binomial series (1+x)^(m), abs(x) <1, m=-1/2
appendix : binomial series
peter.vlasschaert@gmail.com,29/09/2017.
(see : Bessel equations 'other method to find L{jo(t)} ? )

kill(all)$
load("simplify_sum")$

depends([f],x);

f(x):=(1+x)^m

ep:f(x):=(1+x)^m;
eq:ff=(1+x)^m;
er:s(x):=1+sum(a(k)*x^k,k,1,inf);
v:1+sum(a(k)*x^k,k,1,inf);
dv:diff(1,x)+sum(diff(a(k)*x^k,x),k,1,inf);

infinite polynoom

eq3:deftaylor(f1(x),sum((1+x)*k*a(k)*x^k,k,1,inf));
eq4:powerseries(f1(x),x,0);

commands : infinite sum =>  sumcontract   (two sum with same range)
                                                      factcomb         (inside the sum,simplification)
                                                      intosum           (take (x+1) inside sum) see 'eq30'
? expand

you can get also use operator on a function , ""ep: f(x) := (1+x)^m"

eq5:diff(rhs(eq),x);
eq6:diff(part(ep,1),x)=diff(part(ep,2),x);

trick : function eq9 to eq10 to eq11

eq7:part(eq6,2,2,1);
eq8:eq7*eq6;
eq9:ratsubst(q, (x+1)^m, eq8);
eq10:ev(eq9,f(x)=q);
eq11:factor(ratsubst(q, (x+1)^m, eq10));

How to solve : ODE  'eq11'

eq12:lhs(eq11)-rhs(eq11)=0;

radcan : =>         simplify   "e^(()*log())"

eq13:ode2(eq12,q,x);
eq14:ev(eq13,q=f(x));
eq15:solve(eq14,%c)[1];
eq16:radcan(eq15);

find  q , initial conditions  %c =1

eq17:ev(eq13,eq16);

check solution 'eq17' into 'eq12'

eq18:ev(eq12,eq17,diff);

check the value f(x=0) => f(0)=1 same value %c = 1

declare(m,constant)$

eq19:f(0);

because of the above ,power serie s(0)=1

eq25:diff(s(x),x);

eq26:diff(part(er,1),x)=diff(part(er,2),x);
eq27:lhs(eq26) = ev(part(eq26,2),diff);

doesn't work 'eq28' intosum ?
=> differentation outside the sum (must be inside the sum).

eq28:(1+x).eq25=m*s(x);
eq29:(1+x)*dv = m*v;
eq30:intosum(eq29);