Process Calculus

[Issac]

A + B - (C(D-F(G+H))) = B + 2A/C

try that one!

edit: for G

A + B - (C(D-F(G+H))) = B + 2A/C

- C(D-F(G+H)) = B + 2A/C -A - B ................ B-B = 0

D-F(G+H) = (2A/C -A)/-C

-F(G+H) = (2A/C -A)/-C - D

G+H = ((2A/C -A)/-C - D)/-F

G = ((2A/C -A)/-C - D)/-F - H

Now, I would like to make it look a bit nicer by fixing the division with a "negative" unknown... you know.. (a+c)/-b = (-a-c)/b

G = ((2A/C -A)/-C - D)/-F - H
G = (-(2A/C -A)/C - D)/-F - H
G = ((2A/C -A)/C + D)/F - H

There you go

That's what i got....correction?

 

[Issac]

A + B - (C(D-F(G+H))) = B + 2A/C

try that one!

edit: for G

Do you mean B + (2A/C) or (B + 2A)/C?

I mean B + (2A/C).

(multiplication and division goes before plus and minus, unless someone just forgot to use brackets, which I didnt

)

 

[Jamstruth]

I know but, hey ho, maths and text don't go well. I get confused easily.

Right

A+B-(C(D-F(G+H)) = B+ 2A/C
C(D-F(G+H)) = 2A/C - A
D-F(G+H) = 2A - AC
-F(G+H) = A(2-C) - D
G+H = A(2-C)-D+F
G=A(2-C)-D+F-H

NUMBERS EVERYWHERE!!!!

 

[Issac]

I know but, hey ho, maths and text don't go well. I get confused easily.

Right

A+B-(C(D-F(G+H)) = B+ 2A/C
C(D-F(G+H)) = 2A/C - A
D-F(G+H) = 2A - AC
-F(G+H) = A(2-C) - D
G+H = A(2-C)-D+F
G=A(2-C)-D+F-H

NUMBERS EVERYWHERE!!!!

Wrong!

you wrote:
-F * (X) = (Y)
(X) = (Y) + F

should be:
-F * (X) = (Y)
(X) = (Y) / -F

On a little side note: what I just recently studied, having an exam in a few days:

 

[SinHarvest24]

snip.

lol, that hand-written thing you did is alien.
What exams are you studying for? college and/or in what field?




On Topic : I think i'm starting to get the hang of this thing. A few more questions would be welcomed.

 

[Issac]

snip.

lol, that hand-written thing you did is alien.
What exams are you studying for? college and/or in what field?

On Topic : I think i'm starting to get the hang of this thing. A few more questions would be welcomed.

Well, that was just "calculus 2".. a first year exam here at the universities. (I'm in my fifth year and I always black out on exams).
Studying a Master of Science and Engineering in Media Technology at Linköpings University.. (basically, the fields are: computer graphics, vizualisation and ... stuff).

A - (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D

Solve for E

 

[SinHarvest24]

A - (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D

- (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D - A

(B + 1/(C + 1/(D + 1/(E)))) = (F*G*D - A)/1

1/(C + 1/(D + 1/(E))) = (F*G*D - A)/1 - B

C + 1/(D + 1/(E)) = ((F*G*D - A)/1 - B)/-1

1/(D + 1/(E)) = ((F*G*D - A)/1 - B)/-1 - C

D + 1/(E) = (((F*G*D - A)/1 - B)/-1 - C)-1

1/(E) = (((F*G*D - A)/1 - B)/-1 - C)-1 - D

E = ((((F*G*D - A)/1 - B)/-1 - C)-1 - D)-1




Maybe it can be simplified but that is as far my knowledge allows me. Correction?

 

[SifJar]

A-B(C - D) = (E - F)/G original equation
G(A-B)(C-D) = E - F Multiply both sides by G
G(A - B)(C - D) - E = -F Take E from both sides
F = E - G(A - B)(C - D) Multiply both sides by -1
Edit: THIS IS INCORRECT!!! Spot the error and win a bowl of nothing.

The error is that when you multiplied by G, you changed A-B(C-D) to (A-B)(C-D). Your implication is that I will now receive an empty bowl

 

[Jamstruth]

A-B(C - D) = (E - F)/G original equation
G(A-B)(C-D) = E - F Multiply both sides by G
G(A - B)(C - D) - E = -F Take E from both sides
F = E - G(A - B)(C - D) Multiply both sides by -1
Edit: THIS IS INCORRECT!!! Spot the error and win a bowl of nothing.

The error is that when you multiplied by G, you changed A-B(C-D) to (A-B)(C-D). Your implication is that I will now receive an empty bowl

WRONG! It was not an impication but a statement. You now forfeit any right to your prize.

 

[SinHarvest24]

correct my answer's!!!!!!!!!!!!!!!!!!!!!!

 

[Issac]

A - (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D

- (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D - A

(B + 1/(C + 1/(D + 1/(E)))) = (F*G*D - A)/1

snip

Maybe it can be simplified but that is as far my knowledge allows me. Correction?

Nope... this is kinda wrong.

I'll simplify the bolded part:
this is what you wrote:
-1 / x = y
x = y/1

should be:

-1/x = y
x = -1/y

and then continue the same way down the line, and remember to divide and multiply at the right spot (you failed at this too in the last steps):

QUOTE(sinharvest24 @ May 27 2011, 06:33 PM) 1/(E) = (((F*G*D - A)/1 - B)/-1 - C)-1 - D
E = ((((F*G*D - A)/1 - B)/-1 - C)-1 - D)-1

this is what you wrote, simplified:
1 / x = y
x = y -1

it was a REALLY evil exercise though

 

[SinHarvest24]

So you're saying (atleast what i'm interpreting from your small eg.) that it should be something like this-

A - (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D

- (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D - A

(B + 1/(C + 1/(D + 1/(E))))) = -1(F*G*D - A)

or

(B + 1/(C + 1/(D + 1/(E))))) = -1/(F*G*D - A)

Spoiler

What should i do with that minus, does it have any significance as it is before the brackets.
Should i get rid of it by multiplying everything in brackets by -
A - (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D

- (1/(B + 1/(C + 1/(D + 1/(E))))) = F*G*D - A

 

[Cuelhu]

consider a minus as a "-1" multiplying the content inside the brackets. When you take the brackets off, you'll have to change signals based on it.

And you're wrong on something basic. If something is adding, it goes to the other side subtracting and vice-versa. If it's dividing, it multiplies the other side. Same goes if you're changing something on one side, like multiplying by something, you should make this on the other side too.

 

[Zetta_x]

I have an answer to Issac's question, scanning it and will edit this post soon.


Solve for G:

Spoiler

However, if you notice, that C is re-written in a quadratic equation, so I decided to rewrite some things:

Spoiler

 

[mcp2]

lmao Zetta, you should have done:

G= Quadratic/(C^2xF) this is the first line from your quadratic page

=> G/Quadratic = 1/(C^2xF)

=> Quadratic/G = (C^2xF)

=> Quadratic = (C^2xF)G

=> Quadratic number 2, in C. This make finding G a lot easier.

 

[Zetta_x]

Lmao you are totally right =P

 

[mcp2]

So in between this and your stats video, can I assume you're at college majoring in maths or something?

 

[Zetta_x]

I am majoring in applied mathematics with an emphasis on Statistics. Unfortunately pretty much all graduate level mathematics is more of a logic + theoretical route with very little computation (and I live to calculate numbers). I have completed a few courses in Modern (abstract) algebra and it was not that interesting for me to spend the next 4-5 years researching that kind of stuff so I went towards statistics.

In statistics, I have done 3 introductory courses, 2 courses in SAS, a course in time series, 4 courses in statistical and mathematical probability, a course in Markov Chains and Poisson Process, a few discreet mathematical courses, and some courses on regression analysis. I think that's all of them lol.