Child Labor #10

Note that:

$123456789=1(10)^8+2(10)^7+3(10)^6+4(10)^5+5(10)^4+6(10)^3+7(10)^2+8(10)^1+9(10)^0=f(10)$

Thus we find $f(x)$ as:

$\large f(x)=x^8+2x^7+3x^6+4x^5+5x^4+6x^3+7x^2+8x+9$

$\Rightarrow f(1)=(1)^8+2(1)^7+3(1)^6+4(1)^5+5(1)^4+6(1)^3+7(1)^2+8(1)^1+9= \sum_{n=1}^9 n = \dfrac{9\times 10}{2}=\Large{\boxed{45}}$

The only possible way of this polynomial under the condition is $f(x) = \sum_{n=1}^9\:n(x^{9-n})$ which produces $f(10) = 123456789$ . $f(1)$ is then equal to $1+2+3+\dots+9 = 45$

Since f(x) = a0 + a1 x + a2 x^2 + a3 x^3 + a4 x^4 + a5 x^ 5 + a6 x^6 + a7 x^7 + a8 x^8 ought to be its general function series despite Fourier series for representing a function containing discontinuities, it is for function to be continuous throughout.

f(10) = a0 + a1 10 + a2 100 + a3 1000 + a4 10000 + a5 10^ 5 + a6 10^6 + a7 10^7 + a8 10^8 = 123456789

a0 to a8 ought to be 9 to 1.

Therefore, f(1) = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 40 + 5 = 45