Taylorâ€™s series and Maclaurin's series expansion of a function which are infinitely differentiable.

**Series Expansions**

Taylorâ€™s
series and Maclaurin's series expansion of a function which are infinitely
differentiable.

** **

**Theorem
7.5**

Let *f* ( *x*) be a function infinitely
differentiable at *x* = *a* .
Then *f* ( *x*) can be expanded as a series, in an interval ( *x* âˆ’ *a*, *x* + *a*) , of the form

**(b) Maclaurinâ€™s series**

If *a* = 0 , the expansion takes the form

The
series expansion of *f* ( *x*) , in powers of ( *x* âˆ’ *a*) , be given by

Substituting
*x* = *a*
gives *A*_{0} =
*f* (*a*) . Differentiation of (7) gives

Substituting
*x* = *a*
gives *A*_{1} =
*f* â€²(*a*)
. Differentiation of (8) gives

Substituting
*x* = *a*
gives *A*_{2} =
*f* â€²â€²(*a*)
/ 2!. Differentiation of (9) gives

Differentiation
of (10) (*k* âˆ’
3) times gives

Substituting
*x* = *a*
gives

*A _{k}* =

which completes the proof of the
theorem.

In order
to expand a function around a point say *x*
=
*a* , equivalently in powers of ( *x* âˆ’ *a*)
we need to differentiate the given function as many times as the required
powers and evaluate at *x* =
*a* . This will give the value for the
coefficients of the required powers of ( *x*
âˆ’
*a*) .

** **

**Example 7.30**

Expand
log(1+
*x*) as a Maclaurinâ€™s series upto 4
non-zero terms for â€“1 < *x* â‰¤ 1.

**Solution**

Let *f (x)* = log(1+ x) , then the Maclaurin series
of *f (x)* is *f (x)* = , where, various derivatives of
the function *f (x)* evaluated at* x *= 0 are given below:

Substituting
the values and on simplification we get the required expansion of the function
given by,

** **

Expand
tan *x* in ascending powers of *x* upto 5^{th} power for â€“ Ï€/2 <* x *< Ï€/2.

**Solution**

Let *f* (*x*)
=
tan *x* . Then the Mclaurin series of *f* ( *x*)
is

Various
derivatives of the function *f* ( *x*) evaluated at *x* = 0 are given below:

Now,

* f*
â€²(*x*) = *d*/*dx* (tan* x *) = sec^{2}(x)

*fâ€™*â€²(*x*) = *d*/*dx*
(sec^{2} (* x *)) = 2 sec* x *â‹… sec* x *â‹…
tan* x *= 2 sec^{2}*x *â‹… tan *x*

*f* â€²â€²â€²(*x*) = *d*/*dx*
(2sec^{2} (* x *). tan *x*) = 2 sec^{2}*(x *) â‹… sec^{2}*x *+ tan* x *â‹… 4 sec* x *â‹…
sec* x *â‹… tan *x*

= 2 sec^{4} *x* + 4 sec^{2} *x*
â‹… tan^{2} *x*

*f* ^{(iv)}(*x*) = 8sec^{3} *x* â‹… sec *x* â‹… tan *x* + 4 sec^{2} *x* â‹… 2 tan *x* â‹… sec^{2}
*x* + 8sec *x* â‹… sec *x* â‹… tan *x* â‹… tan^{2}
*x*

= 16 sec^{4}
*x* tan *x* + 8sec^{2} *x*
â‹… tan^{3} *x*

*f*^{(v)}( *x*) = 16 sec^{4} *x* â‹… sec^{2} *x* + 64 sec^{3} *x* â‹… sec *x* â‹… tan *x* â‹… tan *x* + 8sec^{2} *x*
â‹… 3 tan^{2} *x* â‹… sec^{2} *x *+16 sec *x* â‹… sec *x* â‹… tan *x* â‹… tan^{3} *x*

= 16 sec^{6} *x* + 88sec^{4} *x*
â‹… tan^{2} *x* + 16 sec^{2} *x* â‹… tan^{4} *x* .

Substituting
the values and on simplification we get the required expansion of the function
as

tan* x *=* x *+ 1/3* x*^{3} + 2/15 x^{5} + .. â€¦ ; âˆ’ Ï€/2 <* x *< Ï€/2

Write
the Taylor series expansion of 1/ *x* about
*x* = 2 by finding the first three
non-zero terms.

**Solution**

Let *f* (*x*)
= 1/*x*, then the Taylor series of *f *(*
x*)* *is

Various
derivatives of the function *f *(*
x*)* *evaluated at* x *=* *2*
*are given below.

Substituting
these values, we get the required expansion of the function as

Tags : Differential Calculus | Mathematics , 12th Maths : UNIT 7 : Applications of Differential Calculus

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12th Maths : UNIT 7 : Applications of Differential Calculus : Series Expansions: Maclaurinâ€™s and Taylorâ€™s Series | Differential Calculus | Mathematics

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