Solution:

Let

S_n=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}.

For 1\le k\le n, we have

n^2+1\le n^2+k\le n^2+n.

Since all quantities are positive, taking square roots and reciprocals reverses the last inequality:

\frac1{\sqrt{n^2+n}} \le \frac1{\sqrt{n^2+k}} \le \frac1n.

Adding these inequalities for k=1,2,\ldots,n,

\frac{n}{\sqrt{n^2+n}} \le S_n \le \frac{n}{n}=1.

Simplify the lower bound:

\frac{n}{\sqrt{n^2+n}} = \frac{1}{\sqrt{1+\frac1n}}.

Therefore

\frac{1}{\sqrt{1+\frac1n}} \le S_n\le1.

Now

\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac1n}}=1, \qquad \lim_{n\to\infty}1=1.

Hence, by the squeeze theorem,

\lim_{n\to\infty}S_n=1.

Final Answer:

\boxed{\displaystyle \lim_{n\to\infty}\left\{ \frac1{\sqrt{n^2+1}}+\cdots+\frac1{\sqrt{n^2+n}} \right\}=1}

Solution:

We are given

f(z)=|z|^2.

To test differentiability at the origin, use the definition of the complex derivative:

f'(0)=\lim_{z\to0}\frac{f(z)-f(0)}{z-0},

provided the limit exists and is independent of the manner in which z approaches zero.

Since f(0)=0,

\frac{f(z)-f(0)}{z} = \frac{|z|^2}{z}.

For z\ne0, use |z|^2=z\overline z:

\frac{|z|^2}{z} = \frac{z\overline z}{z} = \overline z.

Therefore

f'(0) = \lim_{z\to0}\overline z =0.

The limit is the same along every path because |\overline z|=|z|\to0. Hence the derivative exists at the origin.

Remark: Although f is differentiable at z=0, it is not analytic in any neighbourhood of the origin. Indeed, for u=x^2+y^2, v=0, the Cauchy-Riemann equations are satisfied only at (0,0).

Final Answer:

\boxed{f'(0)=0}

Solution:

We require

L^{-1}\left\{\frac{e^{-1/s}}{s}\right\}.

Expand the exponential in its power series:

e^{-1/s} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!s^n}.

Therefore

\frac{e^{-1/s}}{s} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!s^{n+1}}.

Use the standard Laplace-transform pair

L\{t^n\} = \frac{n!}{s^{n+1}}, \qquad L^{-1}\left\{\frac1{s^{n+1}}\right\} = \frac{t^n}{n!}.

Taking the inverse transform term by term gives

\begin{aligned} L^{-1}\left\{\frac{e^{-1/s}}{s}\right\} &= \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\, \frac{t^n}{n!}\\ &= \sum_{n=0}^{\infty} \frac{(-1)^n t^n}{(n!)^2}. \end{aligned}

Recall the Bessel-function expansion

J_0(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left(\frac{x}{2}\right)^{2n}.

Putting x=2\sqrt t,

J_0(2\sqrt t) = \sum_{n=0}^{\infty} \frac{(-1)^n t^n}{(n!)^2}.

Final Answer:

\boxed{ L^{-1}\left\{\frac{e^{-1/s}}{s}\right\} = J_0(2\sqrt t)}

Solution:

Let G=GL(2,\mathbb R), the group of all nonsingular 2\times2 real matrices. The centre is

Z(G)=\{A\in G:AX=XA\text{ for every }X\in G\}.

Let

A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in Z(G).

Since A commutes with every nonsingular matrix, it must commute with

D=\begin{pmatrix}1&0\\0&2\end{pmatrix}.

Compute

AD= \begin{pmatrix} a&2b\\ c&2d \end{pmatrix}, \qquad DA= \begin{pmatrix} a&b\\ 2c&2d \end{pmatrix}.

Since AD=DA,

2b=b,\qquad c=2c,

hence

b=0,\qquad c=0.

Thus A is diagonal:

A=\begin{pmatrix}a&0\\0&d\end{pmatrix}.

Now A must also commute with

S=\begin{pmatrix}0&1\\1&0\end{pmatrix}.

We obtain

AS= \begin{pmatrix}0&a\\d&0\end{pmatrix}, \qquad SA= \begin{pmatrix}0&d\\a&0\end{pmatrix}.

Therefore a=d. Hence

A=aI.

Since A is nonsingular, a\ne0. Conversely, every nonzero scalar matrix aI commutes with every 2\times2 matrix.

Final Answer:

\boxed{ Z(GL(2,\mathbb R)) = \{\lambda I_2:\lambda\in\mathbb R,\ \lambda\ne0\}}

Solution:

Given

lx+my+nz=\phi(x^2+y^2+z^2),

where \phi is arbitrary. Let

p=\frac{\partial z}{\partial x}, \qquad q=\frac{\partial z}{\partial y}.

Differentiate the given relation with respect to x:

l+np = \phi'(x^2+y^2+z^2)(2x+2zp).

Thus

l+np = 2\phi'(x^2+y^2+z^2)(x+zp). \tag{1}

Differentiate with respect to y:

m+nq = 2\phi'(x^2+y^2+z^2)(y+zq). \tag{2}

Eliminate the arbitrary quantity \phi' between (1) and (2):

\frac{l+np}{x+zp} = \frac{m+nq}{y+zq}.

Cross-multiplication gives

(l+np)(y+zq) = (m+nq)(x+zp).

Expanding both sides,

ly+lzq+npy+nzp q = mx+mzp+nqx+nzpq.

The common term nzpq cancels, leaving

ly-mx+lzq-mzp+nyp-nxq=0.

Final Answer:

\boxed{ (l+np)(y+zq)=(m+nq)(x+zp)}

or equivalently

\boxed{ly-mx+lzq-mzp+nyp-nxq=0}.

Solution:

Note on the printed expression: The question writes the terminal differential as “dy”, but because y=y(x), the integrand contains y', and the limits are values of x, the standard intended functional is

J[y] = \int_0^{\pi/2}\left\{(y')^2-y^2+2xy\right\}\,dx.

Euler-Lagrange equation: For

J[y]=\int F(x,y,y')\,dx,

an extremal satisfies

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

Here

F=(y')^2-y^2+2xy.

Therefore

\frac{\partial F}{\partial y} = -2y+2x,

and

\frac{\partial F}{\partial y'}=2y', \qquad \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)=2y''.

The Euler-Lagrange equation becomes

-2y+2x-2y''=0,

or

y''+y=x.

Step 2: Solve the differential equation.

The complementary function of y''+y=0 is

y_c=C_1\cos x+C_2\sin x.

A particular solution of y''+y=x is y_p=x, since y_p''=0. Hence

y=C_1\cos x+C_2\sin x+x.

Use y(0)=0:

C_1=0.

Use y(\pi/2)=0:

C_2+\frac{\pi}{2}=0,

so

C_2=-\frac{\pi}{2}.

Final Answer:

The required extremal is

\boxed{ y=x-\frac{\pi}{2}\sin x}.

Solution:

The contour is the unit circle

z=e^{it},\qquad 0\le t\le2\pi,

traversed once counter-clockwise.

Branch issue: The function \sqrt z has a branch point at z=0. Hence there is no single-valued analytic branch of \sqrt z on a full neighbourhood of a closed curve that winds once around the origin. Therefore the problem is not completely specified unless a determination of the square root along the parametrized curve is chosen.

Using the standard continuous determination along the parametrization beginning with \sqrt1=1, take

\sqrt{z(t)}=e^{it/2}.

Also

dz=ie^{it}\,dt.

Therefore

\begin{aligned} \int_C\sqrt z\,dz &= \int_0^{2\pi} e^{it/2}\,ie^{it}\,dt\\ &= i\int_0^{2\pi}e^{3it/2}\,dt. \end{aligned}

Now

\int e^{3it/2}\,dt = \frac{2}{3i}e^{3it/2}.

Hence

\begin{aligned} \int_C\sqrt z\,dz &= i\left[ \frac{2}{3i}e^{3it/2} \right]_0^{2\pi}\\ &= \frac23(e^{3\pi i}-1)\\ &= \frac23(-1-1)\\ &=-\frac43. \end{aligned}

If the opposite initial determination \sqrt1=-1 were chosen, the integral would change sign. Thus the branch convention is essential.

Final Answer:

For the standard continuous determination \sqrt{e^{it}}=e^{it/2} along 0\le t\le2\pi,

\boxed{\displaystyle \int_C\sqrt z\,dz=-\frac43}

Solution:

Lagrange’s theorem: If H is a subgroup of a finite group G, then |H| divides |G|.

The converse is not true for arbitrary finite groups. We prove that it is true when G is cyclic.

Let

G=\langle a\rangle, \qquad |G|=n.

Let d be any positive divisor of n. Then

n=dm

for some positive integer m=n/d.

Consider the element

a^m=a^{n/d}.

The order of a^m in a cyclic group of order n is

o(a^m) = \frac{n}{\gcd(n,m)}.

Since m=n/d divides n,

\gcd(n,m)=m=\frac nd.

Therefore

o(a^{n/d}) = \frac{n}{n/d} = d.

Hence the cyclic subgroup

H=\left\langle a^{n/d}\right\rangle

has order exactly d.

Thus for every positive divisor d of |G|, there exists a subgroup of G of order d.

Final Answer:

The converse of Lagrange’s theorem holds in a finite cyclic group: if d\mid |G|, then G has a subgroup of order d, namely

\boxed{\left\langle a^{\,|G|/d}\right\rangle}.

Solution:

We construct an example in the dihedral group

D_4=\langle r,s:r^4=e,\ s^2=e,\ srs=r^{-1}\rangle,

the symmetry group of a square.

Take

H=\langle s\rangle=\{e,s\}, \qquad K=\langle r^2s\rangle=\{e,r^2s\}.

Neither subgroup is normal.

Conjugating s by r,

rsr^{-1}=r^2s\notin H.

Hence H is not normal.

Similarly,

r(r^2s)r^{-1}=s\notin K,

so K is not normal.

Now calculate HK.

\begin{aligned} HK &= \{e,s\}\{e,r^2s\}\\ &= \{e,r^2s,s,sr^2s\}. \end{aligned}

Because r^2 is central in D_4 and s^2=e,

sr^2s=r^2.

Therefore

HK=\{e,r^2,s,r^2s\}.

This set is closed under multiplication and is the Klein four subgroup of D_4. Hence HK is a subgroup even though neither H nor K is normal in D_4.

Final Answer:

An example is

\boxed{ H=\langle s\rangle,\quad K=\langle r^2s\rangle \text{ in }D_4, \quad HK=\{e,r^2,s,r^2s\}\le D_4.}

Solution:

The series is

1+\sum_{n=1}^{\infty}\frac{n^n x^n}{n!}.

Let

a_n=\frac{n^n x^n}{n!}.

Apply the ratio test:

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(n+1)^{n+1}x^{n+1}}{(n+1)!} \cdot \frac{n!}{n^n x^n}\\ &= x\frac{(n+1)^n}{n^n}\\ &= x\left(1+\frac1n\right)^n. \end{aligned}

Therefore

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

Hence the series converges absolutely if

ex<1 \quad\Longleftrightarrow\quad 0\le x<\frac1e,

and diverges if

x>\frac1e.

Endpoint x=1/e.

The terms become

a_n=\frac{n^n}{e^n n!}.

Using Stirling’s formula

n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n,

we get

a_n \sim \frac{1}{\sqrt{2\pi n}}.

Since \sum 1/\sqrt n diverges, the limit-comparison test shows that the series diverges at x=1/e.

Final Answer:

For x\in\mathbb R^+,

\boxed{ \text{convergent for }0\le x<\frac1e, \qquad \text{divergent for }x\ge\frac1e.}

Solution:

We have to evaluate

I=\int_0^\infty\frac{x^2}{x^6+1}\,dx

using the residue theorem.

Consider

f(z)=\frac{z^2}{z^6+1}

and integrate over the standard upper-half-plane semicircular contour of radius R. Since f(z)=O(1/z^4), the contribution from the large semicircle tends to zero as R\to\infty.

The poles satisfy

z^6=-1=e^{i(2k+1)\pi}.

The poles in the upper half-plane are

z_1=e^{i\pi/6}, \qquad z_2=e^{i\pi/2}=i, \qquad z_3=e^{i5\pi/6}.

Each pole is simple. At a root z_k of z^6+1,

\operatorname{Res}(f;z_k) = \frac{z_k^2}{6z_k^5} = \frac{1}{6z_k^3}.

For z_1=e^{i\pi/6},

z_1^3=e^{i\pi/2}=i, \qquad \operatorname{Res}(f;z_1)=\frac1{6i}=-\frac{i}{6}.

For z_2=i,

z_2^3=-i, \qquad \operatorname{Res}(f;z_2)=\frac1{-6i}=\frac{i}{6}.

For z_3=e^{i5\pi/6},

z_3^3=e^{i5\pi/2}=i, \qquad \operatorname{Res}(f;z_3)=-\frac{i}{6}.

Hence the sum of the residues in the upper half-plane is

-\frac{i}{6}+\frac{i}{6}-\frac{i}{6} = -\frac{i}{6}.

By the residue theorem,

\int_{-\infty}^{\infty}\frac{x^2}{x^6+1}\,dx = 2\pi i\left(-\frac{i}{6}\right) = \frac{\pi}{3}.

The integrand is even, so

2I=\frac{\pi}{3}.

Final Answer:

\boxed{\displaystyle \int_0^\infty\frac{x^2}{x^6+1}\,dx =\frac{\pi}{6}}

Solution:

Let the required bilinear transformation be

w=\frac{az+b}{cz+d}.

Since z=-i maps to w=\infty, the denominator must vanish at z=-i. We may therefore choose, without loss of generality,

cz+d=z+i.

Thus write

w=\frac{az+b}{z+i}.

Use z=0\mapsto w=5i:

\frac{b}{i}=5i \quad\Longrightarrow\quad b=5i^2=-5.

Now use z=2i\mapsto w=-i/3:

\frac{2ia-5}{3i} = -\frac{i}{3}.

Multiplying by 3i,

2ia-5 = \left(-\frac i3\right)(3i) =1.

Therefore

2ia=6 \quad\Longrightarrow\quad a=-3i.

Hence

\boxed{ w=\frac{-3iz-5}{z+i}}.

Image of the unit circle.

Solve the transformation for z:

w(z+i)=-3iz-5.

Therefore

z(w+3i)=-(5+iw),

so

z=-\frac{5+iw}{w+3i}.

Let w=u+iv. On the boundary |z|=1,

|5+iw|=|w+3i|.

Now

5+iw=(5-v)+iu,

hence

|5+iw|^2=(5-v)^2+u^2.

Also

|w+3i|^2=u^2+(v+3)^2.

Equating,

(5-v)^2+u^2=u^2+(v+3)^2.

Thus

25-10v+v^2=v^2+6v+9,

which gives

v=1.

Therefore |z|=1 maps to the straight line

\operatorname{Im}w=1.

To determine which side corresponds to |z|<1, take the interior point z=0. It maps to w=5i, for which \operatorname{Im}w=5>1. Hence the open unit disc maps to the upper half-plane above that line.

Final Answer:

\boxed{w=\frac{-3iz-5}{z+i}}

and

\boxed{|z|<1\quad\longmapsto\quad\operatorname{Im}w>1}.

Solution:

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

U(P,f)-L(P,f)<\varepsilon.

Since f is continuous on the compact interval [a,b], it is uniformly continuous. Hence, given \varepsilon>0, there exists \delta>0 such that

|x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)| < \frac{\varepsilon}{b-a}.

Choose a partition

P:\ a=x_0<x_1<\cdots<x_n=b

whose mesh is less than \delta, i.e.

x_i-x_{i-1}<\delta \quad\text{for every }i.

On the subinterval I_i=[x_{i-1},x_i], let

M_i=\sup_{I_i}f, \qquad m_i=\inf_{I_i}f.

Because every pair of points in I_i is less than \delta apart, uniform continuity gives

M_i-m_i \le \frac{\varepsilon}{b-a}.

Now

\begin{aligned} U(P,f)-L(P,f) &= \sum_{i=1}^{n}(M_i-m_i)(x_i-x_{i-1})\\ &\le \frac{\varepsilon}{b-a} \sum_{i=1}^{n}(x_i-x_{i-1})\\ &= \frac{\varepsilon}{b-a}(b-a)\\ &=\varepsilon. \end{aligned}

If desired, choose the uniform-continuity bound with \varepsilon/(2(b-a)) to make the final inequality strictly less than \varepsilon. Therefore the Riemann criterion is satisfied.

Final Answer:

\boxed{\text{Every continuous real-valued function on }[a,b]\text{ is Riemann integrable.}}

Solution:

The Newton-Leibniz identity

\int_a^b f'(x)\,dx=f(b)-f(a)

requires suitable hypotheses; differentiability of f alone does not guarantee that f' is Riemann integrable.

Consider the function on [0,1]

f(x)= \begin{cases} x^2\sin(1/x^2),&x>0,\\ 0,&x=0. \end{cases}

Step 1: Show that f is differentiable at 0.

f'(0) = \lim_{h\to0} \frac{h^2\sin(1/h^2)-0}{h} = \lim_{h\to0} h\sin(1/h^2).

Since

\left|h\sin(1/h^2)\right|\le|h|\to0,

we get

f'(0)=0.

For x>0, differentiate directly:

\begin{aligned} f'(x) &= 2x\sin(1/x^2) + x^2\cos(1/x^2)\left(-\frac{2}{x^3}\right)\\ &= 2x\sin(1/x^2) - \frac{2}{x}\cos(1/x^2). \end{aligned}

Step 2: Show that f' is unbounded near zero.

Choose

x_n=\frac1{\sqrt{2\pi n}}.

Then 1/x_n^2=2\pi n, so

\sin(1/x_n^2)=0, \qquad \cos(1/x_n^2)=1.

Hence

f'(x_n) = -\frac{2}{x_n} = -2\sqrt{2\pi n},

whose magnitude tends to infinity.

Thus f' is unbounded on [0,1]. Every Riemann-integrable function on a closed bounded interval must be bounded. Consequently f' is not Riemann integrable, even though f is differentiable everywhere.

Therefore the displayed identity cannot be asserted merely from differentiability; in this example the left-hand Riemann integral is not defined.

Final Answer:

The function

\boxed{ f(x)=x^2\sin(1/x^2)\ (x>0),\quad f(0)=0}

is differentiable on [0,1], but f' is unbounded and therefore not Riemann integrable. Hence the identity is not valid without additional hypotheses.

Solution:

Define

d^*(x,y)=\frac{d(x,y)}{1+d(x,y)}.

We verify the four metric axioms.

1. Non-negativity.

Since d(x,y)\ge0,

d^*(x,y)\ge0.

2. Identity of indiscernibles.

d^*(x,y)=0 \iff d(x,y)=0 \iff x=y.

3. Symmetry.

Because d(x,y)=d(y,x),

d^*(x,y) = \frac{d(x,y)}{1+d(x,y)} = \frac{d(y,x)}{1+d(y,x)} = d^*(y,x).

4. Triangle inequality.

Let

a=d(x,z),\qquad b=d(z,y).

By the triangle inequality for d,

d(x,y)\le a+b.

The function \phi(t)=t/(1+t) is increasing for t\ge0, because

\phi'(t)=\frac1{(1+t)^2}>0.

Hence

d^*(x,y) \le \frac{a+b}{1+a+b}.

It remains to show

\frac{a+b}{1+a+b} \le \frac a{1+a}+\frac b{1+b}.

Subtract the left side from the right side:

\frac a{1+a}+\frac b{1+b} - \frac{a+b}{1+a+b} = \frac{ab(2+a+b)} {(1+a)(1+b)(1+a+b)} \ge0.

Therefore

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

All metric axioms are satisfied.

Final Answer:

\boxed{ d^*(x,y)=\frac{d(x,y)}{1+d(x,y)} \text{ is a metric on }X.}

Solution:

Let

f(x)=x\sin x+\cos x.

We seek a root of f(x)=0. Since

f(2)=2\sin2+\cos2>0,

while

f(3)=3\sin3+\cos3<0,

there is a positive root in (2,3). Take x_0=3.

Newton-Raphson formula:

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

Differentiate:

\begin{aligned} f'(x) &= (\sin x+x\cos x)-\sin x\\ &= x\cos x. \end{aligned}

Therefore

x_{n+1} = x_n- \frac{x_n\sin x_n+\cos x_n} {x_n\cos x_n}.

Equivalently,

x_{n+1}=x_n-\tan x_n-\frac1{x_n}.

Starting from x_0=3, the successive approximations are:

n	x_n	f(x_n)
0	3.000000000	-0.566632472
1	2.809213210	-0.028641559
2	2.798427302	-0.000108719
3	2.798386046	approximately 0

Thus the iteration has converged.

Final Answer:

\boxed{x\approx2.798386046}

Hence, for example, the root correct to four decimal places is \boxed{2.7984}.

Solution:

The data are equally spaced:

\begin{array}{c|cccc} x&0&1&2&3\\ \hline y&1&0&1&10 \end{array}

Form the forward-difference table:

\begin{array}{c|c|c|c|c} x&y&\Delta y&\Delta^2y&\Delta^3y\\ \hline 0&1&-1&2&6\\ 1&0&1&8&\\ 2&1&9&&\\ 3&10&&& \end{array}

Since the third differences are constant, the interpolating polynomial is cubic.

Newton’s forward interpolation formula with x_0=0 and step h=1 is

y = y_0 +u\Delta y_0 +\frac{u(u-1)}{2!}\Delta^2y_0 +\frac{u(u-1)(u-2)}{3!}\Delta^3y_0,

where

u=\frac{x-x_0}{h}=x.

Substitute the differences:

y = 1-x + \frac{x(x-1)}{2}(2) + \frac{x(x-1)(x-2)}{6}(6).

Thus

y = 1-x+x(x-1)+x(x-1)(x-2).

Expand:

x(x-1)=x^2-x,

and

x(x-1)(x-2) = x(x^2-3x+2) = x^3-3x^2+2x.

Therefore

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

Verification:

y(0)=1,\quad y(1)=0,\quad y(2)=1,\quad y(3)=10.

Final Answer:

\boxed{y(x)=x^3-2x^2+1}.

Solution:

For the quadratic equation

ax^2+bx+c=0,\qquad a\ne0,

the discriminant is

D=b^2-4ac.

The nature of the roots is determined as follows:

• If D>0, there are two distinct real roots.
• If D=0, there is one repeated real root.
• If D<0, there are two complex-conjugate roots.

Algorithm:

1. Start.
2. Read a,b,c.
3. If a=0, report that the equation is not quadratic and stop (or solve the resulting linear equation separately).
4. Compute D=b^2-4ac.
5. If D>0, compute x_1=(-b+\sqrt D)/(2a) and x_2=(-b-\sqrt D)/(2a).
6. If D=0, compute x_1=x_2=-b/(2a).
7. If D<0, put E=\sqrt{-D} and compute x_{1,2}=-b/(2a)\pm iE/(2a).
8. Print the roots.
9. Stop.

Flowchart:

              ┌─────────┐
              │  START  │
              └────┬────┘
                   │
                   ▼
            ┌───────────────┐
            │ Read a, b, c  │
            └──────┬────────┘
                   │
                   ▼
              ┌─────────┐
              │ a = 0 ? │
              └──┬───┬──┘
                Yes   No
                 │     │
                 ▼     ▼
        ┌────────────────┐   ┌─────────────────┐
        │ Not quadratic  │   │ D = b² - 4ac    │
        └───────┬────────┘   └────────┬────────┘
                │                     │
                │                     ▼
                │                ┌─────────┐
                │                │ D > 0 ? │
                │                └──┬───┬──┘
                │                  Yes   No
                │                   │     │
                │                   ▼     ▼
                │          ┌────────────┐  ┌─────────┐
                │          │ Two real   │  │ D = 0 ? │
                │          │ roots      │  └──┬───┬──┘
                │          └─────┬──────┘    Yes   No
                │                │            │     │
                │                │            ▼     ▼
                │                │       ┌────────┐ ┌─────────────┐
                │                │       │ Equal  │ │ Complex     │
                │                │       │ roots  │ │ roots       │
                │                │       └───┬────┘ └──────┬──────┘
                │                └───────────┴──────────────┘
                │                            │
                ▼                            ▼
              ┌────────────────────────────────┐
              │       Print root(s)            │
              └──────────────┬─────────────────┘
                             │
                             ▼
                        ┌────────┐
                        │  STOP  │
                        └────────┘
        

Final Answer:

The above algorithm and flowchart determine the roots of ax^2+bx+c=0 for all three possible signs of the discriminant.

Solution:

The PDE is

(p+q)(px+qy)=1,

where

p=\frac{\partial z}{\partial x}, \qquad q=\frac{\partial z}{\partial y}.

Let

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

Since F does not contain z, Charpit’s equations include

\frac{dp}{-F_x} = \frac{dq}{-F_y}.

Now

F_x=p(p+q), \qquad F_y=q(p+q).

Therefore

\frac{dp}{-p(p+q)} = \frac{dq}{-q(p+q)}.

After cancelling the common factor,

\frac{dp}{p} = \frac{dq}{q}.

Integrating,

\log p-\log q=\log a,

where a is an arbitrary constant. Hence

p=aq.

The relation p=aq means

z_x=a z_y.

Thus we may take z as a function of

s=ax+y.

Write z=Z(s). Then

p=aZ'(s), \qquad q=Z'(s).

Substitute in the PDE:

(aZ'+Z') \{x(aZ')+yZ'\} =1.

Therefore

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

Since s=ax+y,

(a+1)s\left(\frac{dZ}{ds}\right)^2=1.

Thus

\frac{dZ}{ds} = \pm\frac1{\sqrt{(a+1)s}}.

Integrating,

Z = \pm\frac{2\sqrt s}{\sqrt{a+1}}+b,

where b is a second arbitrary constant.

Returning to s=ax+y,

z-b = \pm2\sqrt{\frac{ax+y}{a+1}}.

Squaring gives a convenient branch-independent form:

(z-b)^2 = \frac{4(ax+y)}{a+1}, \qquad a\ne-1.

Verification: From the positive branch,

p=\frac{a}{\sqrt{(a+1)(ax+y)}}, \qquad q=\frac{1}{\sqrt{(a+1)(ax+y)}}.

Then

(p+q)(px+qy) = \frac{a+1}{\sqrt{(a+1)(ax+y)}} \cdot \frac{ax+y}{\sqrt{(a+1)(ax+y)}} =1.

Final Answer:

A complete integral is

\boxed{ (z-b)^2=\frac{4(ax+y)}{a+1}}, \qquad a\ne-1,

where a and b are arbitrary constants.