Question

What's wrong with this logic?
t1 = -25\(^0C\)
t2 = 40 \(^0C\)
\(\triangle t = t2-t1= 65^0C = 338.15 K\)
but if I convert the t1, t2 at the beginning, \(\triangle t = 65 K\)
Please, explain m

Answer 1

@Mertsj

Answer 2

You want to know why the change in temperature works either way?

Answer 3

Oh I see the issue here.

Answer 4

I want to know why the results are different

Answer 5

She wants to know why if you convert to Kelvin before finding delta t, you get delta t = 65 degrees kelvin.
If you convert to kelvin after finding delta t, you get 338 degrees kelvin

Answer 6

\[(40+273)-(-25+273)=65 degrees Kelvin\]
\[40-(-25)=65 ^{o}C=65+273=338^{o}K\]

Answer 7

Thank you @Mertsj That's what I mean

Answer 8

Consider \(f(x) = 4x\) and \(g(x) = 4x+10\). Notice \(g(x) = f(x) + 10\)
\(\Delta f= 4(a-b)\) while \(\Delta g= 4(a-b)\).

If you try convert after finding the difference, you would get \(\Delta f =4(a-b) \implies \Delta g =\Delta f + 10 = 4(a-b)+10 \)

Answer 9

So basically, interval lengths don't convert.

Answer 10

I am with you at the last comment only. :)
However, I got it.

Answer 11

Thank you very much @wio

Answer 12

No problem I guess?

Actually, here is how look at it:
The function \(f(x) = x+273\) converts C to K. You are esentially saying: \[
f(x_2)-f(x_1) = f(x_2-x_1)
\]

In reality \[
f(x_2)-f(x_1) = x_2+273 -(x_1+273) = x_2-x_1
\]and \[
f(x_2-x_1) = (x_2-x_1) + 273
\]

Answer 13

Yes, I got it, the gap between the two are the same no matter what the unit is, right?
in this case 65 units in difference for both them (something like the situation of 3hours and 3 o'clock) :)

Answer 14

Yeah, I'm just trying to pinpoint the exact error and in general it is \(\Delta f = f(\Delta x)\) or something.

Answer 15

I have another question about degenerate and non-degenerate, can you help me?

Answer 16

Ok

Answer 17

what's the context of degenerate?

Answer 18

what do they exactly mean? not from definition

Answer 19

What context?

Answer 20

In linear functional, generate can be expressed as span. Can I expand this logic to bilinear form for degenerate ? something likes span the tensor product set?

Answer 21

So \[
f(x,y) = (4-y)g(x,y)
\]would likely be a degenerate form since \(y=4\) would probably zero it out?

Answer 22

in which field?

Answer 23

I'm talking about your definition.

Answer 24

oh the definition from the book:
A bilinear form w on U\(\otimes\)V is degenerate if, as a function of one of its two arguments, it vanishes identically for some non-zero value of its other argument; otherwise it is non-degenerate.

Answer 25

I am not sure. I'm not completely familiar.