User:Egm6321.f12.team1.pb/report5

=Problem 5.5* =

Given:
and

Prove That:

 * $$\phi(x,y,p) = P(x)p + T(x) y + k$$

Solution:
We know

Comparing the only terms containing $$y''$$ in ($$) and ($$), we can obtain
 * $$\phi_p = P(x)$$

Integrating w.r.t $$p$$, we get

Differentiating w.r.t. $$x, y$$, we get
 * $$\phi_x = P'(x)p + \frac {\partial k_1(x,y)} {\partial x}$$

and
 * $$\phi_y = \frac {\partial k_1(x,y)} {\partial y}$$

Substituting $$\phi_x, \phi_y, \phi_p $$ into ($$) and comparing with ($$)
 * $$R(x)y + Q(x) y' + \cancel {P(x) y} = P'(x)p + \frac {\partial k_1(x,y)} {\partial x} + \frac {\partial k_1(x,y)}{\partial y} y' + \cancel {P(x) y}$$
 * $$\Rightarrow R(x)y + Q(x) y'= P'(x)y' + \frac {\partial k_1(x,y)} {\partial x} + \frac {\partial k_1(x,y)}{\partial y} y'$$

Equating coefficients with $$y'$$


 * $$Q(x) = P'(x) + \frac {\partial k_1(x,y)}{\partial y} $$

and we are left with


 * $$R(x)y = \frac {\partial k_1(x,y)} {\partial x} $$

Integrating w.r.t. $$x$$
 * $$k_1(x,y) = \left( \int^{x} R(s)ds \right ) y + k_2(y)$$
 * $$\Rightarrow k_1(x,y) = T(x) y + k_2(y)$$

Substituting into ($$), we have

Differentiating w.r.t $$x$$
 * $$\phi_x = P'(x)p + T'(x) y $$

Differentiating w.r.t $$y$$
 * $$\phi_y = T(x) + k_2'(y) $$

Substituting into ($$)
 * $$P'(x)p + T'(x) y + T(x) + k_2'(y) = R(x)y + Q(x)y'$$
 * $$\Rightarrow T'(x) y + [P'(x) + T(x]y' + k_2'(y)y' = R(x)y + Q(x)y'$$

Equating coefficients of $$y'$$ in both sides, we obtain
 * $$k_2'(y) = 0$$
 * $$\Rightarrow k_2(y) = k (Constant)$$

Substituting into ($$), we have


 * $$\phi(x,y,p) = P(x)p + T(x) y + k\qquad\blacksquare$$

Pavel Bhowmik (talk) 15:45, 31 October 2012 (UTC)

=Problem 5.10 =

Prove That:

 * Use Matlab to plot $$_2 F_1(5,-10;1;x)$$ near $$x=0$$ to display the local maximum (or minimum in this region)

Proof
To prove this equation, we will assume a relation, the proof of which has been left for future work.


 * $$n! \sum\limits_{k=0}^m \dbinom{m}{k} \dbinom{k+n}{n} (-x)^k = \sum\limits_{k=0}^m \dbinom{m}{k} n! \dbinom{k+n}{n} (-x)^k $$
 * $$= \sum\limits_{k=0}^m \dbinom{m}{k} \frac {(k+n)!}{k!} (-x)^k $$
 * $$= \sum\limits_{k=0}^m \dbinom{m}{k}(-1)^k \frac {\mathrm{d}^n}{\mathrm{d}x^n} x^{k+n} $$
 * $$=  \frac {\mathrm{d}^n}{\mathrm{d}x^n} \sum\limits_{k=0}^m (-1)^k\dbinom{m}{k} x^{k+n} $$
 * $$=  \frac {\mathrm{d}^n}{\mathrm{d}x^n} \left[\left( \sum\limits_{k=0}^m \dbinom{m}{k} (-x)^k \right) x^n \right]$$

Similarly

Using ($$), ($$) and ($$), we get


 * $$\frac {\frac {\mathrm{d}^n} {\mathrm{d}x^n} (1-x)^m x^n}{\frac {\mathrm{d}^m} {\mathrm{d}x^m} (1-x)^n x^m} = \frac {n !}{m !} \frac {(1-x)^m}{(1-x)^n}.\quad\blacksquare$$

Proof
The Jacobi polynomials are defined via the hypergeometric function as follows :

Setting $$\alpha = 0, \beta = a+b-1, n=-a$$ we get


 * $$_2F_1(a,b;1;\frac{1-z}{2}) = \frac{(-a)!}{(1)_{-a}} P_{-a}^{(0,a+b-1)}(z)$$

Since $$ (1)_{-a} = (-a)!$$


 * $$_2F_1(a,b;1;\frac{1-z}{2}) = P_{-a}^{(0,a+b-1)}(z)$$

Using Rodrigues' formula as


 * $$P_n^{(\alpha,\beta)} (z) = \frac{(-1)^n}{2^n n!} (1-z)^{-\alpha} (1+z)^{-\beta} \frac{d^n}{dz^n} \left\{ (1-z)^\alpha (1+z)^\beta (1 - z^2)^n \right\} $$

Setting $$\alpha = 0$$


 * $$\Rightarrow\,_2F_1(a,b;1;\frac{1-z}{2}) = \frac{(-1)^{-a}}{2^{-a} (-a)!} (1+z)^{-a-b+1} \frac{d^n}{dz^n} \left\{(1-z)^{-a} (1+z)^{a+b-1 - a } \right\} $$

Setting $$z=1-2x$$


 * $$\Rightarrow \frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}z}\frac{\mathrm{d}z}{\mathrm{d}x}$$
 * $$\Rightarrow \frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}z}\left(-\frac{1}{2}\right)$$

Also and

Substituting ($$), ($$), ($$), ($$) into ($$) and simplifying, we get

Similarly, setting $$\alpha = 0, \beta = -a-b+1, n=b-1, z=1-2x$$ we get


 * $$\Rightarrow \frac {_2F_1(a,b;1;x)} {_2F_1(-b+1,-a+1;1;x)} = (1-x)^{-a-b+1}  \frac {(b-1)!} {(-a)!} \times   \frac {(1-x)^{-a}} {(1-x)^{b-1}}  \times  \frac {\frac {\mathrm{d}^{-a}} {\mathrm{d}x^{-a}} \left[(1-x)^{b-1}x^{-a}\right]} {\frac {\mathrm{d}^{b-1}} {\mathrm{d}x^{b-1}} \left[(1-x)^{-a}x^{b-1}\right]}$$

Setting $$n=-a, m=b+1$$ in ($$)
 * $$\frac {\frac {\mathrm{d}^{-a}} {\mathrm{d}x^{-a}} \left[(1-x)^{b-1}x^{-a}\right]} {\frac {\mathrm{d}^{b-1}} {\mathrm{d}x^{b-1}} \left[(1-x)^{-a}x^{b-1}\right]} = \frac {(-a)!} {(b-1)!} \times \frac {(1-x)^{b-1}} {(1-x)^{-a}} $$
 * $$\Rightarrow \frac {(b-1)!} {(-a)!} \times  \frac {(1-x)^{-a}} {(1-x)^{b-1}}  \times  \frac {\frac {\mathrm{d}^{-a}} {\mathrm{d}x^{-a}} \left[(1-x)^{b-1}x^{-a}\right]} {\frac {\mathrm{d}^{b-1}} {\mathrm{d}x^{b-1}} \left[(1-x)^{-a}x^{b-1}\right]} = 1 $$
 * $$\Rightarrow \frac {_2F_1(a,b;1;x)} {_2F_1(-b+1,-a+1;1;x)} = (1-x)^{-a-b+1}$$
 * $$_2F_1(a,b;1;x) = (1-x)^{-a-b+1}\,_2F_1(-b+1,-a+1;1;x)\qquad \blacksquare$$

Solution:
Using ($$) and setting $$a=5, b=-10$$, we get
 * $$_2F_1(5,-10;1;x) = (1-x)^{-5+10+1}\,_2F_1(-4,11;1;x)$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\,_2F_1(-4,11;1;x)$$

Since in the second term, $$a$$ is a negative integer, the summation will terminate after $$-(-4)+1=5$$ terms
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\sum\limits_{k=0}^4 \frac{(-4)_k(11)_k}{(1)_k}\frac{x^k}{k!}$$

Since $$(1)_k = k!$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\sum\limits_{k=0}^4 (-4)_k(11)_k \frac{x^k}{(k!)^2}$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\left[ (-4)_0(11)_0 \frac{x^0}{(0!)^2} + (-4)_1(11)_1 \frac{x^1}{(1!)^2} + (-4)_2(11)_2 \frac{x^2}{(2!)^2} + (-4)_3(11)_3 \frac{x^3}{(3!)^2} + (-4)_4(11)_4 \frac{x^4}{(4!)^2}\right]$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\left[ 1 + (-4) \times 11 x + (-4)\times(-3)\times 11 \times 12 \times\frac{x^2}{4} + (-4)\times(-3)\times(-2)\times 11 \times 12 \times 13\times \frac{x^3}{36}\right.$$
 * $$\left. + (-4)\times(-3)\times(-2)\times(-1)\times 11 \times 12 \times 13 \times 14 \times \frac{x^4}{(24)^2} \right]$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6 (1 - 44x + 396x^2 - 1144x^3 + 1001x^4)\qquad\blacksquare$$

Since $$_2F_1(5,-10;1;x)$$ is a polynomial with six of its roots are at $$x=1$$, the first derivative of $$_2F_1(5,-10;1;x)$$ will have at least 5 roots at $$x=1$$. Also since $$_2F_1(5,-10;1;x)$$ is of degree 10, the first derivative will have a degree of 9 and a maximum of 9 real roots. Since 5 of the roots are at the same point $$x=1$$, at max we can see 5 distinct points where the first derivative will be 0. So $$_2F_1(5,-10;1;x)$$ can have a maximum of 5 optima combined. The following graph displays all of the 5 points of optima with the blue circles. Pavel Bhowmik (talk) 15:48, 31 October 2012 (UTC)