Power Series:
Most of us will be familiar with the concept of the power series which is stated as :
$$\sum_{n=0}^\infty c_n x^n $$ $$ f(x) = c_0 + c_1x + c_2 x^2 + ... $$
The sum of a power series can be represented by a function with terms involving each power of “x” all the way to infinity.
But we may want to know about all of these coefficients(c0,c1,c2,...). Now that's about differentiation we are ready to expand our concepts for this kind of series which will be known as Taylor Series.
So let's talk about this taylor expansion
Defining Taylor Series:
Consider we are having a power series in such format
$$\sum_{n=0}^\infty c_n(x-a)^n $$ $$ f(x) = c_0 + c_1 (x-a) + c_2 (x-a)^2 + c_3 (x-a)^3 +... $$
It may be described in the form of a function as given but here we have to solve for the coefficients like c0,c1,c2,...
It actually won't be as tricky as you think given the difference x-a in each of these terms.
If we find f(a) all of these binomials become 0 and therefore all of the terms become equal to 0 except for the first coefficient here 0.
$$ f(a) = c_0 + c_1 (a-a) + c_2 (a-a)^2 + c_3 (a-a)^3 +... $$
Therefore,
$$ f(a) = c_0 $$
We get the first coefficient, now how do we get the next one ?
This will be done by differentiation, Let's take the derivative of the function:
$$ f'(x) = 0 + c_1(1) + 2c_2 (x-a) + 3c_3 (x-a)^2 + 4c_4 (x-a)^3 +... $$
Since the derivative of any constant is zero, so c0 will go away and the first derivative of the will takes place
Now, if we plug in x=a or finding f'(a),it will be
$$ f'(a) = 0 + c_1(1) + 2c_2 (a-a) + 3c_3 (a-a)^2 + 4c_4 (a-a)^3 +... $$ $$ f'(a) = c_1 $$
So, the only thing left is the c1 coefficient.
Now, repeat the same process for the second derivative, this will be
$$ f''(x) = 2c_2 + 6c_3(x-a) + 12c_4(x-a)^2 +... $$
Again plugin x=a, we will isolate c2 coefficient
$$ f''(a) = 2c_2 + 6c_3(a-a) + 12c_4(a-a)^2 +... $$ $$ f''(a) = 2c_2 $$
Or,
$$ c_2 = \frac{f''(a)}{2} $$
Now, we might observe a pattern by taking the third derivative and plugging in x=a, we will get the c3 coefficient.
$$ f'''(x) = 6c_3 + 24c_4(x-a)+... $$
So, at x=a
$$ f'''(x) = 6c_3 $$ $$ c_3 = \frac{f'''(x)}{6} $$
Or,
$$ c_3 = \frac{f'''(x)}{3!} $$
Similarly, putting x=a in fourth derivative will give
$$ c_4 = \frac{f^4(x)}{4!} $$
In this way, We can set up an expression to find the nth coefficient of the taylor expansion.
$$ C_n = \frac{f^{(n)} (a)}{n!} \;\;\;\;\; When \; |x-a|= R$$
Where R = Radius of convergence for the series
Taylor Series Formula:
We have a function of Power Series in the form of:
$$\sum_{n=0}^\infty c_n(x-a)^n $$ $$ f(x) = c_0 + c_1 (x-a) + c_2 (x-a)^2 + c_3 (x-a)^3 +... $$
If we generate a new function of the same form but substitute the above formula in for Cn such that the terms in the series look like
$$ f(x) = f(a) + \frac{f'(a)}{1!} (x-a) + \frac{f''(a)}{2!} (x-a)^2 + \frac{f'''(a)}{3!} (x-a)^3 + ... $$
Taylor expansion will expressed as:
$$\sum_{n=0}^\infty \frac{f^{(n)} (a)}{n!} (x-a)^n $$
Where,
n! = Factorial of n
a = Real or Complex Number
f (n)(a) = nth derivative of function f at x=a
We have just generated the Taylor series formula of the function f at x=a which can also be described as f about a or f centered at a.
Well, A Taylor Series is an expanded form of function into an infinite sum of terms, where each term has a larger exponent like x, x2, x3, etc. Taylor Series has a large number of application in physics and mathematics.
Taylor Series Example:
Solving the examples is the best way to polish our concept in any concerned topic. Let's discuss here how to solve taylor expansion in simple example. Then you will be able to solve many taylor series examples by yourself.
Find the Taylor Series of the function: y= ex about x=0.
Solution:
For Taylor Series, we have to find general formula of function for f (n)(a)
For function y= ex,
$$ f^{(n)} (x) = e^x \;\;\;\;\;\;\;\; n=0,1,2,3,... $$
At x = 0,
$$ f^{(n)} (0) = e^0 = 1 \;\;\;\;\;\;\;\; n=0,1,2,3,... $$
Hence, The Taylor Series for y = ex about x = 0 is :
$$ e^x = \sum_{n=0}^\infty \frac{1}{n!} x^n $$
Or,
$$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \;\;\;\;\; (Taylor Series)$$