/sci/ - Science & Math

[Anonymous No.7970438]

Can anyone see what the pattern is here? Or at least how to go about solving it?

[Anonymous No.7970443]

f(x,y,z) = 2x + 4yz

[Anonymous No.7970450]

>>7970438
How are we supposed to see a pattern of z when it's the same every time in your example?

[Anonymous No.7970451]

2x + 8y

[Anonymous No.7970458]

f(x,y,z)=z(x+4y)
Post another

[Anonymous No.7970467]

>>7970438
There are at least infinitely many C-infinity functions that meet all of those constraints.

[Anonymous No.7970469]

thanks guys

>>7970450
sorry im stupid

[Anonymous No.7970481]

>it's an OP horribly tries to hide the fact he's asking us to do his homework episode

[Anonymous No.7970491]

>>7970458
Here you go.

[Anonymous No.7970496]

>>7970469
2x + 4yz again.

[Anonymous No.7970526]

>>7970443
How did you come up with this? Does a non-brute force method exist?

[Anonymous No.7970553]

>>7970491
-9xz -3y
Damn the second one is the trick :(

[Anonymous No.7970570]

>>7970496
thanks it worked :)

[Anonymous No.7970591]

>>7970526
Looking eq 1 and 3, we see the only difference is x changes by 1. This results in a change in f of 2. The same is true of eq 2 and 4.

So our first guess will be f(x,y,z) = g(x,y,z) + 2x.

g = f - 2x.

g = {0, 8, 0, 8}

So we have a new set:

g(0, 0, 2) = 0
g(0, 1, 2) = 8
g(1, 0, 2) = 0
g(1, 1, 2) = 8

We notice that if y is zero, then g is zero. Also, comparing eq 1 and 2 or eq 3 and 4 we see y changes by 1 and g changes by 8.

So we conclude that g(x,y,z) = 8y.

Since z is 2 in all cases, we can replace any appearance of 2 with z.

Thus our solutions thus far are:

g(x,y,z) = 2x + 8y
g(x,y,z) = 2x + 4yz
g(x,y,z) = 2x + 2yz^2
g(x,y,z) = 2x + yz^3
g(x,y,z) = xz + 8y
g(x,y,z) = xz + 4yz
g(x,y,z) = xz + 2yz^2
g(x,y,z) = xz + yz^3

[Anonymous No.7970593]

>>7970526
Looking eq 1 and 3, we see the only difference is x changes by 1. This results in a change in f of 2. The same is true of eq 2 and 4.

So our first guess will be f(x,y,z) = g(x,y,z) + 2x.

g = f - 2x.

g = {0, 8, 0, 8}

So we have a new set:

g(0, 0, 2) = 0
g(0, 1, 2) = 8
g(1, 0, 2) = 0
g(1, 1, 2) = 8

We notice that if y is zero, then g is zero. Also, comparing eq 1 and 2 or eq 3 and 4 we see y changes by 1 and g changes by 8.

So we conclude that g(x,y,z) = 8y.

Since z is 2 in all cases, we can replace any appearance of 2 with z.

Thus our solutions thus far are:

f(x,y,z) = 2x + 8y
f(x,y,z) = 2x + 4yz
f(x,y,z) = 2x + 2yz^2
f(x,y,z) = 2x + yz^3
f(x,y,z) = xz + 8y
f(x,y,z) = xz + 4yz
f(x,y,z) = xz + 2yz^2
f(x,y,z) = xz + yz^3

[Anonymous No.7970605]

This thread turned out surprisingly interesting, cheers lads!

[Cecelia Fitzgerald No.7972195]

So why did you pick the second solution?

[Brad Leach No.7972197]

>>7970438
f could just happen to map those points to those scalars and do something completely different everywhere else

you need more information about f

[Bridget Vega No.7972201]

Why do people post funtions that depend on z if it's always set to 2?

2x+8y

is the simplest solution

[Katrina Kirk No.7972576]

[math]\left(\begin{matrix} 0 & 0 & 2 \\ 0 & 1 & 2 \\ 1 & 0 & 2 \end{matrix} \right)\cdotp a = \left(\begin{matrix}0\\8\\2\end{matrix}\right)[/math]

[eqn]a = \left(\begin{matrix}2\\8\\0\end{matrix}\right) \\f(x,y,z) = 2x+8y+0z[/eqn]

[Patricia Pennington No.7973115]

>>7970438
2x+8y+0z

[Aurora Bell No.7973132]

>>7970451
>>7972201
>>7973115

bingo

[Mary Weber No.7973160]

>>7972576
baby's first LA

[Anonymous No.7973445]

>>7970438
Can you not just do High School simultaneous?

[Anonymous No.7973480]

>>7970438

ac+4bc