Polynomial question

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[jagged juniper]

We do it using remainder theorm. It involves more of logical thinking than math

So, if u analyze it like this, it will work out

Here Q(x) is the quotient and R(x) is the remainder expression. Since we took a 4 degree polynomial along with Q(x), R(x) must be a 3 degree polynomial(a degree lesser than that)

If u analyze properly, u get r(x) as $x^{3}+2 x^{2} + 1$. I think you can figure it out from here.

[tawny epoch]

TON 618

[daring slate]

If you restrict the scope to polynomials of degree less than 4, then sure there should be a unique polynomial P
Otherwise aren't there multiple polynomials fitting any real value for P(40)?

[jagged juniper]

Q(x) can be of any degree. I restricted the expression being multiplied to Q(x).

So its true for all possible polynomials

[daring slate]

Nah, you can add any multiple of (x-1)(x-2)(x-9) and the prerequisite is still satisfied, yet the value of P(40) varies

[gloomy fox]

OMG

ur the best

ty

💙

[daring slate]

Under either restriction, sure, but neither restriction is given in the problem setting

[gloomy fox]

plugging 40 into this the answer is ||67201||

.close

[quaint moss]

Unable to parse the channel name

[gloomy fox]

.solved

how do i close

[daring slate]

I'd advise you not to, yet

[jagged juniper]

Man u need (x-40) there too. Else Q(x) may take a different value.

[daring slate]

Why do I need it?

It does not matter to the values of P(1), P(2) and P(9)

[jagged juniper]

But they've asked P(40). How else will the expression with Q(x) be zero

We don't know Q(x), so the idea is to eliminate it

[daring slate]

Extraneous

[jagged juniper]

https://tenor.com/view/you-good-mate-you-good-australlia-australlian-kangaroo-gif-26064296

[jagged juniper]

how else will u eliminate Q(x)

[daring slate]

Why do we need to...

[gloomy fox]

to make that term 0

[gloomy fox]

u dont know Q(x)

[jagged juniper]

Look he got it. Love u bro

[daring slate]

Exactly, there are infinitely many polynomials that can fit into Q(x), why take 0

[gloomy fox]

love you too

💀

[jagged juniper]

Q(x) is not zero. The terms multiplied to it are

[daring slate]

Fitting different polynomials in can result in different values for P(40), which satisfy the problem setting all the same

[daring slate]

Sorry, the onus is on this:
why pick (x-1)(x-2)(x-9)(x-40) instead of (x-1)(x-2)(x-9)?

[gloomy fox]

https://gonitzoggo.com/problem/637

[daring slate]

In a way you restricted the degree of R(x) to 3, but that isn't required by the problem setting

[jagged juniper]

Coz if we put 40 there, we won't need to worry about the infinitely many values that Q(x) takes

[gloomy fox]

so that when x = 40, (40-40) = 0 and Q(x) is meaningless

[jagged juniper]

That's what I am trying to say for that long

[daring slate]

Sorry I am too busy to really evaluate, but you can construct a degree 3 polynomial P using Lagrange polynomials,
where P(1) = 4, P(2) = 17, P(9) = 892 and P(40) = 0

https://en.wikipedia.org/wiki/Lagrange_polynomial

That would satisfy the problem setting all the same

[jagged juniper]

it's not a 4 degree polynomial. Coz Q(x) can have Any degree.

[daring slate]

I know, I am saying P(40) is not necessarily 67201

[jagged juniper]

It is necessarily 67201

Lemme explain

[daring slate]

@modest stirrup Can you help make a Lagrange interpolation?

[odd current]

Clash of two smart ppl🔥🔥

[daring slate]

Degree 3, P(1) = 4, P(2) = 17, P(9) = 892 and P(40) = 0

[jagged juniper]

Give example

[modest stirrup]

oooh

[daring slate]

https://www.dcode.fr/lagrange-interpolating-polynomial

[gloomy fox]

who did bro just spawn

[daring slate]

\begin{array}{c} \left(\left(14-\frac{21259 (x-9)}{45942}\right) (x-2)+13\right) (x-1)+4 \ = \ -\frac{21259 x^3}{45942}+\frac{149716 x^2}{7657}-\frac{1948829 x}{45942}+\frac{209260}{7657} \end{array}

[tawny epoch]

Rafain
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gloomy fox]

oh no

where is this going

[odd current]

Damn🔥🔥

[modest stirrup]

well its enough to construct a degree 2 polynomial and just multiply it by (x - 40)

[jagged juniper]

Put 40 now

[daring slate]

True

[modest stirrup]

wait no

that'll mess up other points

[daring slate]

No it wouldn't

[daring slate]

I generated it with (40, 0) as an input

So its value at 40 is 0

[modest stirrup]

unless we change their values first

[daring slate]

Oh yeah

Forgot about that

[modest stirrup]

because if we just do it blindly, the value at P(1) will get multiplied by -39

[daring slate]

True true

[modest stirrup]

anyways we do the old method, we find g_i(x) and sum em up

[jagged juniper]

And yeah I am talking about real coefficients ofc. Don't generate one with imaginary coefficients

[daring slate]

What?

Man I am trying to get your eyes open, don't embarrass yourself...

Rationals are real...

[jagged juniper]

I am saying that if u generate another polynomial, do so

[jagged stratus]

There are literally infinitely many polynomials passing through the three given points.

[daring slate]

,w -21259/4594240^3 + 149716/765740^2 - 1948829/45942 * 40 + 209260/7657

[tawny epoch]

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Requested by rafain

[jagged stratus]

The rule is that n points uniquely determine a polynomial of degree n - 1.

[elfin parcel]

sry to interupt but i beleive there can be inf possibilities

[jagged stratus]

But there are infinitely many polynomials of degree n or greater that pass through n points.

[daring slate]

Thanks guys, please take over from here
I am calm enough to return to work, grateful to yall

[modest stirrup]

g_0(x) = (x - 2)(x - 9)(x - 40)/(1 - 2)(1 - 9)(1 - 40) = -1/312 (x^3 - 51x^2 + 458x - 720)
g_1(x) = (x - 1)(x - 9)(x - 40)/(2 - 1)(2 - 9)(2 - 40) = 1/266 (x^3 - 50x^2 + 409x - 360)
g_2(x) = (x - 1)(x - 2)(x - 40)/(9 - 1)(9 - 2)(9 - 40) = -1/1736 (x^3 - 43x^2 + 122x - 80)
g_3(x) = (x - 1)(x - 2)(x - 9)/(40 - 1)(40 - 2)(40 - 9) = 1/45942 (x^3 - 12x + 29x - 18)

f(x) = 4 g0(x) + 17 g1(x) + 892 g2(x) + 0 g3(x)

shouldve told me to do Newton interpolation

[jagged stratus]

What do you mean "if u analyze properly"? The biggest step, the one that's impossible because it yields infinitely many solutions, is the one you just skip over with a hand wave.

[modest stirrup]

wait whats the problem with the question? you are given three points, you can use it to uniquely define a quadratic. then you plug in x = 40

[jagged stratus]

It's not explicitly stated that P is quadratic.

[modest stirrup]

it is implied

[jagged stratus]

Implied how?

[modest stirrup]

well theres only one definitive answer if you make a quadratic, which is the easiest to make out of all other options. for higher degrees anything can be an answer

so the solution is either "its a trick question, anything can be an answer" or you find the quadratic

[jagged stratus]

And the answer is it's a trick question, or at least poorly written. As written there is no solution, and you always always always solve math problems as written, because if you just start assuming they meant something else you could wind up assuming literally anything.

[modest stirrup]

then its probably the best strategy is to cover both bases and say "well the solution is any value because yadda yadda, but to still get marks i will write down the solution i think you wanted to see"

to both be right and give them what they expect to see

[daring slate]

Sure, cover both correctness and prudence

if one's ability allows

[modest stirrup]

welp i guess from here its easy enough to use just Newton or Lagrange interpolation

[daring slate]

Just for instance, adding another point to the generator would yield another polynomial, albeit one of higher degree

[jagged juniper]

I am sorry. I did not pay attention to the possibility that there can be infinite polynomials. We will get a unique polynomial for a degree less than 3 only.

[modest stirrup]

degree equal to 2 specifically

[daring slate]

All good man, learning opportunities everywhere

[jagged juniper]

Yeah a degree lesser

[daring slate]

Yap more, plot one where P(1) = P(2) = P(9) = 0 with degree 2

[modest stirrup]

im just saying that, for example, P(1) = 1, P(2) = 2, P(3) = 3 has a polynomial of degree 1 going through these points

[daring slate]

Yeah got what you meant, just messing around

[modest stirrup]

lol