My Math Forum explicit formula for chebychev's method?

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 October 14th, 2012, 08:08 AM #1 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 explicit formula for chebychev's method? chebychev's method is a recursion formula: $T_{n+1}(x)=2xT_{n}(x)-T_{n-1}(x)$ where: $T_0=1$ $T_1=x$ Thus: $T_2=2x$$x$$-1=2x^2-1$ $T_3=2x$$2x^2-1$$-x=4x^3-3x$ $T_4=2x$$4x^3-3x$$-$$2x^2-1$$=8x^4-8x^2+1$ $T_5=2x$$8x^4-8x^2+1$$-$$4x^3-3x$$=16x^5-20x^3+5x$ $T_6=2x$$16x^5-20x^3+5x$$-$$8x^4-8x^2+1$$=32x^6-48x^4+18x^2-1$ But now i would like to know if there exists a explicit way to get for example $T_1_7$ without calculating $T_1$ to $T_1_6$ first
 October 14th, 2012, 12:01 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,206 Thanks: 513 Math Focus: Calculus/ODEs Re: explicit formula for chebychev's method? Begin with the recurrence: $T_{n+1}=2xT_{n}-T_{n-1}$ This is a homogeneous recurrence whose associated auxiliary equation is: $r^2-2xr+1=0$ Application of the quadratic formula to find the characteristic roots: $r=x\pm\sqrt{x^2-1}$ Thus, the closed form is of the form: $T_n(x)=c_1$$x-\sqrt{x^2-1}$$^n+c_2$$x+\sqrt{x^2-1}$$^n$ Use initial conditions to determine the parameters $c_i$: $T_0=c_1+c_2=1$ $T_1=c_1$$x-\sqrt{x^2-1}$$+c_2$$x+\sqrt{x^2-1}$$=x$ Substituting from the first into the second: $c_1$$x-\sqrt{x^2-1}$$+$$1-c_1$$$$x+\sqrt{x^2-1}$$=x$ $c_1x-c_1\sqrt{x^2-1}+x+\sqrt{x^2-1}-c_1x-c_1\sqrt{x^2-1}=x$ $-2c_1\sqrt{x^2-1}+\sqrt{x^2-1}=0$ $c_1=\frac{1}{2}\:\therefore\:c_2=\frac{1}{2}$ thus, the closed form for $T_n(x)$ is: $T_n(x)=\frac{1}{2}$$\(x-\sqrt{x^2-1}$$^n+$$x+\sqrt{x^2-1}$$^n\)$ Using the binomial theorem, we may state: $T_n(x)=\frac{1}{2}$$\sum_{k=0}^n\({n \choose k}x^{n-k}\(-\sqrt{x^2-1}$$^k\)+\sum_{k=0}^n$${n \choose k}x^{n-k}\(\sqrt{x^2-1}$$^k\)\)$ $T_n(x)=\frac{1}{2}$$\sum_{k=0}^n\({n \choose k}x^{n-k}(-1)^k\(\sqrt{x^2-1}$$^k\)+\sum_{k=0}^n$${n \choose k}x^{n-k}\(\sqrt{x^2-1}$$^k\)\)$ $T_n(x)=\frac{1}{2}\sum_{k=0}^n$$\(1+(-1)^k$${n \choose k}x^{n-k}$$\sqrt{x^2-1}$$^k\)$ $T_n(x)=\sum_{k=0}^{\left\lfloor \frac{n}{2} \right\rfloor}$${n \choose 2k}x^{n-2k}\(x^2-1$$^k\)$
 October 15th, 2012, 11:27 AM #3 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: explicit formula for chebychev's method? what is a homogeneous recurrence and a associated auxiliary equation?
 October 15th, 2012, 11:36 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,206 Thanks: 513 Math Focus: Calculus/ODEs Re: explicit formula for chebychev's method? Read this article for an explanation of the technique I used to get the closed form, before applying the binomial theorem.

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