Login

Welcome, Guest. Please login or register.

March 29, 2024, 12:35:48 am

Author Topic: "Algebra"  (Read 767 times)  Share 

0 Members and 1 Guest are viewing this topic.

Jefferson

  • Forum Regular
  • **
  • Posts: 93
  • Respect: 0
"Algebra"
« on: February 15, 2019, 11:06:47 pm »
0
Could someone help me with this algebra question (in attachment)? I can't seem to follow the solution, especially at the step when they plus and minus the same term, then the negative term disappears. What follows is quite odd too.
Thanks!
« Last Edit: February 15, 2019, 11:13:34 pm by Jefferson »

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: "Algebra"
« Reply #1 on: February 16, 2019, 10:16:41 am »
+3
Equation 2 - A nastier-than-usual chain rule computation.

Equation 3 - Multiplied the terms out.

Equation 4 - Plus/minus in \( bN_0\). Presumably this is where the problems began occurring. Note that the reason for doing this step is because we now have a \( bN_0 + (k-bN_0)e^{-kt}\) appearing in the numerator, which is exactly what's in the denominator (albeit squared). This is actually a fairly common trick in the real world when proving results related to logistic growth - adding and subtracting the extra term.
Note, of course, that due to the extra \( -bN_0\) term we had to introduce, we can't cancel out any numerators against denominators. The next step proceeds to allow this.

Equation 5 - Presumably this is what your main question is. Note that if we (cleverly) expand the numerator,
\[ k^2N_0 [bN_0 + (k-bN_0) e^{-kt} - bN_0] = k^2 N_0[bN_0 + (k-bN_0)e^{-kt}] - bk^2N_0^2 .\] This is what they do here. Note that not only do they expand the numerator, but after expanding they tear the fraction apart into two separate fractions. This is where the \( - \frac{bk^2N_0^2}{[bN_0+(k-bN_0)e^{-kt}]^2}\) term comes from.

Equation 6 - In the first fraction, we can cancel out a \( bN_0 + (k-bN_0)e^{-kt}\) term in the numerator and denominator now, In the second fraction, we just convert into a more appropriate form.

Equation 7 - The final step relies on us going back to the question. We were told that
\[ N = \frac{kN_0}{bN_0 + (k-bN_0) e^{-kt}}. \]
So if we ever see whatever's on the RHS (which, if you look at equation 6 we certainly do), we can sub that back in for \(N\).

Jefferson

  • Forum Regular
  • **
  • Posts: 93
  • Respect: 0
Re: "Algebra"
« Reply #2 on: February 16, 2019, 10:56:03 am »
0
Equation 2 - A nastier-than-usual chain rule computation.

Equation 3 - Multiplied the terms out.

Equation 4 - Plus/minus in \( bN_0\). Presumably this is where the problems began occurring. Note that the reason for doing this step is because we now have a \( bN_0 + (k-bN_0)e^{-kt}\) appearing in the numerator, which is exactly what's in the denominator (albeit squared). This is actually a fairly common trick in the real world when proving results related to logistic growth - adding and subtracting the extra term.
Note, of course, that due to the extra \( -bN_0\) term we had to introduce, we can't cancel out any numerators against denominators. The next step proceeds to allow this.

Equation 5 - Presumably this is what your main question is. Note that if we (cleverly) expand the numerator,
\[ k^2N_0 [bN_0 + (k-bN_0) e^{-kt} - bN_0] = k^2 N_0[bN_0 + (k-bN_0)e^{-kt}] - bk^2N_0^2 .\] This is what they do here. Note that not only do they expand the numerator, but after expanding they tear the fraction apart into two separate fractions. This is where the \( - \frac{bk^2N_0^2}{[bN_0+(k-bN_0)e^{-kt}]^2}\) term comes from.

Equation 6 - In the first fraction, we can cancel out a \( bN_0 + (k-bN_0)e^{-kt}\) term in the numerator and denominator now, In the second fraction, we just convert into a more appropriate form.

Equation 7 - The final step relies on us going back to the question. We were told that
\[ N = \frac{kN_0}{bN_0 + (k-bN_0) e^{-kt}}. \]
So if we ever see whatever's on the RHS (which, if you look at equation 6 we certainly do), we can sub that back in for \(N\).

Ah alright, that makes sense.
Thank you so much!