Solution:

Definition: A non-empty set G with a binary operation is a group if it satisfies closure, associativity, existence of an identity and existence of an inverse for every element.

Here the operation is addition modulo 5, denoted by +_5. The set is

G=\{0,1,2,3,4\}.

1. Closure. For any a,b\in G, the remainder when a+b is divided by 5 is one of 0,1,2,3,4. Hence

a+_5 b\in G.

The complete Cayley table is

+_5	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	0
2	2	3	4	0	1
3	3	4	0	1	2
4	4	0	1	2	3

2. Associativity. Ordinary integer addition is associative. Taking residues modulo 5 preserves this:

(a+_5b)+_5c=a+_5(b+_5c).

3. Identity. The element 0 satisfies

a+_50=0+_5a=a

for every a\in G. Thus 0 is the identity.

4. Inverses. The additive inverses modulo 5 are

0^{-1}=0,\quad 1^{-1}=4,\quad 2^{-1}=3,\quad 3^{-1}=2,\quad 4^{-1}=1,

where the notation means additive inverse. In each case the sum is congruent to zero modulo 5.

Therefore (G,+_5) is a group. It contains exactly five elements, so its order is 5. In fact, the table is symmetric, so the group is also abelian and cyclic, generated by 1.

Final Answer:

\boxed{(G,+_5)\text{ is a finite group and }|G|=5.}

Solution:

The sequence is

x_n=\frac{2n-7}{3n+2},\qquad n\in\mathbb N.

Step 1: Prove monotonic increase.

x_{n+1} = \frac{2(n+1)-7}{3(n+1)+2} = \frac{2n-5}{3n+5}.

Therefore

\begin{aligned} x_{n+1}-x_n &= \frac{2n-5}{3n+5} - \frac{2n-7}{3n+2}\\ &= \frac{(2n-5)(3n+2)-(2n-7)(3n+5)} {(3n+5)(3n+2)}. \end{aligned}

Expand the numerator:

(2n-5)(3n+2)=6n^2-11n-10, (2n-7)(3n+5)=6n^2-11n-35.

Hence

x_{n+1}-x_n = \frac{25}{(3n+5)(3n+2)}>0.

Thus \{x_n\} is strictly increasing.

Step 2: Prove boundedness.

For every n\ge1,

x_n<\frac23 [/katex] <p>because</p> [katex display="true"] 3(2n-7)<2(3n+2) \iff 6n-21<6n+4. [/katex]<p>Thus [katex]2/3 is an upper bound. Since the sequence is increasing and

x_1=\frac{-5}{5}=-1,

we also have x_n\ge-1. Hence the sequence is bounded.

Step 3: Convergence.

A bounded monotone sequence is convergent. Its limit is obtained by dividing numerator and denominator by n:

\lim_{n\to\infty}x_n = \lim_{n\to\infty} \frac{2-\frac7n}{3+\frac2n} = \frac23.

Final Answer:

\boxed{\{x_n\}\text{ is bounded, strictly increasing and converges to }\frac23.}

Solution:

Definition: In a metric space (X,d), the open sphere (open ball) with centre a and radius r>0 is

B(a,r)=\{x\in X:d(x,a)<r\}. [/katex]<p>We must show that every point of [katex]B(a,r) is an interior point.

Let x\in B(a,r). Then

d(x,a)<r.[/katex]<p>Define</p> [katex display="true"] \varepsilon=r-d(x,a).

Since d(x,a)<r[/katex], we have [katex]\varepsilon>0. Consider the open sphere B(x,\varepsilon).

Take any y\in B(x,\varepsilon). Then

d(y,x)<\varepsilon. [/katex]<p>By the triangle inequality,</p> [katex display="true"] d(y,a) \le d(y,x)+d(x,a) <\varepsilon+d(x,a). [/katex]<p>Substitute [katex]\varepsilon=r-d(x,a):

d(y,a)<r-d(x,a)+d(x,a)=r. [/katex]<p>Thus [katex]y\in B(a,r). Therefore

B(x,\varepsilon)\subseteq B(a,r).

So every x\in B(a,r) has an open sphere about it lying wholly inside B(a,r).

Final Answer:

\boxed{\text{Every open sphere in a metric space is an open set.}}

Solution:

Given

Z=f(x+iy)+\phi(x-iy).

Let

u=x+iy,\qquad v=x-iy.

Then

Z=f(u)+\phi(v).

Differentiate twice with respect to x. Since u_x=v_x=1,

Z_x=f'(u)+\phi'(v),

and

Z_{xx}=f''(u)+\phi''(v).

Now differentiate with respect to y. Since u_y=i and v_y=-i,

Z_y=i f'(u)-i\phi'(v).

Differentiating again,

\begin{aligned} Z_{yy} &= i\{f''(u)u_y\} -i\{\phi''(v)v_y\}\\ &= i(i)f''(u)-i(-i)\phi''(v)\\ &= -f''(u)-\phi''(v). \end{aligned}

Therefore

Z_{xx}+Z_{yy} = f''+\phi''-f''-\phi'' =0.

The arbitrary functions have been eliminated.

Final Answer:

\boxed{\frac{\partial^2Z}{\partial x^2}+\frac{\partial^2Z}{\partial y^2}=0}.

Solution:

Let

f(x)=x^3-2x-5.

We have

f(2)=8-4-5=-1<0,[/katex] <p>and</p> [katex display="true"]f(3)=27-6-5=16>0.

Hence a real root lies in (2,3).

Regula-Falsi (false-position) formula: If f(a)f(b)<0[/katex], then the x-intercept of the chord joining [katex](a,f(a))[/katex] and [katex](b,f(b))[/katex] is</p> [katex display="true"] c=\frac{a f(b)-b f(a)}{f(b)-f(a)}.

Starting with a=2, b=3, the successive false-position approximations are:

Iteration	c	f(c)
1	2.058823529	-0.390799919
2	2.081263660	-0.147204060
3	2.089639210	-0.054676503
4	2.092739574	-0.020202866
5	2.093883708	-0.007450506
6	2.094305451	-0.002745673
7	2.094460846	-0.001011574
8	2.094518093	-0.000372653
9	2.094539182	-0.000137246
10	2.094546951	-0.000050569
12	2.094550867	-0.000006862
15	2.094551451	-0.000000343

The values stabilize at

x\approx2.09455148.

Final Answer:

\boxed{x\approx2.09455}

Solution:

Algorithm:

1. Start.
2. Read two numbers A and B.
3. If A>B, print “A is the largest”.
4. Otherwise, if B>A, print “B is the largest”.
5. Otherwise print “Both numbers are equal”.
6. Stop.

Flowchart:

             ┌─────────┐
             │  START  │
             └────┬────┘
                  │
                  ▼
          ┌──────────────┐
          │  Read A, B   │
          └──────┬───────┘
                 │
                 ▼
             ┌───────┐
             │ A > B?│
             └──┬─┬──┘
              Yes No
               │   │
               ▼   ▼
      ┌─────────────┐   ┌───────┐
      │ Print A     │   │ B > A?│
      │ is largest  │   └──┬─┬──┘
      └──────┬──────┘    Yes No
             │            │   │
             │            ▼   ▼
             │   ┌─────────────┐  ┌──────────────┐
             │   │ Print B     │  │ Print equal  │
             │   │ is largest  │  └──────┬───────┘
             │   └──────┬──────┘         │
             └──────────┴────────────────┘
                        │
                        ▼
                   ┌────────┐
                   │  STOP  │
                   └────────┘

Final Answer:

The above flowchart compares the two inputs and prints the larger number, while also handling the equality case.

Solution:

Let a be an element of a group G. We prove that

o(a)=o(a^{-1}).

Case 1: a has finite order.

Suppose

o(a)=n.

By definition, n is the least positive integer for which

a^n=e.

Now

(a^{-1})^n=a^{-n}=(a^n)^{-1}=e^{-1}=e.

Hence the order of a^{-1} divides n. Let

o(a^{-1})=m.

Then

(a^{-1})^m=e.

Taking inverses,

a^m=e.

Since n is the least positive integer with a^n=e, n\le m. But from the first argument m\le n. Therefore

m=n.

Case 2: a has infinite order.

If a^{-1} had finite order m, then (a^{-1})^m=e would imply a^m=e, contradicting the infinite order of a. Thus a^{-1} also has infinite order.

Final Answer:

\boxed{o(a)=o(a^{-1})}

Solution:

Let \phi:G\to H be a group homomorphism. Its homomorphic image is

\operatorname{Im}\phi=\phi(G).

Let

K=\ker\phi=\{g\in G:\phi(g)=e_H\}.

Step 1: K is normal in G.

For g\in G and k\in K,

\phi(gkg^{-1}) = \phi(g)\phi(k)\phi(g)^{-1} = \phi(g)e_H\phi(g)^{-1} =e_H.

Hence gkg^{-1}\in K, so K\trianglelefteq G. Therefore the quotient group G/K is defined.

Step 2: Define a map from the quotient group.

Define

\Psi:G/K\to\operatorname{Im}\phi

by

\Psi(gK)=\phi(g).

Well-defined: If gK=hK, then h^{-1}g\in K. Therefore

\phi(h^{-1}g)=e_H,

which implies

\phi(g)=\phi(h).

Homomorphism:

\Psi((gK)(hK)) = \Psi(ghK) = \phi(gh) = \phi(g)\phi(h) = \Psi(gK)\Psi(hK).

Onto: Every element of \operatorname{Im}\phi has the form \phi(g), and this equals \Psi(gK).

One-to-one: If \Psi(gK)=e_H, then \phi(g)=e_H, so g\in K and gK=K, the identity coset. Thus the kernel of \Psi is trivial.

Hence \Psi is an isomorphism.

Final Answer:

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

This is the First Isomorphism Theorem.

Solution:

Darboux Theorem for derivatives: If f is differentiable on an interval I, then its derivative f' has the intermediate-value property. That is, if x_1<x_2[/katex] and a number [katex]\lambda[/katex] lies between [katex]f'(x_1)[/katex] and [katex]f'(x_2)[/katex], then there exists [katex]c\in(x_1,x_2)[/katex] such that</p> [katex display="true"]f'(c)=\lambda.

Proof.

Suppose, without loss of generality,

f'(x_1)<\lambda<f'(x_2). [/katex]<p>Define</p> [katex display="true"] g(x)=f(x)-\lambda x.

Then

g'(x)=f'(x)-\lambda.

Therefore

g'(x_1)<0,\qquad g'(x_2)>0.

Since g is continuous on [x_1,x_2], it attains a minimum there.

The minimum cannot occur at x_1: because g'(x_1)<0[/katex], for sufficiently small positive [katex]h[/katex],</p> [katex display="true"]g(x_1+h)<g(x_1).[/katex]<p>Similarly, the minimum cannot occur at [katex]x_2: because g'(x_2)>0, for sufficiently small positive h,

g(x_2-h)<g(x_2).[/katex]<p>Hence [katex]g attains its minimum at some interior point c\in(x_1,x_2). By Fermat's theorem,

g'(c)=0.

Thus

f'(c)-\lambda=0 \quad\Longrightarrow\quad f'(c)=\lambda.

The case f'(x_2)<\lambda<f'(x_1)[/katex] is proved similarly by considering a maximum or by reversing signs.</p><p><strong>Final Answer:</strong></p> [katex display="true"] \boxed{\text{Every derivative has the intermediate-value property.}}

Solution:

Let

f(x)=\sin x,\qquad 0\le x\le\frac{\pi}{2}.

The partition is

P_n=\left\{0,h,2h,\ldots,nh\right\}, \qquad h=\frac{\pi}{2n}.

Since \sin x is increasing on [0,\pi/2], on the subinterval [(i-1)h,ih] we have

m_i=\sin((i-1)h),\qquad M_i=\sin(ih).

Therefore the lower Darboux sum is

L(f,P_n) = h\sum_{i=1}^{n}\sin((i-1)h) = h\sum_{j=0}^{n-1}\sin(jh).

Using

\sum_{j=1}^{m}\sin(jh) = \frac{\sin(mh/2)\sin((m+1)h/2)} {\sin(h/2)},

we get

L(f,P_n) = h\, \frac{ \sin((n-1)h/2)\sin(nh/2) }{ \sin(h/2) }.

Since nh=\pi/2,

L(f,P_n) = \frac{h}{\sqrt2} \frac{\sin\left(\frac{\pi}{4}-\frac h2\right)} {\sin(h/2)}.

Similarly, the upper Darboux sum is

U(f,P_n) = h\sum_{i=1}^{n}\sin(ih) = h\, \frac{ \sin(nh/2)\sin((n+1)h/2) }{ \sin(h/2) },

so

U(f,P_n) = \frac{h}{\sqrt2} \frac{\sin\left(\frac{\pi}{4}+\frac h2\right)} {\sin(h/2)}.

A particularly simple difference is obtained directly:

\begin{aligned} U(f,P_n)-L(f,P_n) &= h\sum_{i=1}^{n} \left[\sin(ih)-\sin((i-1)h)\right]\\ &= h(\sin(nh)-\sin0)\\ &= h\\ &= \frac{\pi}{2n}. \end{aligned}

Thus

U(f,P_n)-L(f,P_n)\to0.

By the Darboux criterion, f is Riemann integrable on [0,\pi/2].

Taking n\to\infty (equivalently h\to0) in either explicit sum gives

\lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n) =1.

Therefore the lower and upper Darboux integrals coincide:

\sup_P L(f,P) = \inf_P U(f,P) =1.

This agrees with

\int_0^{\pi/2}\sin x\,dx = [-\cos x]_0^{\pi/2}=1.

Final Answer:

\boxed{ L(f,P_n)= h\frac{\sin((n-1)h/2)\sin(nh/2)}{\sin(h/2)} } \boxed{ U(f,P_n)= h\frac{\sin(nh/2)\sin((n+1)h/2)}{\sin(h/2)} }

where h=\pi/(2n), and

\boxed{\sup_P L(f,P)=\inf_P U(f,P)=1.}

Solution:

The series is

x+\frac12\frac{x^3}{3} +\frac{1\cdot3}{2\cdot4}\frac{x^5}{5} +\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\frac{x^7}{7} +\cdots.

Its general term may be written as

a_n= \frac{1\cdot3\cdot5\cdots(2n-1)} {2\cdot4\cdot6\cdots(2n)} \frac{x^{2n+1}}{2n+1}, \qquad n=0,1,2,\ldots,

with the empty product for n=0 equal to 1.

For n\ge0,

\left|\frac{a_{n+1}}{a_n}\right| = |x|^2 \frac{2n+1}{2n+2} \frac{2n+1}{2n+3}.

Therefore

\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = |x|^2.

By the ratio test, the series converges absolutely for

|x|<1[/katex] <p>and diverges for</p> [katex display="true"]|x|>1.

Endpoint |x|=1.

At x=1, let

b_n= \frac{1\cdot3\cdots(2n-1)} {2\cdot4\cdots(2n)} \frac1{2n+1}.

Then

\frac{b_n}{b_{n+1}} = \frac{(2n+2)(2n+3)}{(2n+1)^2}.

Raabe's test gives

\begin{aligned} \lim_{n\to\infty} n\left(\frac{b_n}{b_{n+1}}-1\right) &= \lim_{n\to\infty} n\frac{6n+5}{(2n+1)^2}\\ &=\frac32>1. \end{aligned}

Hence the series converges at x=1. At x=-1, its absolute values are the same, so it also converges absolutely.

The series is in fact the standard Maclaurin series of \sin^{-1}x.

Final Answer:

\boxed{\text{The series converges for }-1\le x\le1\text{ and diverges for }|x|>1.}

Solution:

Let f,g:A\to\mathbb C be continuous at an arbitrary point x\in A. We use the sequential criterion for continuity in metric spaces.

Let x_n\to x in (A,d). Since f and g are continuous,

f(x_n)\to f(x), \qquad g(x_n)\to g(x).

1. Sum.

Continuity of addition in \mathbb C gives

(f+g)(x_n) = f(x_n)+g(x_n) \to f(x)+g(x) = (f+g)(x).

Hence f+g is continuous.

2. Product.

Continuity of multiplication in \mathbb C gives

(fg)(x_n) = f(x_n)g(x_n) \to f(x)g(x) = (fg)(x).

Thus fg is continuous.

For completeness, the usual estimate behind this assertion is

|f_ng_n-fg| \le |f_n||g_n-g|+|g||f_n-f|,

and a convergent complex sequence \{f_n\} is bounded.

3. Scalar multiple.

Let a\in\mathbb R or a\in\mathbb C. Then

(af)(x_n) = a f(x_n) \to a f(x) = (af)(x).

Hence af is continuous.

Because the point x was arbitrary, all three functions are continuous on A.

Final Answer:

\boxed{f+g,\quad fg,\quad af\text{ are continuous on }(A,d).}

Solution:

Let

f(z)=5\sin2z.

The function \sin2z is entire, because the complex sine function is analytic throughout the complex plane. Therefore f(z) is analytic at every point inside and on the square whose vertices are 1\pm i and -1\pm i.

Cauchy's theorem: If a function is analytic throughout a simply connected region and on its closed boundary C, then

\oint_C f(z)\,dz=0.

Thus directly,

\oint_C5\sin2z\,dz=0.

We can also verify this by exhibiting an antiderivative. Since

\frac{d}{dz}\left(-\frac52\cos2z\right) = 5\sin2z,

the function has the entire primitive

F(z)=-\frac52\cos2z.

For any closed contour, the integral of a derivative is zero:

\oint_C5\sin2z\,dz = \oint_C F'(z)\,dz =0.

Final Answer:

\boxed{\displaystyle\oint_C5\sin2z\,dz=0.}

Hence Cauchy's theorem is verified.

Solution:

The transformation is

w=\frac{2z+3}{z-4}.

We first solve for z in terms of w:

w(z-4)=2z+3.

Thus

z(w-2)=4w+3,

and hence

z=\frac{4w+3}{w-2}.

Let

w=u+iv.

The original circle is

x^2+y^2-4x=0.

Since x^2+y^2=|z|^2 and 4x=2(z+\overline z), its complex form is

z\overline z-2(z+\overline z)=0.

Substitute

z=\frac{4w+3}{w-2}, \qquad \overline z=\frac{4\overline w+3}{\overline w-2}.

After multiplication by |w-2|^2 and simplification,

11(4u+3)=0.

Equivalently, one may check directly that

x^2+y^2-4x = \frac{11(4u+3)} {(u-2)^2+v^2}.

Therefore every finite image point of the circle satisfies

4u+3=0.

The point z=4 lies on the original circle and is mapped to w=\infty, as expected for a generalized straight line in a bilinear transformation.

Final Answer:

\boxed{4u+3=0}

is the image of the circle.

Solution:

The PDE is of Lagrange's form

Pp+Qq=R,

with

P=y+z,\qquad Q=z+x,\qquad R=x+y.

The auxiliary equations are

\frac{dx}{y+z} = \frac{dy}{z+x} = \frac{dz}{x+y} =ds.

We need two independent first integrals.

First integral.

Along a characteristic,

\frac{d}{ds}(x-y) = (y+z)-(z+x) =-(x-y),

and

\frac{d}{ds}(y-z) = (z+x)-(x+y) =-(y-z).

Therefore

\frac{d}{ds}\log(x-y) = \frac{d}{ds}\log(y-z) =-1.

Hence

\frac{x-y}{y-z}=C_1.

Second integral.

Let

S=x+y+z.

Then

\frac{dS}{ds} = (y+z)+(z+x)+(x+y) = 2S.

Also, if A=x-y, then dA/ds=-A. Hence

\frac{d}{ds}(A^2S) = 2A(-A)S+A^2(2S) =0.

Thus

(x-y)^2(x+y+z)=C_2.

The complete general solution of a first-order Lagrange equation is an arbitrary relation between two independent first integrals.

Final Answer:

\boxed{ \Phi\left( \frac{x-y}{y-z}, \;(x-y)^2(x+y+z) \right)=0, }

where \Phi is an arbitrary function.

Solution:

Using the standard notation

D=\frac{\partial}{\partial x}, \qquad D'=\frac{\partial}{\partial y},

the PDE is

(D^2+DD'-6D'^2)z=y\cos x.

Step 1: Complementary function.

Factor the operator:

D^2+DD'-6D'^2 = (D+3D')(D-2D').

For

(D+3D')z=0,

the characteristic equation is

\frac{dy}{dx}=3,

so y-3x is constant. This gives

z_1=\phi_1(y-3x).

For

(D-2D')z=0,

we have

\frac{dy}{dx}=-2,

so y+2x is constant. Thus

z_2=\phi_2(y+2x).

Hence

z_{CF}=\phi_1(y-3x)+\phi_2(y+2x).

Step 2: Particular integral.

We claim

z_p=\sin x-y\cos x.

Compute its derivatives:

z_{p,x} = \cos x+y\sin x, z_{p,xx} = -\sin x+y\cos x, z_{p,y} = -\cos x, \qquad z_{p,xy} = \sin x, \qquad z_{p,yy}=0.

Therefore

\begin{aligned} z_{p,xx}+z_{p,xy}-6z_{p,yy} &= (-\sin x+y\cos x)+\sin x-0\\ &= y\cos x. \end{aligned}

Thus z_p is indeed a particular integral.

Final Answer:

\boxed{ z=\phi_1(y-3x)+\phi_2(y+2x)+\sin x-y\cos x. }

Solution:

We use the definition

L\{f(t)\}(p) = \int_0^\infty e^{-pt}f(t)\,dt, \qquad p>0.

Part I: L\{\sin\sqrt t\}.

Put

t=u^2,\qquad dt=2u\,du.

Then

L\{\sin\sqrt t\} = 2\int_0^\infty u e^{-pu^2}\sin u\,du.

Use the standard Gaussian integral

\int_0^\infty e^{-pu^2}\cos(bu)\,du = \frac12\sqrt{\frac{\pi}{p}} e^{-b^2/(4p)}. \tag{1}

Differentiate (1) with respect to b:

-\int_0^\infty u e^{-pu^2}\sin(bu)\,du = -\frac{b}{4p} \sqrt{\frac{\pi}{p}} e^{-b^2/(4p)}.

Hence

\int_0^\infty u e^{-pu^2}\sin(bu)\,du = \frac{b}{4p} \sqrt{\frac{\pi}{p}} e^{-b^2/(4p)}.

Put b=1 and multiply by 2:

L\{\sin\sqrt t\} = \frac{1}{2p} \sqrt{\frac{\pi}{p}} e^{-1/(4p)}.

Therefore

L\{\sin\sqrt t\} = \frac{\sqrt\pi}{2p^{3/2}} e^{-1/(4p)}.

Part II: Show the stated transform.

Again use t=u^2:

\begin{aligned} L\left\{\frac{\cos\sqrt t}{\sqrt t}\right\} &= \int_0^\infty e^{-pt} \frac{\cos\sqrt t}{\sqrt t}\,dt\\ &= 2\int_0^\infty e^{-pu^2}\cos u\,du. \end{aligned}

Apply (1) with b=1:

2\int_0^\infty e^{-pu^2}\cos u\,du = \sqrt{\frac{\pi}{p}}e^{-1/(4p)}.

Final Answer:

\boxed{ L\{\sin\sqrt t\} = \frac{\sqrt\pi}{2p^{3/2}} e^{-1/(4p)} }

and

\boxed{ L\left\{\frac{\cos\sqrt t}{\sqrt t}\right\} = \sqrt{\frac{\pi}{p}} e^{-1/(4p)}. }

Solution:

The functional is

I[y] = \int_0^{x_2} F(y,y')\,dx, \qquad F= \sqrt{\frac{1+y'^2}{y}}.

The initial point is fixed at (0,0), while the terminal point (x_2,y_2) is movable on the straight line

y_2=x_2+5.

Step 1: Beltrami identity.

Since F does not explicitly contain x, an extremal satisfies

F-y'F_{y'}=C.

Now

F_{y'} = \frac{y'}{\sqrt y\sqrt{1+y'^2}}.

Therefore

\begin{aligned} F-y'F_{y'} &= \frac{\sqrt{1+y'^2}}{\sqrt y} - \frac{y'^2}{\sqrt y\sqrt{1+y'^2}}\\ &= \frac{1}{\sqrt y\sqrt{1+y'^2}}. \end{aligned}

Write the positive constant as 1/\sqrt{2a}. Then

\frac{1}{\sqrt{y(1+y'^2)}} = \frac1{\sqrt{2a}},

or

y(1+y'^2)=2a.

Thus

y'^2=\frac{2a-y}{y}.

Step 2: Parametric form.

Put

y=a(1-\cos\theta).

Then

\frac{2a-y}{y} = \frac{1+\cos\theta}{1-\cos\theta} = \cot^2\frac{\theta}{2}.

The branch compatible with the terminal condition below is

\frac{dy}{dx} = -\cot\frac{\theta}{2}.

Since

\frac{dy}{d\theta}=a\sin\theta,

we have

\frac{dx}{d\theta} = \frac{dy/d\theta}{dy/dx} = -a(1-\cos\theta).

Using x=0 at \theta=0,

x=-a(\theta-\sin\theta).

Hence the extremal belongs to the cycloidal family

x=-a(\theta-\sin\theta), \qquad y=a(1-\cos\theta). \tag{2}

Step 3: Transversality condition at the movable endpoint.

The terminal curve is y=\psi(x)=x+5, so \psi'(x)=1. The transversality condition is

F+(\psi'-y')F_{y'}=0.

Substitute \psi'=1:

F+(1-y')F_{y'}=0.

Using the expressions for F and F_{y'},

\frac{1+y'^2+y'(1-y')} {\sqrt y\sqrt{1+y'^2}} =0.

The numerator reduces to

1+y'=0,

so

y'=-1

at the terminal point.

From (2),

y'=-\cot\frac{\theta}{2}.

Thus at the endpoint

-\cot\frac{\theta_2}{2}=-1 \quad\Longrightarrow\quad \theta_2=\frac{\pi}{2}.

At \theta_2=\pi/2,

y_2=a,

and

x_2 = -a\left(\frac{\pi}{2}-1\right).

Use the terminal-line condition y_2=x_2+5:

a = -a\left(\frac{\pi}{2}-1\right)+5.

Therefore

\frac{\pi a}{2}=5,

so

a=\frac{10}{\pi}.

Final Answer:

The required extremal is the cycloidal arc

\boxed{ x=-\frac{10}{\pi}(\theta-\sin\theta), \qquad y=\frac{10}{\pi}(1-\cos\theta), \quad 0\le\theta\le\frac{\pi}{2}. }

Its terminal point is

\boxed{ x_2=\frac{10}{\pi}-5,\qquad y_2=\frac{10}{\pi}, }

which indeed satisfies y_2=x_2+5.

Solution:

Arrange the data in increasing order of x:

\begin{array}{c|cccc} x&-1&1&2&3\\ \hline y&-21&15&12&3 \end{array}

Form the divided-difference table:

x	f[x]	First divided difference	Second	Third
-1	-21	18	-7	1
1	15	-3	-3
2	12	-9
3	3

Newton's divided-difference interpolation formula gives

\begin{aligned} P(x) =&-21 +18(x+1) -7(x+1)(x-1)\\ &+1(x+1)(x-1)(x-2). \end{aligned}

Expand each part:

18(x+1)=18x+18, -7(x+1)(x-1) = -7(x^2-1) = -7x^2+7,

and

\begin{aligned} (x+1)(x-1)(x-2) &= (x^2-1)(x-2)\\ &= x^3-2x^2-x+2. \end{aligned}

Therefore

\begin{aligned} P(x) &= -21+18x+18-7x^2+7 +x^3-2x^2-x+2\\ &= x^3-9x^2+17x+6. \end{aligned}

The third divided difference is non-zero, so no polynomial of degree less than 3 can interpolate all four data points. Thus this cubic has the lowest possible degree.

Verification:

P(3)=3,\quad P(2)=12,\quad P(1)=15,\quad P(-1)=-21.

Final Answer:

\boxed{P(x)=x^3-9x^2+17x+6}.

Solution:

The initial-value problem is

\frac{dy}{dx}=f(x,y)=x^2+y^2, \qquad y(0)=0.

The step size is

h=0.1.

Euler's formula:

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

Starting with x_0=0, y_0=0:

n	x_n	y_n	f(x_n,y_n)=x_n^2+y_n^2	y_{n+1}
0	0.0	0.000000000	0.000000000	0.000000000
1	0.1	0.000000000	0.010000000	0.001000000
2	0.2	0.001000000	0.040001000	0.005000100
3	0.3	0.005000100	0.090025001	0.014002600
4	0.4	0.014002600	0.160196073	0.030022207

Showing the last step explicitly:

\begin{aligned} y(0.5) &\approx y_5\\ &= y_4+0.1(0.4^2+y_4^2)\\ &= 0.0140026001 +0.1(0.1601960728)\\ &= 0.03002220738. \end{aligned}

Final Answer:

\boxed{y(0.5)\approx0.0300222}

or, to four decimal places, \boxed{y(0.5)\approx0.0300}.