• Work out the Taylor’s Series of the following;

Question 1:

\(f(x)\quad =\quad \frac { 1 }{ 1\quad -\quad x }\)

\(f(x)\quad =\quad \frac { 1 }{ 1\quad -\quad x }\) , \(f(0)\quad\) = 1

\({ f }^{ \prime }(x)\quad =\quad \frac { 1 }{ { (1-x) }^{ 2 } }\) , \({ f }^{ \prime }(0)\) = 1

\({ f }^{ \prime \prime }(x)\quad =\quad \frac { 2 }{ { (1-x) }^{ 3 } }\) , \({ f }^{ \prime \prime }(x)\quad\) = 2

\({ f }^{ \prime \prime \prime}(x)\quad =\quad \frac { 6 }{ { (1-x) }^{ 4 } }\) , \({ f }^{ \prime \prime \prime}(0)\quad\) = 6

\({ f }^{(4)}(x)\quad =\quad \frac { 24 }{ { (1-x) }^{ 5 } }\) , \({ f }^{(4)}(0)\quad\) = 24

\({ f }^{(5)}(x)\quad =\quad \frac { 120 }{ { (1-x) }^{ 6 } }\) , \({ f }^{(5)}(0)\quad\) = 120

Recall from Taylor Series,

\(P(x)\quad =\quad f(0)\quad +\quad f\prime (0)(x)\quad +\quad f\prime \prime (0)\frac { { x }^{ 2 } }{ 2! } \quad +\quad { f }^{ (3) }(0)\frac { { x }^{ 3 } }{ 3! } \quad +\quad ....+\quad { f }^{ (n) }(0)\frac { { x }^{ n } }{ n! }\)

\(\quad \frac { 1 }{ 1\quad -\quad x }\) = \(1\quad +\quad x\quad +\quad \frac { 2{ x }^{ 2 } }{ 2! } \quad +\quad \frac { 6{ x }^{ 3 } }{ 3! } \quad +\quad \frac { 24{ x }^{ 4 } }{ 4! } \quad +\quad \frac { 120{ x }^{ 5 } }{ 5! } + . . .\)

= \(1\quad +\quad x\quad +\quad { x }^{ 2 }\quad +\quad { x }^{ 3 }\quad +\quad x^{ 4 }\quad +\quad { x }^{ 5 }\quad +\quad .\quad .\quad .\quad .\quad .\)

Question 2:

\(f(x)\quad =\quad { e }^{ x }\)

Recall from Taylor Series,

\(P(x)\quad =\quad f(0)\quad +\quad f\prime (0)(x)\quad +\quad f\prime \prime (0)\frac { { x }^{ 2 } }{ 2! } \quad +\quad { f }^{ (3) }(0)\frac { { x }^{ 3 } }{ 3! } \quad +\quad ....+\quad { f }^{ (n) }(0)\frac { { x }^{ n } }{ n! }\)

Let,

\(f(x)\quad =\quad { e }^{ x }\), \(f(0)\quad =\quad { e }^{ 0 }\) = 1

\({ f }^{ \prime }(x)\quad =\quad { e }^{ x }\), \({ f }^{ \prime }(0)\quad =\quad { e }^{ 0 }\) = 1

\({ f }^{ \prime \prime }(x)\quad =\quad { e }^{ x }\) , \({ f }^{ \prime \prime }(0)\quad =\quad { e }^{ 0 }\) = 1

\({ f }^{ \prime \prime \prime }(x)\quad =\quad { e }^{ x }\) , \({ f }^{ \prime \prime \prime }(0)\quad =\quad { e }^{ 0 }\) = 1

\({ f }^{ (4) }(x)\quad =\quad { e }^{ x }\) , \({ f }^{ (4) }(0)\quad =\quad { e }^{ 0 }\) = 1

\({ f }^{ (5) }(x)\quad =\quad { e }^{ x }\) , \({ f }^{ (5) }(0)\quad =\quad { e }^{ 0 }\) = 1

\({ f }^{ (n) }(x)\quad =\quad { e }^{ x }\) , \({ f }^{ (n) }(0)\quad =\quad { e }^{ 0 }\) = 1

Therefore,

\({ e }^{ x }\quad =\quad 1\quad +\quad x\quad +\frac { { x }^{ 2 } }{ 2! } \quad +\quad \frac { { x }^{ 3 } }{ 3! } \quad +\frac { { x }^{ 4 } }{ 4! } \quad +\quad \frac { { x }^{ 5 } }{ 5! } \quad +.\quad .\quad .\quad .\quad +\frac { { x }^{ n } }{ n! }\)

Queestion 3:

\(f(x)\quad =\quad ln(1+x)\)

let \(\quad u=\quad 1+x,\quad du/dx\quad =\quad 1\\ dx\quad =\quad \frac { 1 }{ u } du,\quad therefore,\quad { f }^{ \prime }(x)\quad =\quad \frac { 1 }{ u } .1\\ where\quad u\quad =\quad (1+x)\\ { f }^{ \prime }(x)\quad =\quad \frac { 1 }{ 1+x } \\\)

\({ f }_{ 0 }(x)\quad =\quad ln(1+x)\quad =\quad 0\\ { f\prime }_{ 0 }(x)\quad =\quad ln(1+x)\quad =\quad 1\\ { f\prime \prime }_{ 0 }(x)\quad =\quad ln(1+x)\quad =\quad -1\\ { f\prime \prime \prime }_{ 0 }(x)\quad =\quad ln(1+x)\quad =\quad 2\\ { { f }^{ (4) } }_{ 0 }(x)\quad =\quad ln(1+x)\quad =\quad -6\\ { { f }^{ (5) } }_{ 0 }(x)\quad =\quad ln(1+x)\quad =\quad 24\)

Therefore, it Taylor’s Series is;

\(x\quad -\quad \frac { { x }^{ 2 } }{ 2 } \quad +\quad \frac { { x }^{ 3 } }{ 3 } \quad -\quad \frac { { x }^{ 4 } }{ 4 } \quad +\quad \frac { { x }^{ 5 } }{ 5 } \quad +\quad .\quad .\quad .+\)