v ( x ) v(x)

Algebra Level 2

Let's say v ( x ) = f ( x 2 ) + ( f ( x ) ) 2 v(x)=f(x^2)+(f(x))^2

So are these equations below true? f ( x ) = x + n ⟹ v ( f ( x ) ) = 2 ( x 2 + x n + T n ) f(x)=x+n \implies v(f(x))=2(x^2+xn+T_n) f ( x ) = n x ⟹ v ( f ( x ) ) = 2 T n x 2 f(x)=nx \implies v(f(x))=2T_nx^2 f ( x ) = x n ⟹ v ( f ( x ) ) = 2 x 2 n f(x)=x^n \implies v(f(x))=2x^{2n} where as T n T_n is the Triangular number

Yes No

Gia Hoàng Phạm by Gia Hoàng Phạm , 19, Vietnam

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1 solution

Gia Hoàng Phạm
Aug 31, 2018

The first one is f ( x ) = x + n ⟹ v ( f ( x ) ) = ( x 2 + n ) + ( x + n ) 2 = x 2 + n + x 2 + 2 x n + n 2 = 2 x 2 + 2 x n + ( n + n 2 ) = 2 x 2 + 2 x n + 2 T n = 2 ( x 2 + x n + T n ) f(x)=x+n \implies v(f(x))=(x^2+n)+(x+n)^2=x^2+n+x^2+2xn+n^2=2x^2+2xn+(n+n^2)=2x^2+2xn+2T_n=2(x^2+xn+T_n)

The middle one is f ( x ) = n x ⟹ v ( f ( x ) ) = x 2 n + ( x n ) 2 = x 2 n + x 2 n 2 = ( n + n 2 ) x 2 = 2 T n x 2 f(x)=nx \implies v(f(x))=x^2n+(xn)^2=x^2n+x^2n^2=(n+n^2)x^2=2T_nx^2

The last one is f ( x ) = x n ⟹ v ( f ( n ) ) = x 2 n + ( x n ) 2 = x 2 n + x 2 n = 2 x 2 n f(x)=x^n \implies v(f(n))=x^{2n}+(x^n)^2=x^{2n}+x^{2n}=2x^{2n}

So the answer is True

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