\[ f(x) = \frac{1}{(1-x)} \]
$f(x) = ; f(0) = 1 $
$f’(x) = ; f’(0) = 1 $
$f”(x) = ; f”(0) = 2 $
$f’’‘(x) = ; f’’’(0) = 6 $
$f^4(x) = ; f^4(0) = 24 $
$f^5(x) = ; f^5(0) = 120 $
$1 + x^1 + x^2 + x^3 + x^4 + x^5 + … $
Simplifies to:
\(1 + x + x^2 + x^3 + x^4 + x^5 + ... + x^n\)
In summation form:
$ _n^ x^n$
\(f(x) = e^x\)
Derivatives and evaluation at x = 0:
$f(x) = e^x; f(0) = 1 $
$f’(x) = e^x; f’(0) = 1 $
$f’‘(x) = e^x; f’’(0) = 1 $
$f^3(x) = e^x; f^3(0) = 1 $
$f^4(x) = e^x; f^4(0) = 1 $
$f^5(x) = e^x; f^5(0) = 1 $
Use McClaurin Series formula:
$1 + x^1 + x^2 + x^3 + x^4 + … $
This simplifies to:
\(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + ... + \frac{x^n}{n!}\)
In summation form:
$ _n^ x^n$
\(f(x) = ln(1 + x)\)
Derivatives and evaluation at x = 0.
$f(x) = ln(1 + x); f(0) = 0 $
$f’(x) = ; f’(0) = 1 $
$f’‘(x) = ; f’’(0) = -1 $
\(f^3x = \frac {2}{(x + 1)^2}; f^3(0) = 2\)
\(f^4x = \frac {-6}{(x + 1)^4}; f^4(0) = -6\)
\(f^5x = \frac {24}{(x + 1)^5}; f^5(0) = -24\)
Use McClaurin Series formula:
$0 + x^1 - x^2 + x^3 + x^4 + x^5 + …
Simplifies to:
\(x - \frac {1}{2}x^2 + \frac {1}{3}x^3 - \frac {1}{4}x^4 + \frac {1}{5}x^5 + ... (-1)^{n+1} \frac{1}{n}x^n\)
In summation form:
\(\sum _{n=0}^\infty (-1)^{n+1} \frac{1}{n} x^n\)
\(f(x) = x^\frac{1}{2}\)
Derivative of root x: At x = 0 $ ( ) $ = \(\frac{1}{2} \sqrt{x}; f'(0) = undefined\)
\(f''(x) = - \frac{1}{4x^{\frac{3}{2}}}; f'(0) = undefined\) \(f3(x) = - \frac{3}{8x^{\frac{5}{2}}}; f'(0) = undefined\)