Question

Polynomial powered by 2

Answer 1

A polynomial f(x) has degree 8 and \[\huge f(i)=2^i\] for\[ i=0,1,2,3,4,5,6,7,8\]. Find \\[\huge \color{cyan}{f(9)}\].

Answer 2

@campbell_st @phi @TuringTest @shubhamsrg @sauravshakya @satellite73

Answer 3

Isn't it solved using the lagrange interpolating polynomial?
I'll study it in a couple of days, and I'll be able to help you without googling it xD

Answer 4

f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)/((-1)(-2)(-3)(-4)(-5)(-6)(-7)(-8))+2(x)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)/((1)(-1)(-2)(-3)(-4)(-5)(-6)(-7))+4(x)(x-1)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)/((2)(1)(-1)(-2)(-3)(-4)(-5)(-6))+8(x)(x-1)(x-2)(x-4)(x-5)(x-6)(x-7)(x-8)/((3)(2)(1)(-1)(-2)(-3)(-4)(-5))+16(x)(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)(x-8)/((4)(3)(2)(1)(-1)(-2)(-3)(-4))+32(x)(x-1)(x-2)(x-3)(x-4)(x-6)(x-7)(x-8)/((5)(4)(3)(2)(1)(-1)(-2)(-3))+64(x)(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)(x-8)/((6)(5)(4)(3)(2)(1)(-1)(-2))+128(x)(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-8)/((7)(6)(5)(4)(3)(2)(1)(-1))+256(x)(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)/((8)(7)(6)(5)(4)(3)(2)(1))

Answer 5

man this is hectic you know its for hish school students i bet they have to know lagrange

Answer 6

@shubhamsrg @UnkleRhaukus

Answer 7

Stuck on as this also.

Answer 8

it looks like the equation has to behave like an exponential function 2^i
hence it has to be divisible by 2
@ SqueeSpleen this numbers i=0,1,2...8 are not roots

Answer 9

I got the answer. I have a logic also but it is not perfect, still, I am able to get the correct answer.
Answer is 511.

Answer 10

i only se that the constant is 1
and the closest function that is like 2^x i got is \[f(x)=(\frac{ x }{ 10 }+1)^8\]

Answer 11

Here is what I did
I started with lower degrees,

For each, I took for granted that f(i) = 2^i , where 0<= i <=n (n is the degree)
after finding f(x) with degree n, I found out f(n+1)
Note that this is all symmetrical to what is asked in the actual ques.

for degree 1, let polynomial be ax+b
so a(0) + b =1
and a(1) + b = 2
So our polynomial is x+1 ,
f(2) = 3

for degree 2, let polynomial be ax^2 + bx + c
so a(0) + b(0) + c =1
a(1) + b(1) + c = 2
a(4) + b(2) + c = 4
So our polynomial is (1/2)(x^2 +x +2)
f(3) = 7

Same way when you solve for degree 3,
f(4) = 15

There is a pattern
3 , 7, 15 ,
next term should then be 31 owing to the pattern.
then 63 , 127,255, 511

511 comes when degree is 9. And to my surprise, it is the right answer.

Answer 12

I still don't have any idea why this pattern is followed or, what is the correct procedure for the question.
@mukushla

Answer 13

this is a good method and the fact that it anwsers the question is enough the pattern is very true

Answer 14

The numbers 0,..,8 aren't roots in my equations, it has 9 terms, and when you evaluate the equation in i, the terms 0..,i-1 and i+1,...,8 become zero, but the term i becomes f(i)
I didn't know this was for high school students :P

Answer 15

And I forget to evaluate the polynomial in f(9) xD

Answer 16

Yes, the answer is 511.

Answer 17

The polynomial simplified is:
(40320 + 25584 x + 15980 x^2 - 4340 x^3 + 3969 x^4 - 1064 x^5 +
210 x^6 - 20 x^7 + x^8)/40320
I know the question wasn't what's the polynomial, but what's the value of f[9], but I forget the question because I read a little about it a few days ago xD

Answer 18

@mukushla