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never mind,

it doesnt have odd symmetry

This setup

2A = (1+e^B)e^C + (1+e^(-B))e^(-C)

is similar to drift-diffusion in a transistor .

We couldn't get it into closed form unless we made some assumptions.

is there anything special about

C1, MD/H and MR/H

like MR/H+C1 ≈ C1

this would be nice

**Atled**- Replies: 1

Anyone know how to do regression analysis of a shifted exponential curve.

here is the data

1 0.997706596

2 0.243081214

3 0.079426264

4 0.03319443

5 0.016202707

6 0.00969521

7 0.007108719

8 0.004852038

9 0.003415406

10 0.002462571

11 0.001977777

the equation should look like a^((-x+1)/b) i think.

I cant seem to do it with my calculator it doesn't handle the shift.

Thanks

f(x)= [(x^3) - 3(x^2) + 4]/x^2

divide through by x^2

you get

f(x) = x - 3 + 4x^(-2)

use the power rule f(x) = x^n f'(x) = n*x(n-1)

f'(x) = 1 - 8*x^(-3)

then factor out x^(-3)

[x^3 - 8]*x^(-3) = [(x^3)-8]/x^3

cosh(x) can be approximated with a taylor expansion.

its the same as the expansion for cosine

1 + x²/2 + x^4/ 24+ ....

1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[a]Sin) <- something seems wrong here

Sin needs a variable.

if a = 0 b= pi/4

you get 1/2

but you might be able to get something smaller than that.

You might try taking partial derivatives of Cos[2(a-b)]+Cos[2(a+b)] wrt a & b and look for extrema. You could also try to write an excel spread sheet which tries different combinations of a & b and use the goal seek function.

I am in a hurry but I will try to work on it later.

About myself...

25/m/Ca

EE student at UC Davis

hobbies: cycling, skating, surfing although there are no oceans near davis

A . (A + 3B +7C) = A.A + 3(A.B) + 7(A.C)

A.B and A.C are given

A . (A + 3B +7C) = A.A +(3*3) +(7*-1)

A . (A + 3B +7C) = A.A + 2

whats A.A

X.Y = |X|*|Y| * cos(θ)

A is **PARALLEL** to A there for θ = 0

or

The angle between A and A is 0

cos(0) = 1

A.A = 4

A . (A + 3B +7C) = 4 + 2 =6

hope that helps

edit: previously i said normal

∫ ( ln(x)/x ) dx or ∫ (dx/x) *ln(x)

let u= ln(x)

du/dx = 1/x or du = dx/x

replace dx/x with du

and ln(x) with u

then you are left with ∫ u du == u²/2

thats where the 2 comes from.

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