Solution:

Given: X is a non-empty set and S(X) denotes the set of all bijections from X onto itself. The binary operation is composition of functions.

Required: Prove that (S(X),\circ) is an abelian group if and only if X has either one element or two elements.

Step 1: First verify that (S(X),\circ) is always a group

Let f,g,h\in S(X). Since the composition of two bijections is again a bijection, f\circ g\in S(X). Hence closure holds.

Composition of functions is associative, therefore

(f\circ g)\circ h=f\circ(g\circ h).

The identity map I_X:X\to X, defined by I_X(x)=x for every x\in X, belongs to S(X) and satisfies

f\circ I_X=I_X\circ f=f.

Every f\in S(X) is bijective, so its inverse f^{-1}:X\to X exists and is also a bijection. Thus f^{-1}\in S(X) and

f\circ f^{-1}=f^{-1}\circ f=I_X.

Hence (S(X),\circ) is a group for every non-empty set X. It remains to determine exactly when this group is abelian.

---

Step 2: If |X|=1, then S(X) is abelian

Suppose X=\{a\}. There is only one bijection of X onto itself, namely the identity map:

S(X)=\{I_X\}.

A group containing only one element is trivially abelian.

Step 3: If |X|=2, then S(X) is abelian

Let X=\{a,b\}. There are exactly two bijections:

1. the identity I_X,
2. the transposition \tau which interchanges a and b.

Thus

S(X)=\{I_X,\tau\},\qquad \tau^2=I_X.

This is a cyclic group of order 2, generated by \tau. Every cyclic group is abelian. Hence S(X) is abelian.

---

Step 4: If X has at least three distinct elements, then S(X) is not abelian

Assume that X contains three distinct elements a,b,c. Define two permutations \sigma,\tau\in S(X) as follows:

\sigma interchanges a and b and fixes every other element; \tau interchanges b and c and fixes every other element.

In cycle notation,

\sigma=(ab),\qquad \tau=(bc).

Now compare (\sigma\circ\tau)(a) and (\tau\circ\sigma)(a).

Since \tau fixes a,

(\sigma\circ\tau)(a)=\sigma(\tau(a))=\sigma(a)=b.

On the other hand,

(\tau\circ\sigma)(a)=\tau(\sigma(a))=\tau(b)=c.

Because b\ne c, we have

\sigma\circ\tau\ne\tau\circ\sigma.

Therefore S(X) is not abelian whenever X contains at least three elements. This argument works for both finite and infinite sets.

Final Answer:

(S(X),\circ)\text{ is abelian }\Longleftrightarrow |X|=1\text{ or }|X|=2.

Solution:

Definition: For a non-empty subset A of a metric space (X,d), the distance of a point x\in X from A is defined by

d(x,A)=\inf\{d(x,a):a\in A\}.

The closure \overline A consists of all points x\in X such that every open ball centered at x meets A. Thus

x\in\overline A\Longleftrightarrow B(x,\varepsilon)\cap A\ne\varnothing\quad\text{for every }\varepsilon>0.

We prove both inclusions.

Part I: Show that \overline A\subseteq\{x\in X:d(x,A)=0\}

Let x\in\overline A. Then, by the characterization of closure, for every \varepsilon>0,

B(x,\varepsilon)\cap A\ne\varnothing.

Therefore there exists some a_\varepsilon\in A such that

d(x,a_\varepsilon)<\varepsilon.

Since d(x,A) is the infimum of all numbers d(x,a) with a\in A,

0\le d(x,A)\le d(x,a_\varepsilon)<\varepsilon.

This inequality holds for every \varepsilon>0. The only non-negative real number smaller than every positive \varepsilon is zero. Hence

d(x,A)=0.

Thus

\overline A\subseteq\{x\in X:d(x,A)=0\}.

---

Part II: Show that \{x\in X:d(x,A)=0\}\subseteq\overline A

Let x\in X satisfy

d(x,A)=0.

We must prove that every open ball centered at x intersects A. Let \varepsilon>0 be arbitrary.

Because

\inf\{d(x,a):a\in A\}=0,

there must exist an element a\in A for which

d(x,a)<\varepsilon.

If no such a existed, then d(x,a)\ge\varepsilon for every a\in A, which would imply d(x,A)\ge\varepsilon>0, contradicting d(x,A)=0.

Thus a\in B(x,\varepsilon)\cap A, and hence

B(x,\varepsilon)\cap A\ne\varnothing.

Since \varepsilon>0 was arbitrary, x\in\overline A. Therefore

\{x\in X:d(x,A)=0\}\subseteq\overline A.

Combining the two inclusions gives the required equality.

Final Answer:

\boxed{\overline A=\{x\in X:d(x,A)=0\}}.

Solution:

Given: X=\sqrt{-1}=i. We have to determine X^X=i^i.

Relevant result: Complex logarithm and complex powers

For a non-zero complex number z, the multivalued logarithm is

\log z=\ln|z|+i(\arg z+2\pi n),\qquad n\in\mathbb Z.

A complex power is defined by

z^w=e^{w\log z}.

Step 1: Find all logarithms of i

For i,

|i|=1,\qquad \arg i=\frac{\pi}{2}+2\pi n,\quad n\in\mathbb Z.

Therefore

\log i=\ln 1+i\left(\frac{\pi}{2}+2\pi n\right)=i\left(\frac{\pi}{2}+2\pi n\right).

Step 2: Compute i^i

Using i^i=e^{i\log i},

i^i = e^{\,i\cdot i\left(\frac{\pi}{2}+2\pi n\right)} = e^{-\left(\frac{\pi}{2}+2\pi n\right)}, \qquad n\in\mathbb Z,

because i^2=-1. Thus every value of i^i is real and positive.

Principal value:

The principal argument of i is \pi/2, corresponding to n=0. Hence the principal value is

i^i=e^{-\pi/2}.

Final Answer:

All values are

\boxed{i^i=e^{-(\pi/2+2\pi n)},\quad n\in\mathbb Z}.

The principal value is

\boxed{e^{-\pi/2}}.

Solution:

Let

a_n=(-1)^n+\frac1n.

We determine the limits of the even and odd subsequences.

Even subsequence

Put n=2k. Then

a_{2k}=(-1)^{2k}+\frac1{2k} =1+\frac1{2k}.

Therefore

\lim_{k\to\infty}a_{2k}=1.

Odd subsequence

Put n=2k+1. Then

a_{2k+1}=(-1)^{2k+1}+\frac1{2k+1} =-1+\frac1{2k+1}.

Therefore

\lim_{k\to\infty}a_{2k+1}=-1.

Hence the sequence has exactly two subsequential limit points, namely 1 and -1. The largest subsequential limit is the limit superior and the smallest subsequential limit is the limit inferior.

Equivalently, if

s_N=\sup\{a_n:n\ge N\},\qquad i_N=\inf\{a_n:n\ge N\},

then the largest terms in a sufficiently late tail occur on the even subsequence and approach 1, while the odd terms approach -1. Thus

\lim_{N\to\infty}s_N=1,\qquad \lim_{N\to\infty}i_N=-1.

Final Answer:

\boxed{\limsup_{n\to\infty}\left((-1)^n+\frac1n\right)=1}

and

\boxed{\liminf_{n\to\infty}\left((-1)^n+\frac1n\right)=-1}.

Solution:

Relevant theorem: Lagrange’s theorem

If H is a finite subgroup of a finite group G, then the order of H divides the order of G.

Here G contains subgroups of orders 45 and 75. Therefore

45\mid |G|\qquad\text{and}\qquad 75\mid |G|.

Hence |G| must be a common multiple of 45 and 75.

Factorize:

45=3^2\cdot5,\qquad 75=3\cdot5^2.

Therefore

\operatorname{lcm}(45,75) = 3^2\cdot5^2 = 9\cdot25 = 225.

Thus

225\mid |G|.

The positive multiples of 225 are

225,\ 450,\ 675,\ldots

But the question gives

|G|<400.

Therefore the only possible value is 225.

Final Answer:

\boxed{|G|=225}.

Solution:

Given:

|G|=75,\qquad |H|=45,\qquad |\ker f|=|K|=15.

Relevant theorem: First Isomorphism Theorem

If f:G\to H is a group homomorphism, then

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

For finite groups, taking orders gives

|\operatorname{Im}f|=\frac{|G|}{|\ker f|}.

Substituting the given values,

|f(G)| = |\operatorname{Im}f| = \frac{75}{15} = 5.

This is consistent with Lagrange’s theorem because f(G) is a subgroup of H, and indeed 5\mid45.

Final Answer:

\boxed{|f(G)|=5}.

Solution:

Consider

I(s)=\int_0^\infty e^{-sx}\,dx.

For real s

By the definition of an improper integral,

I(s) = \lim_{R\to\infty}\int_0^R e^{-sx}\,dx.

Case 1: s>0

For s\ne0,

\int e^{-sx}\,dx=-\frac1s e^{-sx}.

Hence

\int_0^R e^{-sx}\,dx = \left[-\frac1s e^{-sx}\right]_0^R = -\frac{e^{-sR}}s+\frac1s = \frac{1-e^{-sR}}s.

If s>0, then e^{-sR}\to0 as R\to\infty. Therefore

I(s)=\frac1s.

Hence the integral converges.

Case 2: s=0

Then the integrand is 1, so

\int_0^R1\,dx=R\to\infty.

Thus the integral diverges.

Case 3: s<0

Write s=-a, where a>0. Then

e^{-sx}=e^{ax},

which grows exponentially, and

\int_0^R e^{ax}\,dx = \frac{e^{aR}-1}{a}\to\infty.

Thus the integral diverges.

Complex parameter: If s is allowed to be complex, then |e^{-sx}|=e^{-(\operatorname{Re}s)x}, so the corresponding condition is \operatorname{Re}s>0.

Final Answer:

For real s,

\boxed{\int_0^\infty e^{-sx}\,dx\text{ converges iff }s>0,\quad\text{and then equals }\frac1s.}

If s is complex, the convergence condition is \boxed{\operatorname{Re}s>0}.

Solution:

Relevant Laplace-transform formula

For \alpha>-1,

\mathcal L\{x^\alpha\} = \int_0^\infty e^{-sx}x^\alpha\,dx = \frac{\Gamma(\alpha+1)}{s^{\alpha+1}}, \qquad s>0.

Here

\alpha=\frac72.

Therefore

\mathcal L\left\{x^{7/2}\right\} = \frac{\Gamma(9/2)}{s^{9/2}}.

Now evaluate \Gamma(9/2) by repeated use of \Gamma(r+1)=r\Gamma(r) and \Gamma(1/2)=\sqrt\pi:

\Gamma\left(\frac92\right) = \frac72\Gamma\left(\frac72\right) = \frac72\cdot\frac52\Gamma\left(\frac52\right) = \frac72\cdot\frac52\cdot\frac32\Gamma\left(\frac32\right) = \frac72\cdot\frac52\cdot\frac32\cdot\frac12\Gamma\left(\frac12\right).

Hence

\Gamma\left(\frac92\right) = \frac{7\cdot5\cdot3\cdot1}{2^4}\sqrt\pi = \frac{105\sqrt\pi}{16}.

Substituting,

\mathcal L\left\{x^{7/2}\right\} = \frac{105\sqrt\pi}{16s^{9/2}}.

Final Answer:

\boxed{\mathcal L\{x^{7/2}\}=\frac{105\sqrt\pi}{16s^{9/2}},\qquad s>0.}

Solution:

Standard interpretation: A right circular cone whose axis is the z-axis and whose vertex is at the origin can be written as

x^2+y^2=z^2\tan^2\alpha,

where \alpha is the semi-vertical angle. The arbitrary parameter to be eliminated is \alpha.

Let

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

Step 1: Differentiate with respect to x

Differentiating

x^2+y^2=z^2\tan^2\alpha

with respect to x, treating \alpha as a constant, gives

2x=2z\,p\,\tan^2\alpha.

Hence

x=zp\tan^2\alpha.\tag{1}

Step 2: Differentiate with respect to y

Similarly,

2y=2z\,q\,\tan^2\alpha,

so

y=zq\tan^2\alpha.\tag{2}

Step 3: Eliminate the parameter

Multiply equation (1) by x and equation (2) by y:

x^2=xzp\tan^2\alpha, y^2=yzq\tan^2\alpha.

Adding,

x^2+y^2 = z\tan^2\alpha(xp+yq).

But from the original cone equation,

x^2+y^2=z^2\tan^2\alpha.

Therefore, wherever the non-degenerate cone equation applies,

z^2\tan^2\alpha = z\tan^2\alpha(xp+yq).

Cancelling the common factor gives

xp+yq=z.

Remark on the wording: The question does not explicitly state that the vertex is fixed at the origin. If the vertex is allowed to move along the z-axis, a two-parameter family may be written as z=a+b\sqrt{x^2+y^2}. Eliminating a,b from the first derivatives gives the rotational condition yp-xq=0. In the usual textbook interpretation of this question, however, the cone has vertex at the origin and the expected PDE is the one derived above.

Final Answer:

Under the standard origin-vertex interpretation,

\boxed{x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z}.

Solution:

Let

u(x,t)=F(x+kt),

where F is sufficiently differentiable and k\ne0. We have to prove that

u_{xx}=\frac1{k^2}u_{tt}.

Put

\xi=x+kt.

Then u(x,t)=F(\xi).

Differentiate with respect to x

By the chain rule,

u_x = F'(\xi)\frac{\partial\xi}{\partial x}.

Since

\frac{\partial\xi}{\partial x}=1,

we obtain

u_x=F'(\xi).

Differentiating once more with respect to x,

u_{xx} = F''(\xi)\frac{\partial\xi}{\partial x} = F''(\xi). \tag{1}

Differentiate with respect to t

Again by the chain rule,

u_t = F'(\xi)\frac{\partial\xi}{\partial t}.

Since

\frac{\partial\xi}{\partial t}=k,

we get

u_t=kF'(\xi).

Differentiating once more with respect to t,

u_{tt} = kF''(\xi)\frac{\partial\xi}{\partial t} = kF''(\xi)\cdot k = k^2F''(\xi). \tag{2}

From (1), F''(\xi)=u_{xx}. Substituting in (2),

u_{tt}=k^2u_{xx}.

Therefore

u_{xx}=\frac1{k^2}u_{tt}.

Final Answer:

\boxed{F(x+kt)\text{ satisfies }F_{xx}=\frac1{k^2}F_{tt}.}

Hence the required wave equation is established.

Solution:

The function is

f(x)= \begin{cases} c,&0\le x\le c,\\ 2c,&c<x\le1. \end{cases}

Since the break point c lies in the interval [0,1], a valid solution must satisfy

0\le c\le1.

We are given

\int_0^1 f(x)\,dx=\frac7{16}.

Step 1: Split the integral at x=c

\int_0^1 f(x)\,dx = \int_0^c c\,dx + \int_c^1 2c\,dx.

Evaluate each integral:

\int_0^c c\,dx = c[x]_0^c = c(c-0) = c^2,

and

\int_c^1 2c\,dx = 2c[x]_c^1 = 2c(1-c).

Therefore

\int_0^1f(x)\,dx = c^2+2c(1-c) = c^2+2c-2c^2 = 2c-c^2.

Using the given value,

2c-c^2=\frac7{16}.

Rearrange:

c^2-2c+\frac7{16}=0.

Multiplying by 16,

16c^2-32c+7=0.

Using the quadratic formula,

c = \frac{32\pm\sqrt{(-32)^2-4(16)(7)}}{2(16)} = \frac{32\pm\sqrt{1024-448}}{32}. c = \frac{32\pm\sqrt{576}}{32} = \frac{32\pm24}{32}.

Thus

c=\frac{56}{32}=\frac74

or

c=\frac8{32}=\frac14.

But c must lie in [0,1]. Hence 7/4 is inadmissible.

Final Answer:

\boxed{c=\frac14}.

Solution:

Relevant result: Darboux sums under refinement

If a partition Q is a refinement of a partition P, then for every bounded function f,

L(P,f)\le L(Q,f)\le U(Q,f)\le U(P,f).

We verify that the present partitions satisfy the hypothesis and then justify each inequality.

The partitions are

P=\{0,0.01,0.02,\ldots,1\}

and

Q=\{0,0.001,0.002,\ldots,1\}.

Every point of P is also a point of Q, because

0.01j=0.001(10j).

Hence

P\subseteq Q,

so Q is a refinement of P.

Step 1: Show L(P,f)\le L(Q,f)

Consider one interval I=[x_{i-1},x_i] of the coarser partition P. The refinement Q divides I into smaller intervals I_1,\ldots,I_m. Let

m_I=\inf_{x\in I}f(x),\qquad m_j=\inf_{x\in I_j}f(x).

Since I_j\subseteq I,

m_I\le m_j

for every j. Therefore

m_I|I| = m_I\sum_{j=1}^m|I_j| \le \sum_{j=1}^m m_j|I_j|.

Summing this inequality over all intervals of P yields

L(P,f)\le L(Q,f).

Step 2: Show U(Q,f)\le U(P,f)

For the same intervals, let

M_I=\sup_{x\in I}f(x),\qquad M_j=\sup_{x\in I_j}f(x).

Since I_j\subseteq I,

M_j\le M_I.

Hence

\sum_{j=1}^mM_j|I_j| \le M_I\sum_{j=1}^m|I_j| = M_I|I|.

Summing over the intervals of P gives

U(Q,f)\le U(P,f).

Step 3: Show L(Q,f)\le U(Q,f)

For every subinterval J of Q,

\inf_{x\in J}f(x)\le\sup_{x\in J}f(x).

Multiplying by the positive length of J and summing over all subintervals gives

L(Q,f)\le U(Q,f).

Combining the three inequalities,

L(P,f)\le L(Q,f)\le U(Q,f)\le U(P,f).

Final Answer:

\boxed{L(P,f)\le L(Q,f)\le U(Q,f)\le U(P,f)}.

The required result follows because Q is a refinement of P.

Solution:

Definition of an open sphere: In a metric space (X,d), the open sphere (open ball) of radius r>0 centered at a\in X is

B(a,r)=\{x\in X:d(x,a)<r\}.

Here X=\mathbb R^2, the center is (0,0), the radius is 1, and the metric is the ordinary Euclidean metric

d(z_1,z_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.

For a point (x,y)\in\mathbb R^2,

d((x,y),(0,0)) = \sqrt{x^2+y^2}.

The point lies in the open sphere of unit radius precisely when

\sqrt{x^2+y^2}<1.

Squaring both sides,

x^2+y^2<1.

Thus the required open sphere is the entire interior of the unit circle x^2+y^2=1, while the circular boundary itself is not included.

Schematic figure:

                 y
                 ^
              . . . . .
           .             .
         .       inside    .
        .      x²+y²<1      .
-------.---------O-----------.------> x
        .                   .
         .                 .
           .             .
              . . . . .

Dotted circumference: x²+y²=1 (not included)

Final Answer:

\boxed{B((0,0),1)=\{(x,y)\in\mathbb R^2:x^2+y^2<1\}}.

Solution:

Definition of a closed sphere: In a metric space (X,d), the closed sphere (closed ball) of radius r>0 centered at a\in X is

\overline B(a,r)=\{x\in X:d(x,a)\le r\}.

Here the metric is the taxicab (or \ell^1) metric

d(z_1,z_2)=|x_1-x_2|+|y_1-y_2|.

For the center (0,0),

d((x,y),(0,0)) = |x|+|y|.

Therefore a point (x,y) belongs to the closed sphere of unit radius exactly when

|x|+|y|\le1.

Description of the boundary

The boundary is

|x|+|y|=1.

In the four quadrants this becomes four straight-line segments:

x+y=1,\quad -x+y=1,\quad -x-y=1,\quad x-y=1.

These segments join the four vertices

(1,0),\quad(0,1),\quad(-1,0),\quad(0,-1).

Hence the taxicab closed unit sphere is a diamond-shaped region, including its boundary.

Schematic figure:

                 y
                 ^
                 (0,1)
                   /\
                  /  \
                 /    \
        (-1,0)  /  O   \  (1,0)
---------------<-------->---------------> x
                 \      /
                  \    /
                   \  /
                   \/
                 (0,-1)

        Entire diamond, including edges, is included.

Final Answer:

\boxed{\overline B((0,0),1)=\{(x,y)\in\mathbb R^2:|x|+|y|\le1\}}.

Solution:

Consider the positive-term series

\sum_{n=1}^{\infty}\frac1{\sqrt{n!}}.

Let

a_n=\frac1{\sqrt{n!}}.

Use the ratio test

The ratio test states that for a series of positive terms \sum a_n, if

L=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1,

then the series converges.

Compute the ratio:

\frac{a_{n+1}}{a_n} = \frac{1/\sqrt{(n+1)!}}{1/\sqrt{n!}} = \frac{\sqrt{n!}}{\sqrt{(n+1)!}}.

Since

(n+1)!=(n+1)n!,

we obtain

\frac{a_{n+1}}{a_n} = \frac{\sqrt{n!}}{\sqrt{(n+1)n!}} = \frac1{\sqrt{n+1}}.

Therefore

L = \lim_{n\to\infty}\frac1{\sqrt{n+1}} = 0.

Since

0<1,

the ratio test proves that the series is convergent.

Final Answer:

\boxed{\displaystyle\sum_{n=1}^{\infty}\frac1{\sqrt{n!}}\text{ converges.}}

Solution:

Consider

\sum_{n=1}^{\infty}\left(1+\frac1n\right)^{-n^2}.

Let

a_n=\left(1+\frac1n\right)^{-n^2}.

Use Cauchy’s root test

The root test states that if

L=\limsup_{n\to\infty}\sqrt[n]{|a_n|}<1,

then \sum a_n converges absolutely.

Here a_n>0, and

\sqrt[n]{a_n} = \left[\left(1+\frac1n\right)^{-n^2}\right]^{1/n} = \left(1+\frac1n\right)^{-n}.

Use the standard limit

\lim_{n\to\infty}\left(1+\frac1n\right)^n=e.

Taking reciprocals gives

\lim_{n\to\infty}\left(1+\frac1n\right)^{-n} = \frac1e.

Therefore

L=\frac1e.

Since

\frac1e<1,

Cauchy’s root test gives convergence.

Final Answer:

\boxed{\displaystyle\sum_{n=1}^{\infty}\left(1+\frac1n\right)^{-n^2}\text{ converges.}}

Solution:

Trapezoidal rule

Let

I=\int_a^b f(x)\,dx.

Take a uniform step size h such that

n=\frac{b-a}{h}

is a positive integer. Define

x_i=a+ih,\qquad i=0,1,\ldots,n.

The composite trapezoidal rule is

I \approx \frac h2 \left[ f(x_0)+f(x_n) + 2\sum_{i=1}^{n-1}f(x_i) \right].

Algorithm

1. Start.
2. Read a,b,h.
3. Compute n=(b-a)/h.
4. Set S=f(a)+f(b).
5. Set i=1.
6. While i\le n-1, compute x=a+ih and update S=S+2f(x).
7. Increase i by 1 and repeat Step 6 until all interior ordinates have been used.
8. Compute I=(h/2)S.
9. Print I.
10. Stop.

Equivalent pseudocode

START
Read a, b, h
n = (b - a) / h
S = f(a) + f(b)
i = 1

WHILE i <= n - 1
    x = a + i*h
    S = S + 2*f(x)
    i = i + 1
END WHILE

I = (h/2)*S
Print I
STOP

Flow chart

           ┌─────────┐
           │  START  │
           └────┬────┘
                │
                v
      ┌──────────────────┐
      │ Read a, b and h  │
      └────────┬─────────┘
               │
               v
      ┌──────────────────┐
      │ n=(b-a)/h        │
      │ S=f(a)+f(b)      │
      │ i=1              │
      └────────┬─────────┘
               │
               v
          ┌────────────┐
          │ i <= n-1 ?│
          └─────┬──────┘
           Yes  │   No
                │───────────────┐
                v               v
      ┌──────────────────┐   ┌──────────────┐
      │ x=a+i*h          │   │ I=(h/2)*S   │
      │ S=S+2*f(x)       │   └──────┬───────┘
      │ i=i+1            │          │
      └────────┬─────────┘          v
               │              ┌──────────────┐
               └─────────────>│   Print I    │
                    loop      └──────┬───────┘
                                     │
                                     v
                                ┌─────────┐
                                │  STOP   │
                                └─────────┘

Final Answer:

The required trapezoidal algorithm evaluates

\boxed{I\approx\frac h2\left[f(a)+f(b)+2\sum_{i=1}^{n-1}f(a+ih)\right],\quad n=\frac{b-a}{h}.}

Solution:

Euler’s method

For the initial-value problem

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

Euler’s method replaces the curve locally by its tangent. With step size h, the recurrence is

x_{i+1}=x_i+h, y_{i+1}=y_i+h\,f(x_i,y_i).

Suppose the required point is x=x_n and

n=\frac{x_n-x_0}{h},

where n is a non-negative integer.

Algorithm

1. Start.
2. Read the initial values x_0,y_0, the target value x_n, and the step size h.
3. Set x=x_0, y=y_0.
4. While x<x_n, compute the slope m=f(x,y).
5. Compute the next value y_{\text{new}}=y+h\,m.
6. Update x=x+h.
7. Update y=y_{\text{new}}.
8. Repeat until x=x_n.
9. Print y, which is the Euler approximation to y(x_n).
10. Stop.

Equivalent pseudocode

START
Read x0, y0, xn, h
x = x0
y = y0

WHILE x < xn
    y = y + h*f(x,y)
    x = x + h
END WHILE

Print y
STOP

Important implementation note: In an actual program, if y is updated before evaluating f(x,y) for the same step, one must preserve the old x,y values. A safe form is

slope = f(x,y)
ynew  = y + h*slope
x     = x + h
y     = ynew

Flow chart

          ┌─────────┐
          │  START  │
          └────┬────┘
               │
               v
   ┌─────────────────────────┐
   │ Read x0, y0, xn and h   │
   └───────────┬─────────────┘
               │
               v
      ┌────────────────┐
      │ x=x0, y=y0     │
      └───────┬────────┘
              │
              v
         ┌───────────┐
         │ x < xn ? │
         └─────┬─────┘
          Yes  │  No
               │───────────────┐
               v               v
   ┌──────────────────────┐  ┌────────────┐
   │ m=f(x,y)             │  │  Print y   │
   │ ynew=y+h*m           │  └─────┬──────┘
   │ x=x+h                │        │
   │ y=ynew               │        v
   └──────────┬───────────┘   ┌─────────┐
              │               │  STOP   │
              └──────────────>└─────────┘
                   loop

Final Answer:

Euler’s recurrence for finding the value at x=x_n is

\boxed{x_{i+1}=x_i+h,\qquad y_{i+1}=y_i+h\,f(x_i,y_i),\quad i=0,1,\ldots,n-1.}