2=e.

Such an element is either the identity e, of order 1, or a nonidentity element of order 2.

Let N be the number of elements of order 2. Then the number of self-inverse elements is N+1. Therefore

|G| = (\text{an even number})+(N+1).

Since |G| is even, N+1 must be even. Hence N is odd.

In particular N\ne0, so at least one element of order 2 exists.

Final Answer:

\boxed{\text{A finite group of even order has an odd number of elements of order }2.}

Solution:

Abel’s test for improper integrals

If \int_a^\infty f(x)\,dx converges and g(x) is monotonic and bounded on [a,\infty), then

\int_a^\infty f(x)g(x)\,dx

converges.

Now consider

I=\int_0^\infty\frac{\sin x}{\log(x+2)}\,dx.

For this integral the direct choice f(x)=\sin x does not satisfy Abel’s first hypothesis, because

\int_0^R\sin x\,dx=1-\cos R

has no limit as R\to\infty. The appropriate convergence test for the displayed integral is Dirichlet’s test.

Dirichlet’s test: If \int_a^t f(x)\,dx is bounded and g(x) decreases to zero, then \int_a^\infty f(x)g(x)\,dx converges.

Here

F(t)=\int_0^t\sin x\,dx=1-\cos t,

so |F(t)|\le2. Also

g(x)=\frac1{\log(x+2)}, \qquad g'(x) = -\frac1{(x+2)\{\log(x+2)\}^2} <0,

and

\lim_{x\to\infty}g(x)=0.

Thus Dirichlet’s test applies and the integral converges.

Final Answer:

\boxed{\displaystyle\int_0^\infty\frac{\sin x}{\log(x+2)}\,dx \text{ converges.}}

Solution:

Logarithmic test

For a positive-term series \sum u_n, if

L=\lim_{n\to\infty}\frac{\log(1/u_n)}{\log n}

exists, then the series converges if L>1, diverges if L<1, and the test is inconclusive if L=1.

Now suppose \{n^2a_n\} is convergent. Every convergent real sequence is bounded, so there is M>0 such that

|n^2a_n|\le M \qquad(n\ge1).

Hence

|a_n|\le\frac{M}{n^2}.

The p-series

\sum_{n=1}^\infty\frac1{n^2}

converges because p=2>1. Therefore

\sum_{n=1}^{\infty}\frac{M}{n^2}

also converges. By comparison,

\sum_{n=1}^{\infty}|a_n|

converges.

Final Answer:

\boxed{\{n^2a_n\}\text{ convergent }\Longrightarrow \sum_{n=1}^{\infty}a_n\text{ is absolutely convergent.}}

Solution:

Second Mean Value Theorem of Integral Calculus

If f is monotonic on [a,b] and g is integrable, then there exists \xi\in[a,b] such that

\int_a^b f(x)g(x)\,dx = f(a)\int_a^\xi g(x)\,dx + f(b)\int_\xi^b g(x)\,dx.

Now suppose f:[0,1]\to\mathbb R is continuous and

\int_0^1|f(x)|\,dx=0.

Assume, for contradiction, that f(c)\ne0 for some c\in[0,1]. Then |f(c)|>0. Put

\varepsilon=\frac{|f(c)|}{2}.

By continuity of |f| at c, there exists \delta>0 such that

|x-c|<\delta \quad\Longrightarrow\quad |f(x)|>\frac{|f(c)|}{2}.

Let J=(c-\delta,c+\delta)\cap[0,1]. This interval has positive length. Therefore

\int_0^1|f(x)|\,dx \ge \int_J|f(x)|\,dx \ge \frac{|f(c)|}{2}\,|J| >0,

contradicting the hypothesis. Hence no such c exists.

Final Answer:

\boxed{f(x)=0\quad\text{for every }x\in[0,1].}

Solution:

Let

z=x+iy,\qquad w=u+iv, \qquad w=\frac{z-1}{z+1}.

The line through 0 and 1+i is

y=x.

Invert the transformation:

w(z+1)=z-1 \quad\Longrightarrow\quad z=\frac{1+w}{1-w}.

Substituting w=u+iv,

z=\frac{1+u+iv}{1-u-iv}.

Multiplying numerator and denominator by 1-u+iv,

z= \frac{1-u^2-v^2+2iv}{(1-u)^2+v^2}.

Thus

x=\frac{1-u^2-v^2}{(1-u)^2+v^2}, \qquad y=\frac{2v}{(1-u)^2+v^2}.

Since y=x,

2v=1-u^2-v^2.

Hence

u^2+v^2+2v=1 \quad\Longrightarrow\quad u^2+(v+1)^2=2.

This is the circle with centre (0,-1) and radius \sqrt2. The point w=1 is the image of the point at infinity on the line and is not attained by a finite point of L.

Final Answer:

\boxed{u^2+(v+1)^2=2}

Solution:

Let the required bilinear transformation be

w=\frac{az+b}{cz+d}, \qquad ad-bc\ne0.

The given mappings are

0\mapsto2i,\qquad i\mapsto-1,\qquad 1+i\mapsto0.

Since a common nonzero multiplier in a,b,c,d does not change the transformation, take d=1.

From z=0\mapsto w=2i,

b=2i.

From z=1+i\mapsto w=0, the numerator vanishes:

a(1+i)+2i=0 \quad\Longrightarrow\quad a=-\frac{2i}{1+i}.

Rationalizing,

a = -\frac{2i(1-i)}{2} =-1-i.

Now use z=i\mapsto w=-1:

-1 = \frac{(-1-i)i+2i}{ci+1}.

Since

(-1-i)i=1-i,

the numerator equals 1+i. Therefore

-1=\frac{1+i}{1+ci} \quad\Longrightarrow\quad 1+ci=-1-i.

Thus

ci=-2-i, \qquad c=\frac{-2-i}{i}=-1+2i.

Hence

w= \frac{(-1-i)z+2i}{(-1+2i)z+1}.

Verification

For z=0, w=2i. For z=1+i,

(-1-i)(1+i)+2i=-2i+2i=0,

so w=0. For z=i,

w=\frac{1+i}{-1-i}=-1.

Final Answer:

\boxed{w=\frac{(-1-i)z+2i}{(-1+2i)z+1}.}

Solution:

Liouville’s Theorem: Every bounded entire function is constant.

Proof

Let f be entire and suppose

|f(z)|\le M \qquad(z\in\mathbb C)

for some M>0. Fix z_0\in\mathbb C. Since f is entire, Cauchy’s integral formula for the derivative gives, on the circle |z-z_0|=R,

f'(z_0) = \frac1{2\pi i} \int_{|z-z_0|=R} \frac{f(z)}{(z-z_0)^2}\,dz.

Taking absolute values and using the ML-estimate,

|f'(z_0)| \le \frac1{2\pi} \cdot \frac{M}{R^2} \cdot 2\pi R = \frac{M}{R}.

This is true for every R>0. Letting R\to\infty,

|f'(z_0)|=0,

hence f'(z_0)=0. Since z_0 was arbitrary, f'\equiv0, so f is constant.

Is \sin z bounded?

Take z=iy, where y\in\mathbb R. Then

\sin(iy) = \frac{e^{-y}-e^y}{2i} = i\sinh y.

Therefore

|\sin(iy)|=|\sinh y| = \left|\frac{e^y-e^{-y}}2\right|.

As y\to\infty, this tends to infinity. Hence \sin z is not bounded on \mathbb C.

Final Answer:

\boxed{\text{Every bounded entire function is constant, but }\sin z \text{ is unbounded on }\mathbb C.}

Solution:

Definition: A metric space (X,d) is connected if it cannot be written as

X=U\cup V

where U and V are nonempty, disjoint, open subsets of X. Equivalently, the only subsets of X that are both open and closed are \varnothing and X.

We prove that [a,b] is connected.

If a=b, the interval is a singleton and is connected. Assume a<b. Suppose, for contradiction, that

[a,b]=U\cup V,

where U,V are nonempty, disjoint and relatively open. Then each is also relatively closed because it is the complement of the other.

Choose u\in U and v\in V. Interchange them if necessary so that u<v. Let

S=U\cap[u,v].

The set S is nonempty and bounded above by v. Let

c=\sup S.

Because U is closed in [a,b] and points of S approach c, we have c\in U. Since U is relatively open, there exists \varepsilon>0 such that

(c-\varepsilon,c+\varepsilon)\cap[a,b]\subset U.

If c<v, choose \eta>0 with

\eta<\min\{\varepsilon,v-c\}.

Then c+\eta\in S, contradicting the fact that c is an upper bound of S. Hence c=v. But c\in U, so v\in U, while by choice v\in V, contradicting U\cap V=\varnothing.

Therefore no separation exists.

Final Answer:

\boxed{\text{Every closed interval }[a,b]\subset\mathbb R\text{ is connected.}}

Solution:

Definition: Let (X,d_X) and (Y,d_Y) be metric spaces. A function f:X\to Y is continuous at x_0\in X if, for every \varepsilon>0, there exists \delta>0 such that

d_X(x,x_0)<\delta \quad\Longrightarrow\quad d_Y(f(x),f(x_0))<\varepsilon.

Let X be compact and f:X\to\mathbb R continuous. We prove that f is bounded.

For each x\in X, apply continuity with \varepsilon=1. There exists \delta_x>0 such that

d(y,x)<\delta_x \quad\Longrightarrow\quad |f(y)-f(x)|<1.

Hence, whenever y\in B(x,\delta_x),

|f(y)| \le |f(y)-f(x)|+|f(x)| <1+|f(x)|.

The balls \{B(x,\delta_x):x\in X\} form an open cover of X. By compactness, choose a finite subcover

X\subset B(x_1,\delta_{x_1})\cup\cdots\cup B(x_n,\delta_{x_n}).

Set

M=1+\max_{1\le i\le n}|f(x_i)|.

For any y\in X, y lies in some B(x_j,\delta_{x_j}). Therefore

|f(y)| <1+|f(x_j)| \le M.

Thus f is bounded on X.

Final Answer:

\boxed{\text{Every real-valued continuous function on a compact metric space is bounded.}}

Solution:

The data are

\begin{array}{c|ccccc} x & -3 & -1 & 0 & 3 & 5\\ \hline f(x) & -30 & -22 & -12 & 330 & 3458 \end{array}

Newton’s divided-difference formula

\begin{aligned} P_n(x)={}&f[x_0] +(x-x_0)f[x_0,x_1]\\ &+(x-x_0)(x-x_1)f[x_0,x_1,x_2]+\cdots. \end{aligned}

Take x_0=-3,\ x_1=-1,\ x_2=0,\ x_3=3,\ x_4=5.

First divided differences

f[x_0,x_1]=\frac{-22+30}{2}=4, \qquad f[x_1,x_2]=\frac{-12+22}{1}=10, f[x_2,x_3]=\frac{330+12}{3}=114, \qquad f[x_3,x_4]=\frac{3458-330}{2}=1564.

Second divided differences

f[x_0,x_1,x_2]=\frac{10-4}{3}=2, f[x_1,x_2,x_3]=\frac{114-10}{4}=26, \qquad f[x_2,x_3,x_4]=\frac{1564-114}{5}=290.

Third divided differences

f[x_0,x_1,x_2,x_3]=\frac{26-2}{6}=4, f[x_1,x_2,x_3,x_4]=\frac{290-26}{6}=44.

Fourth divided difference

f[x_0,x_1,x_2,x_3,x_4] = \frac{44-4}{8}=5.

Therefore

\begin{aligned} P_4(x)={}&-30+4(x+3)+2(x+3)(x+1)\\ &+4(x+3)(x+1)x\\ &+5(x+3)(x+1)x(x-3). \end{aligned}

Expanding,

P_4(x)=5x^4+9x^3-27x^2-21x-12.

At x=3.5,

\begin{aligned} P_4(3.5) ={}&-30+4(6.5)+2(6.5)(4.5)\\ &+4(6.5)(4.5)(3.5)\\ &+5(6.5)(4.5)(3.5)(0.5). \end{aligned}

Now

-30+26=-4,\quad 2(6.5)(4.5)=58.5, 4(6.5)(4.5)(3.5)=409.5, 5(6.5)(4.5)(3.5)(0.5)=255.9375.

Hence

P_4(3.5)=-4+58.5+409.5+255.9375=719.9375.

Final Answer:

\boxed{P_4(x)=5x^4+9x^3-27x^2-21x-12,\qquad f(3.5)=719.9375.}

Solution:

The parabola is y=x^2 and the line is x-y-5=0.

Take a general point P=(t,t^2) on the parabola. The perpendicular distance from (x_1,y_1) to Ax+By+C=0 is

d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.

Therefore

d(t) = \frac{|t-t^2-5|}{\sqrt2}.

Now

t-t^2-5 = -\left(t^2-t+5\right) = -\left[\left(t-\frac12\right)^2+\frac{19}{4}\right]<0.

Hence

d(t)=\frac{t^2-t+5}{\sqrt2}.

Differentiate:

d'(t)=\frac{2t-1}{\sqrt2}.

For an extremum,

2t-1=0 \quad\Longrightarrow\quad t=\frac12.

Also

d''(t)=\sqrt2>0,

so this gives the minimum distance.

Therefore

\begin{aligned} d_{\min} &= \frac{\frac14-\frac12+5}{\sqrt2}\\ &= \frac{19}{4\sqrt2} = \frac{19\sqrt2}{8}. \end{aligned}

Final Answer:

\boxed{d_{\min}=\frac{19}{4\sqrt2}=\frac{19\sqrt2}{8}.}

Solution:

The initial-value problem is

y''-3y'-4y=x^2, \qquad y(0)=2,\quad y'(0)=1.

Let \mathcal L\{y\}=Y(s). We use

\mathcal L\{y'\}=sY-y(0), \qquad \mathcal L\{y''\}=s^2Y-sy(0)-y'(0),

and

\mathcal L\{x^2\}=\frac{2}{s^3}.

Taking transforms,

(s^2Y-2s-1)-3(sY-2)-4Y=\frac2{s^3}.

Hence

(s^2-3s-4)Y-2s+5=\frac2{s^3},

so

(s^2-3s-4)Y=2s-5+\frac2{s^3}.

Since s^2-3s-4=(s-4)(s+1),

Y(s) = \frac{2s^4-5s^3+2}{s^3(s-4)(s+1)}.

Resolving into partial fractions,

Y(s) = \frac{97}{160}\frac1{s-4} +\frac95\frac1{s+1} -\frac{13}{32}\frac1s +\frac38\frac1{s^2} -\frac12\frac1{s^3}.

Taking inverse transforms term by term,

\begin{aligned} y(x) ={}& \frac{97}{160}e^{4x} +\frac95e^{-x} -\frac{13}{32} +\frac38x -\frac12\left(\frac{x^2}{2}\right)\\ ={}& \frac{97}{160}e^{4x} +\frac95e^{-x} -\frac{13}{32} +\frac38x -\frac{x^2}{4}. \end{aligned}

Verification of the initial conditions

y(0)=\frac{97}{160}+\frac95-\frac{13}{32} =\frac{97+288-65}{160}=2.

Also

y'(x)=\frac{97}{40}e^{4x}-\frac95e^{-x}+\frac38-\frac x2,

so

y'(0)=\frac{97}{40}-\frac95+\frac38 =\frac{97-72+15}{40}=1.

Final Answer:

\boxed{ y(x)= \frac{97}{160}e^{4x} +\frac95e^{-x} -\frac{13}{32} +\frac38x -\frac{x^2}{4}. }

Solution:

The function has period T=2\pi. For a periodic function,

\mathcal L\{f(x)\} = \frac{\int_0^T e^{-sx}f(x)\,dx}{1-e^{-sT}}.

Therefore

F(s) = \frac1{1-e^{-2\pi s}} \left[ \int_0^\pi xe^{-sx}\,dx + \int_\pi^{2\pi}(2\pi-x)e^{-sx}\,dx \right].

First integral

\int xe^{-sx}\,dx = -\frac{x}{s}e^{-sx}-\frac1{s^2}e^{-sx}.

Hence

I_1 = \int_0^\pi xe^{-sx}\,dx = \frac1{s^2} - e^{-\pi s}\left(\frac{\pi}{s}+\frac1{s^2}\right).

Second integral

An antiderivative is

\int(2\pi-x)e^{-sx}\,dx = \frac{x-2\pi}{s}e^{-sx} +\frac1{s^2}e^{-sx}.

Therefore

I_2 = e^{-\pi s}\left(\frac{\pi}{s}-\frac1{s^2}\right) +\frac1{s^2}e^{-2\pi s}.

Adding,

I_1+I_2 = \frac{1-2e^{-\pi s}+e^{-2\pi s}}{s^2} = \frac{(1-e^{-\pi s})^2}{s^2}.

Thus

F(s) = \frac{(1-e^{-\pi s})^2} {s^2(1-e^{-2\pi s})}.

Using

1-e^{-2\pi s} = (1-e^{-\pi s})(1+e^{-\pi s}),

we obtain

F(s)= \frac{1-e^{-\pi s}} {s^2(1+e^{-\pi s})}.

Final Answer:

\boxed{ \mathcal L\{f(x)\} = \frac{1-e^{-\pi s}}{s^2(1+e^{-\pi s})} = \frac1{s^2}\tanh\left(\frac{\pi s}{2}\right). }

Solution:

The PDE is

D(D^2-D'+1)(D+2D'+1)^3z=0,

where D=\partial/\partial x and D'=\partial/\partial y. Since the operators have constant coefficients, their contributions may be treated separately.

1. Factor D

Dz_1=0 \quad\Longrightarrow\quad \frac{\partial z_1}{\partial x}=0,

so

z_1=\phi_1(y).

2. Factor D^2-D'+1

We require

z_{xx}-z_y+z=0,

or

z_y=z_{xx}+z.

Put z_2=e^yU(x,y). Then

(z_2)_y=e^y(U+U_y), \qquad (z_2)_{xx}=e^yU_{xx}.

Substitution yields

U_y=U_{xx}.

Thus U is any solution of the one-dimensional heat equation. A standard spectral representation is

U(x,y)=\int_\Gamma A(\lambda)e^{\lambda x+\lambda^2y}\,d\lambda,

so

z_2 = \int_\Gamma A(\lambda)e^{\lambda x+(\lambda^2+1)y}\,d\lambda.

3. Repeated factor (D+2D'+1)^3

First solve

z_x+2z_y+z=0.

The characteristic equations are

\frac{dx}{1}=\frac{dy}{2}=\frac{dz}{-z}.

From the first two ratios,

y-2x=C.

From the first and third,

\frac{dz}{z}=-dx \quad\Longrightarrow\quad z=e^{-x}\phi(y-2x).

Because the factor occurs three times, the repeated-factor rule introduces polynomial multipliers of degrees 0,1,2 in x. Hence

z_3 = e^{-x} \left[ \phi_2(y-2x) +x\phi_3(y-2x) +x^2\phi_4(y-2x) \right].

Combining the three parts,

\begin{aligned} z(x,y)={}& \phi_1(y) + \int_\Gamma A(\lambda)e^{\lambda x+(\lambda^2+1)y}\,d\lambda\\ &+ e^{-x}\left[ \phi_2(y-2x)+x\phi_3(y-2x)+x^2\phi_4(y-2x) \right]. \end{aligned}

Final Answer:

\boxed{ \begin{aligned} z={}&\phi_1(y) +\int_\Gamma A(\lambda)e^{\lambda x+(\lambda^2+1)y}\,d\lambda\\ &+e^{-x}\left[ \phi_2(y-2x)+x\phi_3(y-2x)+x^2\phi_4(y-2x) \right]. \end{aligned}}

Solution:

The PDE is

xp^2+yq^2=z, \qquad p=z_x,\quad q=z_y.

Write

F=xp^2+yq^2-z=0.

Charpit’s equations

\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{dz}{pF_p+qF_q} = \frac{-dp}{F_x+pF_z} = \frac{-dq}{F_y+qF_z}.

Here

F_p=2xp,\quad F_q=2yq,\quad F_x=p^2,\quad F_y=q^2,\quad F_z=-1.

Therefore

\frac{dx}{2xp} = \frac{dy}{2yq} = \frac{dz}{2xp^2+2yq^2} = \frac{-dp}{p^2-p} = \frac{-dq}{q^2-q}.

Using the x,p pair,

\frac{dx}{2xp} = -\frac{dp}{p(p-1)} = \frac{dp}{p(1-p)}.

Canceling p,

\frac{dx}{2x}=\frac{dp}{1-p}.

Integrating,

\frac12\log x=-\log|1-p|+C,

hence

\sqrt{x}(1-p)=a.

Thus

p=1-\frac{a}{\sqrt x}.

Similarly, from the y,q pair,

q=1-\frac{b}{\sqrt y}.

Since p=z_x,

z_x=1-\frac{a}{\sqrt x}.

Integrating with respect to x,

z=x-2a\sqrt x+\phi(y).

Now q=z_y=\phi'(y), so

\phi'(y)=1-\frac{b}{\sqrt y}.

Integrating,

\phi(y)=y-2b\sqrt y+C.

Hence

z=x+y-2a\sqrt x-2b\sqrt y+C.

Now

xp^2 = x\left(1-\frac a{\sqrt x}\right)^2 = x-2a\sqrt x+a^2,

and similarly

yq^2=y-2b\sqrt y+b^2.

Thus the PDE requires

C=a^2+b^2.

Therefore

z=x+y-2a\sqrt x-2b\sqrt y+a^2+b^2 =(\sqrt x-a)^2+(\sqrt y-b)^2.

Final Answer:

\boxed{z=(\sqrt x-a)^2+(\sqrt y-b)^2,}

where a,b are arbitrary constants.

Solution:

Let

f(x)=xe^x-\cos x.

We first locate a sign change:

f(0.5)\approx-0.05322193, \qquad f(0.6)\approx0.26793567.

Hence a root lies in (0.5,0.6).

Regula-Falsi formula

For f(a)f(b)<0,

x=\frac{af(b)-bf(a)}{f(b)-f(a)}.

Iteration 1

x_1 = \frac{0.5(0.26793567)-0.6(-0.05322193)} {0.26793567+0.05322193} \approx0.51657190. f(x_1)\approx-0.00360274.

Iteration 2

Using a=0.51657190, b=0.6,

x_2\approx0.51767882, \qquad f(x_2)\approx-0.00023893.

Iteration 3

x_3\approx0.51775216, \qquad f(x_3)\approx-1.5824\times10^{-5}.

Iteration 4

x_4\approx0.51775702, \qquad f(x_4)\approx-1.0479\times10^{-6}.

Iteration 5

x_5\approx0.51775734, \qquad f(x_5)\approx-6.94\times10^{-8}.

The successive values agree to six decimal places.

Final Answer:

\boxed{x\approx0.517757.}

Solution:

We have to approximate

I=\int_0^1\frac{dx}{1+x}

with step size h=0.25. The number of subintervals is

n=\frac{1-0}{0.25}=4.

The tabulated values are

\begin{array}{c|ccccc} x & 0 & 0.25 & 0.50 & 0.75 & 1\\ \hline f(x)=\frac1{1+x} &1 &0.8 &0.66666667 &0.57142857 &0.5 \end{array}

(i) Composite trapezoidal rule

I_T = \frac h2 \left[ f_0+f_4+2(f_1+f_2+f_3) \right].

Substituting the values,

\begin{aligned} I_T &= \frac{0.25}{2} \left[ 1+0.5 +2(0.8+0.66666667+0.57142857) \right]\\ &= 0.125 \left[ 1.5+2(2.03809524) \right]\\ &= 0.125(5.57619048)\\ &= 0.69702381. \end{aligned}

Hence, correct to three decimal places,

I_T\approx0.697.

(ii) Simpson’s one-third rule

Since n=4 is even,

I_S = \frac h3 \left[ f_0+f_4+4(f_1+f_3)+2f_2 \right].

Therefore

\begin{aligned} I_S &= \frac{0.25}{3} \left[ 1+0.5+4(0.8+0.57142857)+2(0.66666667) \right]\\ &= \frac1{12} \left[ 1.5+5.48571428+1.33333334 \right]\\ &= \frac{8.31904762}{12}\\ &= 0.69325397. \end{aligned}

Hence, correct to three decimal places,

I_S\approx0.693.

For reference, the exact value is \log2\approx0.693147, so Simpson’s approximation is much closer.

Final Answer:

\boxed{ I_{\text{trapezoidal}}\approx0.697,\qquad I_{\text{Simpson }1/3}\approx0.693. }

Solution:

The functional is

v[y]=\int_0^1\left(2y+(y'')^2\right)\,dx,

subject to

y(0)=0,\quad y'(0)=1,\quad y(1)=1,\quad y'(1)=1.

Since the integrand contains y'', we use the Euler-Poisson equation.

Euler-Poisson equation

For J=\int F(x,y,y',y'')\,dx, an extremal satisfies

\frac{\partial F}{\partial y} - \frac d{dx}\left(\frac{\partial F}{\partial y'}\right) + \frac{d^2}{dx^2}\left(\frac{\partial F}{\partial y''}\right) =0.

Here

F=2y+(y'')^2,

so

\frac{\partial F}{\partial y}=2,\qquad \frac{\partial F}{\partial y'}=0,\qquad \frac{\partial F}{\partial y''}=2y''.

Hence

2+\frac{d^2}{dx^2}(2y'')=0 \quad\Longrightarrow\quad y^{(4)}=-1.

Integrate four times

y'''=-x+C_1, y''=-\frac{x^2}{2}+C_1x+C_2, y'=-\frac{x^3}{6}+\frac{C_1x^2}{2}+C_2x+C_3, y=-\frac{x^4}{24}+\frac{C_1x^3}{6}+\frac{C_2x^2}{2}+C_3x+C_4.

From y(0)=0, C_4=0. From y'(0)=1, C_3=1. Thus

y=-\frac{x^4}{24}+\frac{C_1x^3}{6}+\frac{C_2x^2}{2}+x.

Using y(1)=1,

-\frac1{24}+\frac{C_1}{6}+\frac{C_2}{2}+1=1,

hence

4C_1+12C_2=1. \tag{1}

Also

y'=-\frac{x^3}{6}+\frac{C_1x^2}{2}+C_2x+1.

Using y'(1)=1,

-\frac16+\frac{C_1}{2}+C_2+1=1,

so

3C_1+6C_2=1. \tag{2}

Twice (2) gives 6C_1+12C_2=2. Subtracting (1),

2C_1=1 \quad\Longrightarrow\quad C_1=\frac12.

Then from (2),

\frac32+6C_2=1 \quad\Longrightarrow\quad C_2=-\frac1{12}.

Therefore

y = -\frac{x^4}{24} +\frac{x^3}{12} -\frac{x^2}{24} +x.

Factoring the polynomial correction,

y=x-\frac{x^2(x-1)^2}{24}.

Final Answer:

\boxed{y(x)=x-\frac{x^2(x-1)^2}{24}.}

Solution:

For the initial-value problem

\frac{dy}{dx}=f(x,y),\qquad y(x_0)=y_0,

the classical fourth-order Runge-Kutta method uses four increments.

RK4 formula

k_1=h f(x_n,y_n), k_2=h f\left(x_n+\frac h2,\ y_n+\frac{k_1}{2}\right), k_3=h f\left(x_n+\frac h2,\ y_n+\frac{k_2}{2}\right), k_4=h f(x_n+h,\ y_n+k_3).

Then

y_{n+1} = y_n+\frac{k_1+2k_2+2k_3+k_4}{6}, \qquad x_{n+1}=x_n+h.

Flow chart

START
  |
  v
Read x0, y0, h, xn
  |
  v
Set x = x0, y = y0
  |
  v
Is x < xn ?
 / \
No  Yes
 |    |
 |    v
 |  k1 = h*f(x,y)
 |  k2 = h*f(x+h/2, y+k1/2)
 |  k3 = h*f(x+h/2, y+k2/2)
 |  k4 = h*f(x+h, y+k3)
 |    |
 |    v
 |  y = y + (k1+2*k2+2*k3+k4)/6
 |  x = x + h
 |    |
 |    +----------------------+
 |                           |
 +---------------------------+
  |
  v
Print x, y
  |
  v
STOP
        

C-language form corresponding to the flow chart

#include <stdio.h>
#include <math.h>

double f(double x, double y)
{
    /* Replace by the given differential equation y' = f(x,y). */
    return y;
}

int main(void)
{
    double x, y, h, xn;
    double k1, k2, k3, k4;

    printf("Enter x0, y0, h and final x: ");
    scanf("%lf %lf %lf %lf", &x, &y, &h, &xn);

    while (x < xn)
    {
        if (x + h > xn)
            h = xn - x;

        k1 = h * f(x, y);
        k2 = h * f(x + h/2.0, y + k1/2.0);
        k3 = h * f(x + h/2.0, y + k2/2.0);
        k4 = h * f(x + h, y + k3);

        y = y + (k1 + 2.0*k2 + 2.0*k3 + k4)/6.0;
        x = x + h;
    }

    printf("x = %.6f, y = %.6f\n", x, y);
    return 0;
}
        

The four slopes are calculated in sequence, combined in the weights 1:2:2:1, and the process is repeated until the final value of x is reached.

Final Answer:

\boxed{ y_{n+1}=y_n+\frac{k_1+2k_2+2k_3+k_4}{6}. }