코딩 수학


sympy ¸ðµâÀ» ºÒ·¯¿À°í, »ç¿ëÇÒ ±âÈ£ º¯¼ö¸¦ ¼±¾ðÇÑ´Ù. ¸ËÇ÷Ը³ ¸ðµâÀ» ºÒ·¯¿Â´Ù.

In [1]:
from sympy import *
init_printing()          

x, y, z = symbols('x y z')
a, b, c, t = symbols('a b c t')

%matplotlib inline 

극한

limit( ) ÇÔ¼ö·Î ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$ \lim _ {x \to 1 } \frac {x^3-1}{x-1} = 3 $$

In [2]:
limit( ( x**3 - 1 ) / ( x - 1 ), x, 1 )
Out[2]:
$$3$$

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$\lim _ {x \to \infty } \frac {1+x}{x} = 1 $$

¹«ÇÑ´ë´Â oo ¿Í °°ÀÌ ¿µ¾î ¼Ò¹®ÀÚ o µÎ °³¸¦ °ãÃļ­ ³ªÅ¸³½´Ù.

In [3]:
limit( ( 1 + x ) / x, x, oo )
Out[3]:
$$1$$

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$\lim _ {x \to 1} \frac {\sqrt{x+3}-2} {x-1} = \frac 1 4 $$

In [4]:
limit( (sqrt(x+3) - 2 ) / (x-1), x, 1 )
Out[4]:
$$\frac{1}{4}$$

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$ \begin{align} \lim _ {x \to 1+0 } \frac 1 {x-1} &= \infty \\ \lim _ {x \to 1-0 } \frac 1 {x-1} &= - \infty \end{align} $$

¿ì±ØÇѰªÀ» ±¸ÇÏ·Á¸é Ãß°¡ÀûÀÎ ÀÎÀÚ·Î '+' ¸¦ ³Ö°í, Á±ØÇѰªÀ» ±¸ÇÏ·Á¸é '-' ¸¦ ³Ö´Â´Ù.

In [5]:
limit( 1/( x - 1 ) / x, x, 1, '+' )
Out[5]:
$$\infty$$
In [6]:
limit( 1/( x - 1 ) / x, x, 1, '-' )
Out[6]:
$$-\infty$$

삼각함수의 극한

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$ \begin{align} \lim _ {x \to \frac {\pi} 2 +0 } \tan x &= -\infty \\ \lim _ {x \to \frac {\pi} 2 -0 } \tan x &= \infty \end{align} $$

In [7]:
limit( tan(x), x, pi/2, '+' )
Out[7]:
$$-\infty$$
In [8]:
limit( tan(x), x, pi/2, '-' )
Out[8]:
$$\infty$$
In [9]:
plot( tan(x), xlim=(-3.14,3.14), ylim=(-6,6) )
Out[9]:
<sympy.plotting.plot.Plot at 0x7442f30>

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$\lim _ {x \to 0 } \frac {\sin x} x = 1 $$

In [10]:
limit( sin(x)/x, x, 0 )
Out[10]:
$$1$$
In [11]:
plot( sin(x)/x, ylim=(-2,2) )
Out[11]:
<sympy.plotting.plot.Plot at 0x86ba330>

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$\lim _ {x \to 0 } \, x \sin \frac 1 x = 0 $$

In [12]:
limit( x * sin(1/x), x, 0 )
Out[12]:
$$0$$
In [13]:
plot( x * sin(1/x), xlim=(-2,2) )
Out[13]:
<sympy.plotting.plot.Plot at 0x87263d0>

지수함수, 로그함수의 극한

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$\lim _ {x \to \infty } \, \frac {2^x-2^{-x}} {2^x+2^{-x}} = 1 $$

In [14]:
limit( ( 2**x - 2**(-x) ) / ( 2**x + 2**(-x) ), x, oo )
Out[14]:
$$1$$

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$\lim _ {x \to \infty } \, \left \{ \log _ 2 (x+1) - \log _ 2 x \right \} = 0 $$

In [15]:
limit( log(x+1,2) - log(x,2), x, oo )
Out[15]:
$$0$$
In [16]:
plot ( log(x,2), log(x+1,2), xlim=(-4,6), ylim=(-4,4))
Out[16]:
<sympy.plotting.plot.Plot at 0x89b76b0>

ÀÚ¿¬·Î±×ÀÇ ¹Ø $$ e = \lim _ {x \to \infty } \, \left ( 1 + \frac 1 x \right ) ^x = \lim _ {x \to 0 } \, \left ( 1 + x \right ) ^{\frac 1 x} $$

In [17]:
limit( ( 1 + 1/x )**x, x, oo )
Out[17]:
$$e$$
In [18]:
N( _ )
Out[18]:
$$2.71828182845905$$

´ÙÀ½ ±ØÇѰªÀ» ±¸ÇÑ´Ù. $$ \lim _ {x \to \infty } \, \left ( 1 + \frac 2 x \right ) ^x = e ^ 2 $$

In [19]:
limit( ( 1 + 2/x )**x, x, oo )
Out[19]:
$$e^{2}$$

Áö¼öÇÔ¼ö¿Í ·Î±×ÇÔ¼öÀÇ ±ØÇÑ

$$ \lim _ {x \to 0} \, \frac {e^x-1} x = 1, \quad \lim _ {x \to 0} \, \frac {\ln (1+x)} x = 1 $$
In [20]:
limit( ( exp(x) - 1 ) / x, x, 0 )
Out[20]:
$$1$$
In [21]:
limit( ln( 1 + x ) / x, x, 0 )
Out[21]:
$$1$$