Daniel's polynomial

I'd like to start with a thanks to Brilliant.org for accepting and improving my problem. Now, for the solving!

We notice for when $x=1\rightarrow7$ , we have $f(x)=x^2+1$ . This suggests that we find a polynomial $g$ that uses this handy fact. If we set $g(x)=f(x)-(x^2+1)$ , we see that $x=1\rightarrow7$ are the roots of $g(x)$ !

Therefore, $g(x)=a(x-1)(x-2)\cdots (x-6)(x-7)$ for some constant $a$ . Plugging in $0$ and $8$ in for $x$ gives that $g(0)=-5040a$ and $g(8)=5040a$ .

Therefore, $f(0)=-5040a+(0^2+1)=-5040a+1$ and $f(8)=5040a+(8^2+1)=5040a+65$ . We add $f(0)$ and $f(8)$ together, and find that the $-5040a$ and the $5040a$ conveniently cancel out, giving us a final sum of $1+65=\boxed{66}$ .

With a little thought we see there is a 2nd degree polynomial which satisfies the values given, namely

f(x) = x^2 + 1

But we are told it is a 7th degree polynomial. What form must it have then?

f(x) = c(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7) + x^2 + 1

Now

f(0) + f(8) = -5040c + 1 + 5040c + 65

The two unknown terms cancel, and we're left with 66 as the answer.

Because we have:

$2= 1^{2} + 1$

$5= 2^{2} + 1$

...

$50= 7^{2} + 1$

Let define $P(x) = f(x) - (x^{2} + 1)$ where $degree(P) = degree(f)$ = 7.

Now we see $x= 1...7$ are all the 7 roots of $P(x)$ so

$P(x) = a_{7} \times (x-1) \times (x-2) \times ... \times (x-7)$ , where $a_{7} <> 0$ is the coefficient of $x^{7}$ in $f(x)$ (and $P(x)$ )

Now we have:

1. $P(0) = a_{7} \times (-1) \times (-2) \times ... \times (-7) = -7! \times a_{7}$
2. $P(8) = a_{7} \times 7 \times 6 \times ... \times 1 = 7! \times a_{7}$

and thus:

1. $f(0) = -7! \times a_{7} + 1$
2. $f(8) = 7! \times a_{7} + 65$

So, finally $f(0) + f(8) = -7! \times a_{7} + 1 + 7! \times a_{7} + 65 = \boxed{66}$

f(0) = 1 f(1) = 1 + 1 f(2) = 1 + 1 + 3 f(3) = 1 + 1 + 3 + 5 f(4) = 1 + 1 + 3 + 5 + 7...... and so on. Hence, f(8) = 1 + 1 + 3 + ... + 15 = 65. So, 65 + 1 = 66.

If we take difference of consecutive terms we'll find that they are in AP f(2)-f(1)=3
f(3)-f(2)=5
f(4)-f(3)=7 f(5)-f(4)=9 f(6)-f(5)=11 f(7)-f(6)=13
this an AP with common difference 2.
Therefore
* f(1)-f(0) =1 *
2-f(0)=1
* f(0)=1 *
similarly,
f(8)-f(7)=15
* f(8)-50=15 *
f(8)=65
therefore
* f(0)+f(8)=1+65=66 *

Here , f(1) , f(2 ) , f ( 3 ) ....f ( 7 ) follow a pattern a^2 +1 hence f(0 ) = 1 and f ( 8 ) =65 hence answer = 1 + 65 = 66

f(1) =2 f(2) =5 . . . f(7) =50 so we can see that f(1)=2, so it can be 1+1 or 1^2 +1 or 1^3 +1, than we have 3 formula ->f(X)= X+1 ,f(X)= X^2 +1,f(X)= X^3 +1 => put in 2 => 2 +1 = 4 , 2^2 +1 =5, 2^3 +1 = 9 so the second fotmula is correct put in 0 => 0^2 +1 =1 , 8 => 8^2 +1 =65 s0 65+1 = 66

since,

f(2) -f(1) =3,

and f(3) -f(2) =5

,f(4) -f(3) =7

this sequence shows that, all the polynomial has a difference of increasing odd no.s

so f(8) =50+15=65 and, f(0) =1,

therefore,

f(8) +f(0) =65+1 =66

f(x)=x+1 f(1)=(1x1)+1 f(1)=2

f(0)=(0x0)+1 f(0)=1

f(8)=(8x8)+1 f(8)=65

f(0)+f(8)=1+65=66

f(2) - f(1) = 3

f(3) - f(2) = 5

f(4) - f(3) = 7

f(5) - f(4) = 9

f(6) - f(5) = 11

f(7) - f(6) = 13

so by observing the pattern

f(8) - f(7) = 15

f(8) = 15 + f(7) = 15 + 50 = 65

and

f(1) - f(0) = 1

f(0) = f(1) - 1 = 2 - 1 = 1

f(0) + f(8) = 1 + 65 = 66

This kind of question is very easy. Just see the trend followed by the polynomial when we put 0,1,2,3,4,5,6,7. we notice that the remainder produced is a simple A.P. Thus we can find what will we get as we put 8 & 0 in poly. So, f(0)+f(8) =1+65 =66

2,5,10,17 .... are in series with diff as consecutive odd numbers so o and 8 are 1 and 65

For this problem, it's unnecessary to even come up with the polynomial. The values for the function follow a specific pattern: for every x + 1, delta(y) goes up by 2, such that the difference between f(1) and f(2) is 3, and the difference between f(2) and f(3) is 5, and so on. Following this pattern, we know the difference between f(7) and f(8) should by 15, meaning f(8) = f(7) + 15 = 65. Additionally, we know the difference between f(0) and f(1) should be 1, meaning f(0) = f(1) - 1 = 1

From there, it's just arithmetic: f(0) + f(8) = 1 + 65 = 66