# How to make a polynomial so that f(i) = 1/(2^i)?

I know that, sequence has formula $$f(n) = \dfrac{1}{2^n}$$ satifying the conditions $$f(1)=\dfrac{1}{2}$$, $$f(2)=\dfrac{1}{4}$$, $$f(3)=\dfrac{1}{8}$$, $$f(4)=\dfrac{1}{16}$$. Now I am trying to find a polynominal that also satisfies these conditions.

f[x_] = a x^3 + b x^2 + c x + d;
Solve[{f[1] == 1/2, f[2] == 1/4, f[3] == 1/8, f[4] == 1/16 }, {a, b,
c, d}]


{{a -> -(1/96), b -> 1/8, c -> -(53/96), d -> 15/16}}

With more conditions, $$f(1)=\dfrac{1}{2}$$, $$f(2)=\dfrac{1}{4}$$, $$f(3)=\dfrac{1}{8}$$, $$f(4)=\dfrac{1}{16}$$, $$\ldots$$, $$f(10)=\dfrac{1}{2^{10}}$$. How can I make a polynomial without solving a system of equations?

• I can't correct the typo because an edit must be at least 6 characters. I think the last expression before ". How can" should begin with f(10), not f(4). Nov 29, 2023 at 23:01

InterpolatingPolynomial[Table[{i, 2^-i}, {i, 4}], x] // Expand
(*    15/16 - 53 x/96 + x^2/8 - x^3/96    *)

InterpolatingPolynomial[Table[{i, 2^-i}, {i, 10}], x] // Expand
(*    1023/1024 - 254437 x/368640 + 348803 x^2/1474560 -
2458693 x^3/46448640 + 5561 x^4/655360 - 17653 x^5/17694720 +
83 x^6/983040 - 43 x^7/8847360 + x^8/5898240 - x^9/371589120    *)


The $$i$$-values of the data points are actually the default assumed values, and so we can code-golf this last example down:

InterpolatingPolynomial[2^-Range[10], x] // Expand
(*    1023/1024 - 254437 x/368640 + 348803 x^2/1474560 -
2458693 x^3/46448640 + 5561 x^4/655360 - 17653 x^5/17694720 +
83 x^6/983040 - 43 x^7/8847360 + x^8/5898240 - x^9/371589120    *)