Solution:

The permutation is

A= \begin{pmatrix} 1&2&3&4&5\\ 2&3&1&5&4 \end{pmatrix}.

Thus

1\mapsto2,\quad 2\mapsto3,\quad 3\mapsto1,

and

4\mapsto5,\quad 5\mapsto4.

Hence in disjoint-cycle notation,

A=(1\,2\,3)(4\,5).

Since the cycles are disjoint, their powers may be taken independently:

A^3=(1\,2\,3)^3(4\,5)^3.

Now

(1\,2\,3)^3=e

and, because a transposition has order 2,

(4\,5)^3=(4\,5).

Therefore

A^3=(4\,5).

In two-row notation,

A^3= \begin{pmatrix} 1&2&3&4&5\\ 1&2&3&5&4 \end{pmatrix}.

Order of A.

The order of a permutation written as a product of disjoint cycles is the least common multiple of their lengths. Here the cycle lengths are 3 and 2. Therefore

o(A)=\operatorname{lcm}(3,2)=6.

Final Answer:

\boxed{ A^3= \begin{pmatrix} 1&2&3&4&5\\ 1&2&3&5&4 \end{pmatrix}, \qquad o(A)=6.}

Solution:

Consider

I=\int_0^\infty e^{-x^2}\,dx.

Split the improper integral:

I= \int_0^1e^{-x^2}\,dx + \int_1^\infty e^{-x^2}\,dx.

The first integral is proper because e^{-x^2} is continuous on [0,1]. Hence it is finite.

For x\ge1,

x^2\ge x.

Therefore

0<e^{-x^2}\le e^{-x}. [/katex]<p>But</p> [katex display="true"] \int_1^\infty e^{-x}\,dx = \left[-e^{-x}\right]_1^\infty = e^{-1} <\infty. [/katex]<p>Hence, by the comparison test for improper integrals,</p> [katex display="true"] \int_1^\infty e^{-x^2}\,dx

converges. Consequently the original integral converges.

Its well-known exact value may also be obtained by squaring the integral and using polar coordinates:

\left(\int_{-\infty}^{\infty}e^{-x^2}\,dx\right)^2 = \iint_{\mathbb R^2}e^{-(x^2+y^2)}\,dx\,dy = 2\pi\int_0^\infty re^{-r^2}\,dr = \pi.

Thus

\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt\pi.

Since the integrand is even,

\int_0^\infty e^{-x^2}\,dx=\frac{\sqrt\pi}{2}.

Final Answer:

\boxed{\displaystyle \int_0^\infty e^{-x^2}\,dx\text{ converges and equals }\frac{\sqrt\pi}{2}.}

Solution:

Let \{x_n\} be a Cauchy sequence in a metric space (X,d).

Definition of Cauchy sequence: For every \varepsilon>0, there exists N such that

m,n\ge N \quad\Rightarrow\quad d(x_m,x_n)<\varepsilon. [/katex]<p>Choose [katex]\varepsilon=1. Then there exists N such that

m,n\ge N \quad\Rightarrow\quad d(x_m,x_n)<1. [/katex]<p>Fix [katex]n=N. Then for every m\ge N,

d(x_m,x_N)<1. [/katex]<p>Thus all terms [katex]x_N,x_{N+1},\ldots lie in the ball B(x_N,1).

Only finitely many terms remain:

x_1,x_2,\ldots,x_{N-1}.

Let

M= \max\{1,\, d(x_1,x_N),\ldots,d(x_{N-1},x_N)\}.

Then for every n,

d(x_n,x_N)\le M.

Hence the entire sequence lies in a finite-radius ball about x_N.

Final Answer:

\boxed{\text{Every Cauchy sequence is bounded.}}

Solution:

Let

d(x,y)=|x-y|, \qquad x,y\in\mathbb R.

We verify the four metric axioms.

1. Non-negativity.

For all x,y\in\mathbb R,

d(x,y)=|x-y|\ge0.

2. Identity of indiscernibles.

d(x,y)=0 \iff |x-y|=0 \iff x-y=0 \iff x=y.

3. Symmetry.

d(x,y) = |x-y| = |-(y-x)| = |y-x| = d(y,x).

4. Triangle inequality.

For x,y,z\in\mathbb R,

x-z=(x-y)+(y-z).

Using the ordinary triangle inequality for absolute values,

|x-z| \le |x-y|+|y-z|.

Thus

d(x,z)\le d(x,y)+d(y,z).

All metric axioms are satisfied.

Final Answer:

\boxed{(\mathbb R,d),\quad d(x,y)=|x-y|,\text{ is a metric space.}}

Solution:

Let

F(x,y,z,p,q) = p^2+q^2-2px-2qy+1=0.

The equation does not contain z explicitly.

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=2(p-x),\quad F_q=2(q-y), F_x=-2p,\quad F_y=-2q,\quad F_z=0.

Therefore

\frac{dx}{2(p-x)} = \frac{dy}{2(q-y)} = \frac{dp}{2p} = \frac{dq}{2q} =\cdots

Use

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

Then

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

Thus

\frac{dx}{dp}+\frac{x}{p}=1.

The integrating factor is p, so

\frac{d}{dp}(px)=p.

Integrating,

px=\frac{p^2}{2}+\frac{a}{2},

or

2px-p^2=a. \tag{1}

Similarly,

2qy-q^2=b. \tag{2}

But the original equation gives

(p^2-2px)+(q^2-2qy)+1=0.

Using (1) and (2),

-a-b+1=0,

so

b=1-a.

Equation (1) is quadratic in p:

p^2-2xp+a=0.

Hence

p=x\pm\sqrt{x^2-a}.

Similarly,

q=y\pm\sqrt{y^2-(1-a)}.

Since p=z_x and q=z_y, integrate:

\int\sqrt{x^2-a}\,dx = \frac12 \left[ x\sqrt{x^2-a} -a\log\left|x+\sqrt{x^2-a}\right| \right],

and similarly for the y-term.

Thus a complete integral is

\begin{aligned} z={}& \frac{x^2+y^2}{2}\\ &\pm\frac12 \left[ x\sqrt{x^2-a} -a\log\left|x+\sqrt{x^2-a}\right| \right]\\ &\pm\frac12 \left[ y\sqrt{y^2-(1-a)} -(1-a)\log\left|y+\sqrt{y^2-(1-a)}\right| \right] +C, \end{aligned}

where a and C are arbitrary constants and either consistent square-root branch may be chosen on the domain considered.

Final Answer:

\boxed{ \begin{aligned} z={}& \frac{x^2+y^2}{2} \pm\frac12\!\left[ x\sqrt{x^2-a} -a\log\left|x+\sqrt{x^2-a}\right| \right]\\ &\pm\frac12\!\left[ y\sqrt{y^2-1+a} -(1-a)\log\left|y+\sqrt{y^2-1+a}\right| \right] +C. \end{aligned}}

Solution:

Algorithm:

1. Start.
2. Set N=2.
3. Print N.
4. Set N=N+2.
5. If N\le50, return to Step 3.
6. Otherwise stop.

Flowchart:

           ┌─────────┐
           │  START  │
           └────┬────┘
                │
                ▼
           ┌─────────┐
           │  N = 2  │
           └────┬────┘
                │
                ▼
          ┌────────────┐
          │  Print N   │
          └─────┬──────┘
                │
                ▼
          ┌────────────┐
          │  N = N + 2 │
          └─────┬──────┘
                │
                ▼
            ┌────────┐
            │ N ≤ 50?│
            └──┬──┬──┘
             Yes  No
              │    │
              │    ▼
              │ ┌────────┐
              │ │  STOP  │
              │ └────────┘
              │
              └──────────────► Print N

The output is

2,4,6,\ldots,48,50.

Final Answer:

The above flowchart prints all even integers from 1 to 50.

Solution:

Cayley's Theorem — Statement: Every group G is isomorphic to a group of permutations. In particular, if |G|=n, then G is isomorphic to a subgroup of the symmetric group S_n.

Proof.

For each a\in G, define a map

L_a:G\to G

by

L_a(x)=ax.

Step 1: L_a is a permutation of G.

If

L_a(x_1)=L_a(x_2),

then

ax_1=ax_2.

By left cancellation,

x_1=x_2.

Thus L_a is one-to-one.

For any y\in G, choose

x=a^{-1}y.

Then

L_a(x) = a(a^{-1}y) = y.

Hence L_a is onto. Therefore it is a permutation of G.

Step 2: Define the representation.

Let \operatorname{Sym}(G) denote the group of all permutations of the set G. Define

\Phi:G\to\operatorname{Sym}(G)

by

\Phi(a)=L_a.

Step 3: \Phi is a homomorphism.

For a,b\in G,

\begin{aligned} \Phi(ab)(x) &=L_{ab}(x)\\ &=(ab)x\\ &=a(bx)\\ &=L_a(L_b(x)). \end{aligned}

Therefore

\Phi(ab)=\Phi(a)\circ\Phi(b).

Step 4: \Phi is one-to-one.

Suppose

\Phi(a)=\Phi(b).

Then L_a=L_b. Evaluating both at the identity e,

a=L_a(e)=L_b(e)=b.

Hence \Phi is injective.

Therefore G is isomorphic to the subgroup \Phi(G) of \operatorname{Sym}(G). If |G|=n, then \operatorname{Sym}(G)\cong S_n.

Final Answer:

\boxed{G\cong\Phi(G)\le\operatorname{Sym}(G).}

Hence every group is isomorphic to a permutation group.

Solution:

Let I=(\mathbb Z,+) and

H=\{mx:x\in\mathbb Z\}=m\mathbb Z.

Since \mathbb Z is abelian, every subgroup is normal. Hence the quotient group \mathbb Z/H is defined.

For m>0, the distinct cosets are

H,\quad 1+H,\quad 2+H,\quad \ldots,\quad (m-1)+H.

Indeed, two integers r,s determine the same coset exactly when

r-s\in m\mathbb Z,

i.e. when r\equiv s\pmod m.

Thus

\mathbb Z/m\mathbb Z = \{H,1+H,\ldots,(m-1)+H\}.

For m=5:

H=5\mathbb Z,

and the quotient group contains

H,\quad1+H,\quad2+H,\quad3+H,\quad4+H.

The group operation is coset addition:

(a+H)+(b+H)=(a+b)+H.

+	H	1+H	2+H	3+H	4+H
H	H	1+H	2+H	3+H	4+H
1+H	1+H	2+H	3+H	4+H	H
2+H	2+H	3+H	4+H	H	1+H
3+H	3+H	4+H	H	1+H	2+H
4+H	4+H	H	1+H	2+H	3+H

Final Answer:

\boxed{ \mathbb Z/m\mathbb Z = \{H,1+H,\ldots,(m-1)+H\}.}

For m=5, the above table is the composition table of the quotient group.

Solution:

Let f be bounded on [a,b].

Darboux criterion: The function f is Riemann integrable on [a,b] if and only if, for every \varepsilon>0, there exists a partition P such that

0\le U(f,P)-L(f,P)<\varepsilon. [/katex]<p><strong>Necessity.</strong></p><p>Assume [katex]f is Riemann integrable. Let its integral be I. Then

\underline{\int_a^b}f = \overline{\int_a^b}f = I.

Given \varepsilon>0, by the definition of the upper integral there is a partition P_1 such that

U(f,P_1)<I+\frac{\varepsilon}{2}. [/katex]<p>Similarly, by the definition of the lower integral there is a partition [katex]P_2 such that

L(f,P_2)>I-\frac{\varepsilon}{2}.

Let P=P_1\cup P_2 be their common refinement. Refinement can only decrease upper sums and increase lower sums, so

U(f,P)\le U(f,P_1),

and

L(f,P)\ge L(f,P_2).

Therefore

\begin{aligned} 0 \le U(f,P)-L(f,P) &< \left(I+\frac{\varepsilon}{2}\right) - \left(I-\frac{\varepsilon}{2}\right)\\ &=\varepsilon. \end{aligned} [/katex]<p><strong>Sufficiency.</strong></p><p>Conversely, suppose that for every [katex]\varepsilon>0 there exists a partition P such that

U(f,P)-L(f,P)<\varepsilon. [/katex]<p>Let</p> [katex display="true"] U^*=\overline{\int_a^b}f, \qquad L^*=\underline{\int_a^b}f.

For every partition P,

L(f,P)\le L^*\le U^*\le U(f,P).

Hence

0\le U^*-L^* \le U(f,P)-L(f,P) <\varepsilon. [/katex]<p>Since this holds for every [katex]\varepsilon>0,

U^*-L^*=0.

Therefore

U^*=L^*,

so f is Riemann integrable.

Final Answer:

\boxed{ f\in R[a,b] \iff \forall\varepsilon>0,\ \exists P: U(f,P)-L(f,P)<\varepsilon.} [/katex]</div> </div><div id="q3b" class="math-qna-wrapper"> <div class="question-box"> <h2>HPAS 2013 Maths Optional Paper-2 Question 3(b)</h2> <p>For the functions:[katex] f(x)=x , g(x)=e^{x} ,
then verify the second mean value theorem in the interval [-1, 1].

Solution:

Second Mean Value Theorem (Bonnet form): If f is monotonic on [a,b] and g is integrable and of one sign there, then there exists c\in[a,b] such that

\int_a^b f(x)g(x)\,dx = f(a)\int_a^c g(x)\,dx + f(b)\int_c^b g(x)\,dx.

Here

f(x)=x,\qquad g(x)=e^x,\qquad [a,b]=[-1,1].

The function f(x)=x is increasing, and g(x)=e^x>0. Hence the theorem applies.

First compute the left-hand side:

\begin{aligned} \int_{-1}^{1}xe^x\,dx &= [e^x(x-1)]_{-1}^{1}\\ &=0-\left(-\frac{2}{e}\right)\\ &=\frac{2}{e}. \end{aligned}

The theorem requires some c\in[-1,1] such that

\frac2e = (-1)\int_{-1}^{c}e^x\,dx + (1)\int_c^1e^x\,dx.

Evaluate the two integrals:

\int_{-1}^{c}e^x\,dx = e^c-e^{-1}, \int_c^1e^x\,dx = e-e^c.

Hence

\frac2e = -(e^c-e^{-1})+(e-e^c) = e+\frac1e-2e^c.

Therefore

2e^c = e-\frac1e,

so

e^c = \frac{e-e^{-1}}2 = \sinh1.

Thus

c=\log(\sinh1)\approx0.1614.

Since -1<c<1[/katex], such a point lies inside the required interval.</p><p><strong>Final Answer:</strong></p> <p>The Second Mean Value Theorem is verified, with</p> [katex display="true"] \boxed{c=\log(\sinh1)\approx0.1614.}

Solution:

The series is

x^2 +\frac{2^2}{3\cdot4}x^4 +\frac{2^2\cdot4^2}{3\cdot4\cdot5\cdot6}x^6 +\cdots.

Let u_n denote the term containing x^{2n}. Then

u_n = \left[ \prod_{k=1}^{n-1} \frac{(2k)^2}{(2k+1)(2k+2)} \right] x^{2n}.

Hence

\left|\frac{u_{n+1}}{u_n}\right| = |x|^2 \frac{(2n)^2}{(2n+1)(2n+2)}.

Taking the limit,

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

Therefore, by the ratio test:

• the series converges absolutely if |x|<1[/katex],</li> <li>the series diverges if [katex]|x|>1.

Endpoint |x|=1.

At x=\pm1, all even powers equal 1. Let the positive coefficient be a_n. Then

\frac{a_n}{a_{n+1}} = \frac{(2n+1)(2n+2)}{4n^2}.

Use Raabe's test:

\begin{aligned} \lim_{n\to\infty} n\left(\frac{a_n}{a_{n+1}}-1\right) &= \lim_{n\to\infty} n \frac{(2n+1)(2n+2)-4n^2}{4n^2}\\ &= \lim_{n\to\infty} \frac{6n+2}{4n}\\ &=\frac32. \end{aligned}

Since 3/2>1, the endpoint series converges. Thus both x=1 and x=-1 are included.

Final Answer:

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

Solution:

A metric space is complete if every Cauchy sequence in it converges to a point of the space.

Let \{x_n\} be an arbitrary Cauchy sequence in (\mathbb R,d), where

d(x,y)=|x-y|.

From the general theorem proved for Cauchy sequences, every Cauchy sequence is bounded. Hence there exists M>0 such that

|x_n|\le M

for every n.

By the Bolzano-Weierstrass theorem, every bounded sequence of real numbers has a convergent subsequence. Therefore there is a subsequence \{x_{n_k}\} and some L\in\mathbb R such that

x_{n_k}\to L.

We now show that the whole Cauchy sequence converges to L.

Given \varepsilon>0, because \{x_n\} is Cauchy there exists N_1 such that

m,n\ge N_1 \quad\Rightarrow\quad |x_m-x_n|<\frac{\varepsilon}{2}. [/katex]<p>Since [katex]x_{n_k}\to L, there exists k so large that

n_k\ge N_1

and

|x_{n_k}-L|<\frac{\varepsilon}{2}. [/katex]<p>Then for every [katex]n\ge N_1,

\begin{aligned} |x_n-L| &\le |x_n-x_{n_k}| + |x_{n_k}-L|\\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2}\\ &=\varepsilon. \end{aligned} [/katex]<p>Thus [katex]x_n\to L, and L\in\mathbb R.

Final Answer:

\boxed{(\mathbb R,|\cdot|)\text{ is a complete metric space.}}

Solution:

Let

f(z)=u+iv.

We are given

u-v=e^x(\cos y-\sin y). \tag{1}

Consider the analytic function

(1+i)f(z).

Since

(1+i)(u+iv) = (u-v)+i(u+v),

its real part is exactly u-v.

Also

e^z=e^x(\cos y+i\sin y).

Therefore

(1+i)e^z = e^x \left[ (\cos y-\sin y) +i(\sin y+\cos y) \right].

Its real part is

e^x(\cos y-\sin y),

which is the given expression.

Hence the two analytic functions (1+i)f(z) and (1+i)e^z have the same real part. Their difference is therefore an analytic function with zero real part, and so must be a purely imaginary constant:

(1+i)f(z)-(1+i)e^z=iC,

where C is real.

Thus

f(z) = e^z+\frac{iC}{1+i}.

Now

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

Absorbing the factor 1/2 into the arbitrary real constant,

f(z)=e^z+C(1+i).

Final Answer:

\boxed{f(z)=e^z+C(1+i),\qquad C\in\mathbb R.}

Solution:

Let

w=f(z)=u(x,y)+iv(x,y), \qquad z=x+iy.

For a regular conformal transformation, assume u and v possess continuous first partial derivatives. Conformality means that infinitesimal angles and their orientation are preserved.

Consider at a point z_0 two infinitesimal directions: the positive x-direction and the positive y-direction. Their angle is \pi/2.

The corresponding tangent vectors in the w-plane are

w_x=u_x+iv_x

and

w_y=u_y+iv_y.

Because the map is conformal and orientation-preserving, rotation of the original tangent by +\pi/2 must correspond locally to multiplication of its image tangent by i. Hence

w_y=i\,w_x.

Therefore

u_y+iv_y = i(u_x+iv_x) = -v_x+i u_x.

Equating real and imaginary parts gives

u_y=-v_x,

and

v_y=u_x.

Thus the Cauchy-Riemann equations hold:

u_x=v_y, \qquad u_y=-v_x.

Because the first partial derivatives are continuous, the Cauchy-Riemann equations imply that f is complex differentiable at every point of D. Therefore it is analytic in D.

Moreover, conformality requires the local scale factor to be non-zero, so f'(z)\ne0 throughout the conformal region.

Final Answer:

\boxed{\text{A regular orientation-preserving conformal transformation }w=f(z)\text{ is analytic in }D.}

Solution:

Let

F(x,y,z,p,q) = pxy+pq+qy-yz=0.

For Charpit's method,

F_p=xy+q,\qquad F_q=p+y, F_x=yp,\qquad F_y=xp+q-z,\qquad F_z=-y.

One of Charpit's equations is

\frac{dp}{-F_x-pF_z}.

But

-F_x-pF_z = -yp-p(-y) =0.

Therefore

dp=0,

and so

p=a,

where a is an arbitrary constant.

Since p=z_x=a, integrate with respect to x:

z=ax+\phi(y),

where \phi is at this stage a function of y.

Then

q=z_y=\phi'(y).

Substitute in the PDE:

axy + a\phi' + y\phi' = y(ax+\phi).

Cancel axy:

(a+y)\phi'=y\phi.

Thus

\frac{\phi'}{\phi} = \frac{y}{y+a} = 1-\frac{a}{y+a}.

Integrating,

\log|\phi| = y-a\log|y+a|+\log|b|,

where b is another arbitrary constant.

Hence

\phi(y) = b\,e^y(y+a)^{-a}.

Therefore a complete integral is

z = ax+ b\,e^y(y+a)^{-a}.

Verification:

p=a, \qquad q= \phi'(y) = \frac{y}{y+a}\phi.

Then

pq+qy = q(a+y) = y\phi,

so

pxy+pq+qy = axy+y\phi = y(ax+\phi) = yz.

Final Answer:

\boxed{ z=ax+b\,e^y(y+a)^{-a}, }

where a,b are arbitrary constants.

Solution:

Using the standard notation

p=z_x,\quad q=z_y,\quad r=z_{xx},\quad s=z_{xy},

the equation is

pq=x(ps-qr). \tag{1}

Assume first that p\ne0. Put

u=\frac{q}{p}.

Differentiate with respect to x:

u_x = \frac{p q_x-q p_x}{p^2} = \frac{ps-qr}{p^2}.

From (1),

pq=x p^2u_x.

Since q=pu, the left side is p^2u. Hence

p^2u=xp^2u_x.

Cancel p^2:

u=xu_x.

For fixed y,

\frac{du}{u}=\frac{dx}{x}.

Therefore

u=xA(y),

where A(y) is arbitrary. Thus

\frac{q}{p}=xA(y).

Write

A(y)=\frac{G'(y)}{G(y)}

for an arbitrary non-zero differentiable function G. Then

q = x\frac{G'}{G}p.

Hence

z_y - x\frac{G'}{G}z_x=0. \tag{2}

The characteristic equations of (2) are

\frac{dx}{-xG'/G} = \frac{dy}{1} = \frac{dz}{0}.

From the first two ratios,

\frac{dx}{dy} = -x\frac{G'}{G}.

Thus

\frac{d}{dy}(xG)=0.

Hence xG(y) is constant along a characteristic, while z is also constant. Therefore

z=F(xG(y)),

where F is an arbitrary function.

Verification.

Let u=xG(y), z=F(u). Then

p=F'G, \qquad q=F'xG', r=F''G^2,

and

s=F''xGG'+F'G'.

Therefore

ps-qr = F'^2GG'.

Hence

x(ps-qr) = xF'^2GG' = pq.

So the family satisfies the PDE.

The degenerate branch p=0 gives the additional special solutions z=\Psi(y).

Final Answer:

\boxed{z=F(xG(y))}

where F and G are arbitrary differentiable functions; additionally, z=\Psi(y) is the degenerate p=0 branch.

Solution:

Given Rodrigues' formula

L_n(x) = \frac{e^x}{n!} \frac{d^n}{dx^n} (e^{-x}x^n).

We require

\mathcal L\{L_n(x);p\} = \int_0^\infty e^{-px}L_n(x)\,dx, \qquad p>1.

Substitute Rodrigues' formula:

\mathcal L\{L_n;p\} = \frac1{n!} \int_0^\infty e^{-(p-1)x} \frac{d^n}{dx^n}(e^{-x}x^n)\,dx.

Let

h(x)=e^{-x}x^n.

Integrate by parts once:

\int_0^\infty e^{-(p-1)x}h^{(n)}(x)\,dx = \left[e^{-(p-1)x}h^{(n-1)}(x)\right]_0^\infty + (p-1) \int_0^\infty e^{-(p-1)x}h^{(n-1)}(x)\,dx.

The boundary term vanishes. Repeating integration by parts n times gives

\int_0^\infty e^{-(p-1)x}h^{(n)}(x)\,dx = (p-1)^n \int_0^\infty e^{-(p-1)x}h(x)\,dx.

Therefore

\mathcal L\{L_n;p\} = \frac{(p-1)^n}{n!} \int_0^\infty e^{-(p-1)x}e^{-x}x^n\,dx.

Hence

\mathcal L\{L_n;p\} = \frac{(p-1)^n}{n!} \int_0^\infty e^{-px}x^n\,dx.

Using

\int_0^\infty e^{-px}x^n\,dx = \frac{n!}{p^{n+1}},

we obtain

Final Answer:

\boxed{ \mathcal L\{L_n(x);p\} = \frac{(p-1)^n}{p^{n+1}}, \qquad p>1.}

Solution:

The functional is

I[y,z] = \int_0^{\pi/2} F(y,z,y',z')\,dx,

where

F=(y')^2+(z')^2+2yz.

For two dependent functions, the Euler-Lagrange equations are

\frac{\partial F}{\partial y} - \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) =0,

and

\frac{\partial F}{\partial z} - \frac{d}{dx} \left( \frac{\partial F}{\partial z'} \right) =0.

For y,

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

Therefore

2z-2y''=0,

so

y''=z. \tag{1}

Similarly,

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

which gives

z''=y. \tag{2}

Differentiate (1) twice and use (2):

y^{(4)}=z''=y.

Thus

y^{(4)}-y=0.

The auxiliary equation is

m^4-1=0 = (m-1)(m+1)(m^2+1).

Hence

y=A\cosh x+B\sinh x+C\cos x+D\sin x.

Using z=y'',

z=A\cosh x+B\sinh x-C\cos x-D\sin x.

Apply y(0)=0:

A+C=0. \tag{3}

Apply z(0)=0:

A-C=0. \tag{4}

Equations (3) and (4) give

A=C=0.

Therefore

y=B\sinh x+D\sin x, z=B\sinh x-D\sin x.

At x=\pi/2,

B\sinh\frac{\pi}{2}+D=-1,

and

B\sinh\frac{\pi}{2}-D=1.

Add the equations:

2B\sinh\frac{\pi}{2}=0,

so

B=0.

Then

D=-1.

Final Answer:

\boxed{y(x)=-\sin x,\qquad z(x)=\sin x.}

Solution:

Let

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

Then

f'(x)=2x-5.

Newton-Raphson formula:

x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)} = x_n- \frac{x_n^2-5x_n+2}{2x_n-5}.

The quadratic has one root between 0 and 1 and another between 4 and 5.

Smaller root: take x_0=0.

n	x_n
0	0.000000000
1	0.400000000
2	0.438095238
3	0.438447157
4	0.438447187

Hence the smaller root is

x\approx0.4384.

Larger root: take x_0=5.

n	x_n
0	5.000000000
1	4.600000000
2	4.561904762
3	4.561552843
4	4.561552813

Hence the larger root is

x\approx4.5616.

Final Answer:

\boxed{x_1\approx0.4384,\qquad x_2\approx4.5616.}

Solution:

We have

I=\int_0^1\frac{dx}{1+x^2}.

Since both Simpson's 1/3 rule and Simpson's 3/8 rule are required, choose n=6 equal subintervals. Then

h=\frac{1-0}{6}=\frac16.

Let

f(x)=\frac1{1+x^2}.

i	x_i	f_i=f(x_i)
0	0	1.000000000
1	1/6	0.972972973
2	1/3	0.900000000
3	1/2	0.800000000
4	2/3	0.692307692
5	5/6	0.590163934
6	1	0.500000000

1. Simpson's 1/3 rule.

The formula is

I \approx \frac h3 \left[ f_0+f_6 +4(f_1+f_3+f_5) +2(f_2+f_4) \right].

Substitute the tabulated values:

\begin{aligned} I_{1/3} &= \frac1{18} \big[ 1+0.5\\ &\quad+4(0.972972973+0.8+0.590163934)\\ &\quad+2(0.9+0.692307692) \big]\\ &\approx0.785397945. \end{aligned}

But

\int_0^1\frac{dx}{1+x^2} = [\tan^{-1}x]_0^1 = \frac{\pi}{4}.

Hence

\pi \approx 4I_{1/3} = 3.141591781.

2. Simpson's 3/8 rule.

For n=6,

I \approx \frac{3h}{8} \left[ f_0+f_6 +3(f_1+f_2+f_4+f_5) +2f_3 \right].

Thus

\begin{aligned} I_{3/8} &= \frac1{16} \big[ 1+0.5\\ &\quad+3(0.972972973+0.9+0.692307692+0.590163934)\\ &\quad+2(0.8) \big]\\ &\approx0.785395862. \end{aligned}

Therefore

\pi \approx 4I_{3/8} = 3.141583450.

Final Answer:

Using Simpson's 1/3 rule,

\boxed{I\approx0.785397945,\qquad \pi\approx3.141591781.}

Using Simpson's 3/8 rule,

\boxed{I\approx0.785395862,\qquad \pi\approx3.141583450.}