A Polynomial Equation With A Surprise At The End

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[@agsantiago22]

The c1 *i^x ... comes from solving the equation as a difference equation (the discrete version of a differential equation). The values of C1 and C2 are unknown because there are no initial conditions.

[@SyberMath]

😍

[@armacham]

Take the derivative twice, you get p''(x) + p''(x-2) = 0
if p''(x) = 0 always, then you have a linear equation
what if it's not 0?
then you have p''(x) = -p''(x-2)
Can you think of any equations like that? Sine and cosine come to mind
A sine wave with a period of 4 would fit right in
p(x) = 3x + 5 + m*sin( pi*(x+a)/2 )
m and a can be any real numbers.
this will be true because m*sin(pi*(x+a)/2 ) + m*sin(pi*(x - 2 +a)/2 ) will always be equal to zero

[@GreenMeansGOF]

Another way for P(x) and P(x+2) to be of the same form is if P(x)=Ae^(Bx). P(x)+P(x+2)=0 implies that B is an odd multiple if iπ/2 which gives the two independent solutions i^x and (-i)^x. As to how to derive these solutions without seeing the answer ahead of time, I do not know.

[@pokemonjourneysfan5925]

Well, you didn’t tell wolframalpha that P(x) is a polynomial, so it interprets it as a function, this then becomes a reccurence relation similar to diff equation.

[@GreenMeansGOF]

@@pokemonjourneysfan5925 yeah. It’s a 0th order differential equation. It makes me wonder if there is a general theory around this.

[@seanfraser3125]

Since P(x) + P(x-2) must be of the same degree as P, P is linear.
Let P(x) = ax+b. Then ax+b + a(x-2)+b = 6x+4.
Combining like terms we have 2ax + 2(b-a) = 6x+4. Clearly a=3. So 2(b-3)=4, and thus b=5.
So P(x) = 3x+5

[@d-8664]

Why do you repeat the calculation?

[@seanfraser3125]

@@d-8664 I like to do the problem myself in the comments when I can. Sometimes I use the same approach as SyberMath, sometimes I use a different one.

[@giuseppemalaguti435]

3x+5

[@mcwulf25]

The complex part is mad. It works, because we can show that
i^x + i^(x-2) = i^(x-1)(i + 1/i) = 0
Same with the powers of (-i).
Mad because how did Wolfram "think" to put those in in the first place?

[@dugong369]

Can we add on c3*sin(πx/2) + c4*cos(πx/2)? To Wolfram Alpha's answer?

[@JohnSmith-mz7dh]

We can also differentiate twice. We get p’’(x)+p’’(x+2)=0. Ok, the only polynomial which obeys this rule is p’’(x)=0.
This is since these two polynomials add up to another polynomial P(x) which is equal to zero. This means that all coefficients must cancel out for p’’(x) and p’’(x+2) (which cannot possible happened unless all the coefficients are zero.)
We therefore know that p(x) is a linear polynomial (integrate once we get p’(x)=c, and twice we get p(x)=cx+d. p(x)+p(x-2)=2cx-2c+2d=6x+4. So c=3 (to cancel the x terms) and therefore 2d=4+2c=10, so d=5.
p(x)=3x+5

[@scottleung9587]

I also got P(x)=3x+5 as the answer.

[@kianmath71]

P(x) = 3x + 5

[@honestadministrator]

Here in P ( x) = a x + b
a x + b + a ( x -2) + b = 6 x +4
a = 3 , 2 b - 6 = 4 i. e. b = 5

[@rob876]

Add solution to P(x) + P(x+2) = 0
let P(x) = cr^x
then cr^x + cr^(x+2) = 0
1 + r^2 = 0
r = ±i
P(x) = C1(i)^x + C2(-i)^x

[@SyberMath]

Wow!

[@SimonClarkstone]

Not a polynomial, but interesting anyway.

[@user-gu6dc7yu1m]

3x+5+c1(-i)^x+c2 i^x