Given:
f n (x)= 1/2．(a^n+b^n) (a,b are the roots of t^2-2xt+1=0,where -1≦x≦1 )
and
2x．f n (x)=f n+1 (x)+f n-1 (x)
show that for n≧1 , f n (x) is a polynomail in x,of degree n and with leading
coefficient 2^(n-1)
Prove by M.I.
Firstly, note that a+b = 2x and ab = 1.
When n=1, f1(x)= (a+b)/2 = 2x/2 = x, which is a polynomial in x of degree 1 with leading coefficient 1=2^0.
When n=2, f2(x)=(a^2+b^2)/2 = [(a+b)^2-2ab]/2 = (4x^2 - 2)/2 = 2x^2 - 1, which is a polynomial in x of degree 2 with leading coefficient 2 = 2^1.
The proposition is true for n=1 and n=2.
Assume the proposition is true for n = k-1 and n=k, i.e.
Let fk-1(x) = 2^(k-2) x^(k-1) +Q(x), where Q(x) is polynomial in x with degree less than k-1, and
fk(x) = 2^(k-1)x^k + R(x), where R(x) is polynomial in x with degree less than k.
When n=k+1, fk+1(x) = 2xfk(x) - fk-1(x)
=2x．〔2^(k-1)x^k + R(x)〕 - 〔2^(k-2) x^(k-1) +Q(x)〕
= 2^kx^(k+1) + 2xR(x)- 2^(k-2) x^(k-1)-Q(x), which is polynomial in x with degree k+1 and leading coefficient 2^k.
The proposition is true for n = k+1.
By M.I. the proposition is true for all n>=1
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Verified answer
Prove by M.I.
Firstly, note that a+b = 2x and ab = 1.
When n=1, f1(x)= (a+b)/2 = 2x/2 = x, which is a polynomial in x of degree 1 with leading coefficient 1=2^0.
When n=2, f2(x)=(a^2+b^2)/2 = [(a+b)^2-2ab]/2 = (4x^2 - 2)/2 = 2x^2 - 1, which is a polynomial in x of degree 2 with leading coefficient 2 = 2^1.
The proposition is true for n=1 and n=2.
Assume the proposition is true for n = k-1 and n=k, i.e.
Let fk-1(x) = 2^(k-2) x^(k-1) +Q(x), where Q(x) is polynomial in x with degree less than k-1, and
fk(x) = 2^(k-1)x^k + R(x), where R(x) is polynomial in x with degree less than k.
When n=k+1, fk+1(x) = 2xfk(x) - fk-1(x)
=2x．〔2^(k-1)x^k + R(x)〕 - 〔2^(k-2) x^(k-1) +Q(x)〕
= 2^kx^(k+1) + 2xR(x)- 2^(k-2) x^(k-1)-Q(x), which is polynomial in x with degree k+1 and leading coefficient 2^k.
The proposition is true for n = k+1.
By M.I. the proposition is true for all n>=1