Taylor polynomial is a fractional sum of a Taylor series.Taylor series is a demonstration of functions as an infinite sum of conditions which are calculated from the values of it's derivatives at a single point. Taylor series can be regarded as the limit of the taylor polynomials. If the series is centered at nil, then the series is also called as Maclaurin series.

For a function f which is differentiable any number of times, the series of expansion of f(a + h) about a point a, very very large n is given by $f(a + h) = f(a) + \frac{f'(a)}{1!}h + \frac{f''(a)}{2!}h^{2} + ...... + \frac{f^{n}(a)}{n!}h^{n} + ..... $ is known as Taylor's Series. We use the above formula to solve the taylor series online.

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Taylor Polynomial is given by the formula given below:

P_{n}(x) = f(a) + (x - a) f'(a) / 1! + (x - a)^{2} f"(a) / 2! + ...... + (x - a)^{n} f^{n}(a) / n!

For a function y = f(x) by a polynomial near x = a of degree n:

P_{n}(x) = a_{0} + a_{1}(x - a) + a_{2}(x - a)^{2} + ...... + a_{n}(x - a)^{n}

If a = 1, then this Taylor polynomial is known as Maciaurin polynomial.

The answer is not 10%. In fact, since you will get 1 dollar for each 0.90 dollar you invest. You get back 1/0.90 $\approx$ $1.11 for each dollar that you invest, giving you an 11% profit.

Now, in another way, we were being offered a discount of x. In this case, *x *= 0.10.

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

We would like to find an easier way to compute approximation to f(x), to see why f(0.90) $\approx$
0.11, and to see quickly what will happen for other discount rates. In
fact, we shall see very soon that a good approximation is f(x) $\approx$ x + x^{2}.

If x = 0.10, f(x) $\approx$ 0.10 + 0.01 = 0.11

f(x) can be approximated by a polynomial. This is nice because polynomials are the only easiest functions to compute and manipulate.

The second degree polynomial of the function f(x) at any point x = a is as follows:

$f(x) \approx P_{2}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^{2}$

Here, the first and second derivative of P

For better and better approximation, we use more derivatives and getting higher degree polynomials. The Taylor's polynomial of n

$ P_{n}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^{2} + \frac{f'''(a)}{3 \times 2}(x - a)^{3} + \frac{f''''(a)}{4 \times 3 \times 2}(x - a)^{4} + ...... + \frac{f^{n}(a)}{n \times (n - 1) \times (n - 2) \times ...... 3 \times 2}(x - a)^{n}$

The n

The third degree Taylor polynomial is $P_{n}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^{2} + \frac{f'''(a)}{3 \times 2}(x - a)^{3}$

We have f(x) = sin x and given that a = 0.

Now, f'(x) = cos x, f''(x) = - sin x and f'''(x) = - cos x

Now, the second degree Taylor polynomial is

$f(x) \approx P_{2}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^{2}$

= $\sin 0 + \cos 0 (x - 0) + \frac{\sin 0}{2}(x - 0)^{2}$

= 0 + x + 0 = x

The third degree Taylor polynomial is

$ P_{n}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^{2} + \frac{f'''(a)}{3 \times 2}(x - a)^{3}$

= $\sin 0 + \cos 0 (x - 0) - \frac{\sin 0}{2}(x - 0)^{2} - \frac{\cos 0}{3 \times 2}(x - 0)^{3}$

= $x - \frac{x^{3}}{3.2}$

Now, f'(x) = cos x, f''(x) = - sin x and f'''(x) = - cos x

Now, the second degree Taylor polynomial is

$f(x) \approx P_{2}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^{2}$

= $\sin 0 + \cos 0 (x - 0) + \frac{\sin 0}{2}(x - 0)^{2}$

= 0 + x + 0 = x

The third degree Taylor polynomial is

$ P_{n}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^{2} + \frac{f'''(a)}{3 \times 2}(x - a)^{3}$

= $\sin 0 + \cos 0 (x - 0) - \frac{\sin 0}{2}(x - 0)^{2} - \frac{\cos 0}{3 \times 2}(x - 0)^{3}$

= $x - \frac{x^{3}}{3.2}$

$f(x)= f(a)+ f'(a)(x-a)+ \frac{f''(a)}{2!}(x-a)^{2}+ \frac{f'''(a)}{3!}(x-a)^{3}+ .....+ \frac{f^{n}(a)}{n!}(x-a)^{n}+ R_{n}$

f(x) = T

Where, $R_{n} = \int_{a}^{x}\frac{f^{n+1}(t)}{n!}(x-t)^{n}dt$ is known as the remainder term or error term of the Taylor's polynomial.

T

and n! is called as n factorial which is equal to n! = n(n - 1)(n - 2)(n - 3) ....... 3 . 2 . 1

Then,

f(x) = T

Given below are some of the examples on taylor polynomials.

$f(x) = \sin x f(\pi) = \sin \frac{\pi}{2} = 1$

$f'(x) = \cos x f'(\frac{\pi}{2}) = \cos \frac{\pi}{2} = 0$

$f''(x) = -\sin x f''(\frac{\pi}{2}) = -\sin \frac{\pi}{2} = -1$

$f'''(x) = -\cos x f'''(\frac{\pi}{2}) = -\cos \frac{\pi}{2} = 0$

$f(x) = \sin x = f(\frac{\pi}{2}) + f'(\frac{\frac{\pi}{2}}{1!})(x - \frac{\pi}{2}) + f''(\frac{\frac{\pi}{2}}{2!})(x - \frac{\pi}{2})^{2}$

$\Rightarrow f(x) = 1 - 0(x - \frac{\pi}{2}) + \frac{-1}{2!}(x - \frac{\pi}{2})^{2}$

= $1 - \frac{1}{2!}(x - \frac{\pi}{2})^{2} + \frac{1}{4!}(x - \frac{\pi}{2})^{4} - .....$

$f'(x) = \cos x f'(\frac{\pi}{2}) = \cos \frac{\pi}{2} = 0$

$f''(x) = -\sin x f''(\frac{\pi}{2}) = -\sin \frac{\pi}{2} = -1$

$f'''(x) = -\cos x f'''(\frac{\pi}{2}) = -\cos \frac{\pi}{2} = 0$

$f(x) = \sin x = f(\frac{\pi}{2}) + f'(\frac{\frac{\pi}{2}}{1!})(x - \frac{\pi}{2}) + f''(\frac{\frac{\pi}{2}}{2!})(x - \frac{\pi}{2})^{2}$

$\Rightarrow f(x) = 1 - 0(x - \frac{\pi}{2}) + \frac{-1}{2!}(x - \frac{\pi}{2})^{2}$

= $1 - \frac{1}{2!}(x - \frac{\pi}{2})^{2} + \frac{1}{4!}(x - \frac{\pi}{2})^{4} - .....$

The function is $f(x) = \log x f(1) = \log 1 = 0$

$f'(x) = \frac{1}{x}$, then $f'(1) = \frac{1}{1} = 1$

$f''(x) = \frac{-1}{x^{2}}$, then $f''(1) = \frac{-1}{1} = -1$

$f'''(x) = \frac{2}{x^{3}}$, then $f'''(1) = \frac{2}{1} = 2$

$f(x) = \log x = f(1) + \frac{f'(1)}{1!}(x - 1) + \frac{f''(1)}{2!}(x - 1)^{2} + \frac{f'''(1)}{3!}(x - 1)^{3} + .....$

= $0 + \frac{1}{1!}(x - 1) + \frac{-1}{2!}(x - 1)^{2} + \frac{2}{3!}(x - 1)^{3} + ......$

= $0 + (x - 1) - \frac{(x - 1)^{2}}{2} + \frac{(x - 1)^{3}}{3} - ........$

$\log x = (x - 1) - \frac{(x - 1)^{2}}{2} + \frac{(x - 1)^{3}}{3} - ........$

$f'(x) = \frac{1}{x}$, then $f'(1) = \frac{1}{1} = 1$

$f''(x) = \frac{-1}{x^{2}}$, then $f''(1) = \frac{-1}{1} = -1$

$f'''(x) = \frac{2}{x^{3}}$, then $f'''(1) = \frac{2}{1} = 2$

$f(x) = \log x = f(1) + \frac{f'(1)}{1!}(x - 1) + \frac{f''(1)}{2!}(x - 1)^{2} + \frac{f'''(1)}{3!}(x - 1)^{3} + .....$

= $0 + \frac{1}{1!}(x - 1) + \frac{-1}{2!}(x - 1)^{2} + \frac{2}{3!}(x - 1)^{3} + ......$

= $0 + (x - 1) - \frac{(x - 1)^{2}}{2} + \frac{(x - 1)^{3}}{3} - ........$

$\log x = (x - 1) - \frac{(x - 1)^{2}}{2} + \frac{(x - 1)^{3}}{3} - ........$

The function is f(x) = e^{x}. f(1) = e

f'(x) = e^{x}

f'(1) = e

f''(x) = e^{x}

f''(1) = e

f'''(x) = e^{x}

f'''(1) = e

$f(x) = e^{x} = f(1) + \frac{f'(1)}{1!}(x - 1) + \frac{f''(1)}{2!}(x - 1)^{2} + ......$

$ \Rightarrow e^{x} = e + \frac{e}{1!}(x - 1) + \frac{e}{2!}(x - 1)^{2} + ......$

= $e + e(x - 1) + \frac{e}{2}(x - 1)^{2} + ......$

f'(x) = e

f'(1) = e

f''(x) = e

f''(1) = e

f'''(x) = e

f'''(1) = e

$f(x) = e^{x} = f(1) + \frac{f'(1)}{1!}(x - 1) + \frac{f''(1)}{2!}(x - 1)^{2} + ......$

$ \Rightarrow e^{x} = e + \frac{e}{1!}(x - 1) + \frac{e}{2!}(x - 1)^{2} + ......$

= $e + e(x - 1) + \frac{e}{2}(x - 1)^{2} + ......$

The function is $f(x) = \frac{1}{x + 1}$.

Then, $f(2) = \frac{1}{2 + 1} = \frac{1}{3}$

$f'(x) = \frac{-1}{(x + 1)^{2}}$, then $f''(2) = \frac{-1}{(2 + 1)^{2}} = \frac{-1}{9}$

$f''(x) = \frac{1.2}{(x + 1)^{3}} = \frac{2!}{(x + 1)^{3}}$, then $f''(2) = \frac{2!}{(2 + 1)^{3}} = \frac{2!}{27}$

$f(x) = \frac{1}{1 + x} = f(2) + \frac{f'(2)}{1!}(x - 2) + \frac{f''(2)}{2!}(x - 2)^{2} + ......$

$\Rightarrow \frac{1}{1 + x} = \frac{1}{3} + \frac{\frac{-1}{9}}{1}(x - 2) + \frac{\frac{2!}{27}}{2!}(x - 2)^{2} + .....$

= $\frac{1}{3} - \frac{1}{9}(x - 2) + \frac{1}{27}(x - 2)^{2} - ......$

Then, $f(2) = \frac{1}{2 + 1} = \frac{1}{3}$

$f'(x) = \frac{-1}{(x + 1)^{2}}$, then $f''(2) = \frac{-1}{(2 + 1)^{2}} = \frac{-1}{9}$

$f''(x) = \frac{1.2}{(x + 1)^{3}} = \frac{2!}{(x + 1)^{3}}$, then $f''(2) = \frac{2!}{(2 + 1)^{3}} = \frac{2!}{27}$

$f(x) = \frac{1}{1 + x} = f(2) + \frac{f'(2)}{1!}(x - 2) + \frac{f''(2)}{2!}(x - 2)^{2} + ......$

$\Rightarrow \frac{1}{1 + x} = \frac{1}{3} + \frac{\frac{-1}{9}}{1}(x - 2) + \frac{\frac{2!}{27}}{2!}(x - 2)^{2} + .....$

= $\frac{1}{3} - \frac{1}{9}(x - 2) + \frac{1}{27}(x - 2)^{2} - ......$

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