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gerryliyana

  • one year ago

W- Lambert function

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  1. abb0t
    • one year ago
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    U ok? If it's a medical emergency, dial 9-1-1.

  2. gerryliyana
    • one year ago
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    I want to ask u about how to derive the 3rd Equation (On picture below), becomes the4th equation (on picture below) by using W-Lambert function ???

  3. abb0t
    • one year ago
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    I used to know this when I took an enzyme kinetics course and this derivation is long. About half a page - 1 page typed out.

  4. abb0t
    • one year ago
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    Maybe you can read about it on the article you found it on? http://mvputz.iqstorm.ro/upload/LogisticEnzymeKinetics_Paper1.pdf

  5. abb0t
    • one year ago
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    It's a good paper :)

  6. gerryliyana
    • one year ago
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    i have been reading this paper, and I did not find what i want.., would u kindly help me?? do you still have a reference about this derivation (complete with W-Lambert function)

  7. abb0t
    • one year ago
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    I will look through my notes from my kinetics class. If I find it, I'll post it.

  8. gerryliyana
    • one year ago
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    ok thank u :)

  9. gerryliyana
    • one year ago
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    @UnkleRhaukus

  10. UnkleRhaukus
    • one year ago
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    hmm , i haven't heard of the W- Lambert function before

  11. gerryliyana
    • one year ago
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    do u have any idea @oldrin.bataku ??

  12. oldrin.bataku
    • one year ago
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    W(x) is defined such that W(x) e^W(x) = x

  13. gerryliyana
    • one year ago
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    yes.., i know.., i did read it on wikipedia. but I'm a little bit confused how to derive it (3rd equations) :)

  14. gerryliyana
    • one year ago
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    I'm looking for some easy derivation of the above, step by step., :)

  15. oldrin.bataku
    • one year ago
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    $$V_{max}t=([S_0]-[S](t))-K_M\ln\left(\frac{[S](t)}{[S_0]}\right)\\\ln\left(\frac{[S](t)}{[S_0]}\right)=\frac{[S_0]-[S](t)-V_{max}t}{K_M}\\\frac{[S](t)}{[S_0]}=\exp\left(\frac{[S_0]-[S](t)-V_{max}t}{K_M}\right)=\frac{e^\frac{{[S_0]-V_{max}t}}{K_M}}{\frac{e^{[S](t)}}{K_M}}$$ $$[S](t)e^{[S](t)}=\frac{S_0}{K_M}e^{\frac{[S_0]-V_{max}t}{K_m}}=\frac{S_0}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_m}\right)$$ $$[S](t)=W\left(\frac{S_0}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_m}\right)\right)$$

  16. oldrin.bataku
    • one year ago
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    It's just algebra :-p

  17. abb0t
    • one year ago
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    It is just algebra :) intense algebra.

  18. gerryliyana
    • one year ago
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    i thank all of u guys :)

  19. gerryliyana
    • one year ago
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    thank u so much @oldrin.bataku :)

  20. oldrin.bataku
    • one year ago
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    $$V_{max}t=([S_0]-[S](t))-K_M\ln\left(\frac{[S](t)}{[S_0]}\right)\\\ln\left(\frac{[S](t)}{[S_0]}\right)=\frac{[S_0]-[S](t)-V_{max}t}{K_M}\\\frac{[S](t)}{[S_0]}=\exp\left(\frac{[S_0]-[S](t)-V_{max}t}{K_M}\right)=\frac{e^\frac{{[S_0]-V_{max}t}}{K_M}}{e^\frac{[S](t)}{K_M}}$$$$\frac{[S](t)}{K_M}e^\frac{[S](t)}{K_M}=\frac{[S_0]}{K_M}e^{\frac{[S_0]-V_{max}t}{K_M}}=\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_M}\right)$$$$\frac{[S](t)}{K_M}=W\left[\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_M}\right)\right]$$$$[S](t)=K_MW\left[\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]-V_{max}t}{K_M}\right)\right]$$Sorry, I made a few really dumb errors the first time! http://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics#Determination_of_constants

  21. gerryliyana
    • one year ago
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    ok.., no problem, thank u @oldrin.bataku ;)

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