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# Taylor Polynomial

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 Formula

Taylor Polynomial is given by the formula given below:

Pn(x) = f(a) + (x - a) f'(a) / 1! + (x - a)2 f"(a) / 2! + ...... + (x - a)n fn(a) / n!

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

Pn(x) = a0 + a1(x - a) + a2(x - a)2 + ...... + an(x - a)n

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

### Solved Example

Question: A broker offers you a bond at 90% of their face value. When you cash them in later at their full face value, what percent of profit will you make?
Solution:
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$x + x2. If x = 0.10, f(x)$\approx$0.10 + 0.01 = 0.11 f(x) can be approximated by a polynomial. This is nice because polynomials are the only easiest functions to compute and manipulate. ## Second Degree Taylor Polynomial Back to Top To get the better approximation of any function, we try to approximate that function by the use of quadratic polynomial. The other thing is to find out a polynomial that has the same value as the function has at any point like a, the same derivative at the same point a as well as the same second derivative at that point. The second degree polynomial of the function f(x) at any point x = a is as follows:$f(x) \approx P_{2}(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^{2}$Here, the first and second derivative of P2(x) is the same as the function f(x) does at the point x = a. ## Third Degree Taylor Polynomial Back to Top For better and better approximation, we use more derivatives and getting higher degree polynomials. The Taylor's polynomial of nth degree at any point near by x = a is as follows:$ 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}$The nth derivative of the function f(x) and the nth degree taylor polynomial Pn(x) has the same value at x = a. The third degree Taylor polynomial is$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}$### Solved Example Question: Find the second and third degree taylor polynomial for f(x) = sin x near 0. Solution: We have f(x) = sin x and given that a = 0. Now, f'(x) = cos x, f''(x) = - sin x and f'''(x) = - cos x Now, the second degree Taylor polynomial is$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}$## Taylor Polynomial Remainder Back to Top If we have a function f(x), which is n times differentiable, then we can express f(x) as$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}$f(x) = Tn(x) + Rn(x) Where,$R_{n} = \int_{a}^{x}\frac{f^{n+1}(t)}{n!}(x-t)^{n}dt$is known as the remainder term or error term of the Taylor's polynomial. Tn(x) =$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}$and n! is called as n factorial which is equal to n! = n(n - 1)(n - 2)(n - 3) ....... 3 . 2 . 1 Then, f(x) = Tn(x) + Rn(x) and$\lim_{n \rightarrow \infty}R_{n} = 0$## Taylor Polynomial Examples Back to Top Given below are some of the examples on taylor polynomials. ### Solved Examples Question 1: Find the Taylor's series of expansion for$f(x) = \sin x$about$x = \frac{\pi}{2}$. Solution:$f(x) = \sin x f(\pi) = \sin \frac{\pi}{2} = 1f'(x) = \cos x f'(\frac{\pi}{2}) = \cos \frac{\pi}{2} = 0f''(x) = -\sin x f''(\frac{\pi}{2}) = -\sin \frac{\pi}{2} = -1f'''(x) = -\cos x f'''(\frac{\pi}{2}) = -\cos \frac{\pi}{2} = 0f(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}$=$1 - \frac{1}{2!}(x - \frac{\pi}{2})^{2} + \frac{1}{4!}(x - \frac{\pi}{2})^{4} - .....$Question 2: Find the Taylors' series of expansion for$f(x) = \log x$about$x = 1$Solution: The function is$f(x) = \log x f(1) = \log 1 = 0f'(x) = \frac{1}{x}$, then$f'(1) = \frac{1}{1} = 1f''(x) = \frac{-1}{x^{2}}$, then$f''(1) = \frac{-1}{1} = -1f'''(x) = \frac{2}{x^{3}}$, then$f'''(1) = \frac{2}{1} = 2f(x) = \log x = f(1) + \frac{f'(1)}{1!}(x - 1) + \frac{f''(1)}{2!}(x - 1)^{2} + \frac{f'''(1)}{3!}(x - 1)^{3} + .....$=$0 + \frac{1}{1!}(x - 1) + \frac{-1}{2!}(x - 1)^{2} + \frac{2}{3!}(x - 1)^{3} + ......$=$0 + (x - 1) - \frac{(x - 1)^{2}}{2} + \frac{(x - 1)^{3}}{3} - ........\log x = (x - 1) - \frac{(x - 1)^{2}}{2} + \frac{(x - 1)^{3}}{3} - ........$Question 3: Find the Taylor series for f(x) = ex about the point x = 1 Solution: The function is f(x) = ex. f(1) = e f'(x) = ex f'(1) = e f''(x) = ex f''(1) = e f'''(x) = ex f'''(1) = e$f(x) = e^{x} = f(1) + \frac{f'(1)}{1!}(x - 1) + \frac{f''(1)}{2!}(x - 1)^{2} + ...... \Rightarrow e^{x} = e + \frac{e}{1!}(x - 1) + \frac{e}{2!}(x - 1)^{2} + ......$=$e + e(x - 1) + \frac{e}{2}(x - 1)^{2} + ......$Question 4: Find the Taylor series for the function$f(x) = \frac{1}{x + 1}$about$x = 2$Solution: 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} - ......\$

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