Question

An easy one. Let P be a six degree polynomial.
If P(1) = 1, P(2) = 2 ........till P(6) = 6.
Find P(7) = ?

Answer 1

And NO. Its not 7.

Answer 2

The answer can be any number
\[
P[(x)= \frac{x^6}{360}-\frac{7
x^5}{120}+\frac{35
x^4}{72}-\frac{49
x^3}{24}+\frac{203
x^2}{45}-\frac{39 x}{10}+2
\]
P[7]=9

Here is another one
\[\frac{x^6}{180}-\frac{7
x^5}{60}+\frac{35
x^4}{36}-\frac{49
x^3}{12}+\frac{406
x^2}{45}-\frac{44 x}{5}+4

\]
P[7] =11
All depends on how the constant term is defined.

Answer 3

In your explanation, your expression isnt getting the conditions satisfied.
P(1) = 1
P(2) = 2
P(3) = 3
P(4) = 4
P(5) = 5
P(6) = 6

Answer 4

Yes it is. Which one is not?

Answer 5

Here is the general polynomial in terms of the conatant term a0
\[
P(x)= \frac{\text{a0}
x^6}{720}-\frac{7 \text{a0}
x^5}{240}+\frac{35
\text{a0}
x^4}{144}-\frac{49
\text{a0}
x^3}{48}+\frac{203
\text{a0}
x^2}{90}+\frac{1}{20}
(20-49 \text{a0})
x+\text{a0}
\]
\[
P[7} =7+ a0
\]

Answer 6

P(7) =7+ a0

Answer 7

A polynomial of degree 6 depends on 7 coefficients. You are given 6 conditions, so you can choose p(0) to be any thing and solve in temrs of a0. and

P(7)= a0+7

Answer 8

i think a matrix would help out

Answer 9

we have 1 free variable making this infinite number of polys that fit as elias pointed out

Answer 10

\[\begin{vmatrix}0\\0\\0\\0\\0\\1\\0\end{vmatrix}+k\begin{vmatrix}1/720\\-7/240\\35/144\\-49/48\\203/90\\-49/20\\1\end{vmatrix}\]

Answer 11

@eliassaab That's really good. How did you think of those expressions btw?

Answer 12

I mean my first answer was,
\[P(x) = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6) + x\]

Answer 13

I solved 6 equations in six unknowns using a symbolic manipulator.

Answer 14

That is one of the infinite possibilities. Probably the nicest one.

Answer 15

Haha. True that. Good work. Thanks :)

Answer 16

This would work too
P(x)=K (x−1)(x−2)(x−3)(x−4)(x−5)(x−6)+x
Where K is any constant.

Answer 17

Yes. And that easily proves that it can have infinite permutations. Nicccce.