Algebra: Solving for an Immediate Value To Calculate A Desired Result
Introduction
Considering the following problem:
We have to equations:
F = f(X)
A = a(X)
Assume that X is close to zero (0) as possible. Therefore, we are only considering one root.
We are given the value of F and we are tasked to find A. In order to solve for A, we must solve for F first.
Example 1:
f(X) = 2 *X
a(X) = 1/(X^3 + 3)
F = -0.04
Solving for X:
-0.04 = 2 * X
-0.02 = X
Calculating A:
A = 1/((-0.02)^3 + 3)
A = 0.3333342222
Example 2:
f(X) = e^(X^2 - 0.5)
a(X) = ln(X + 1)/(ln X + 1)
F = 10
10 = e^(X^2 - 0.5)
ln 10 = X^2 -0.5
X^2 = ln 10 + 0.5
(we'll only use the principal root in this example)
X = √(ln 10 + 0.5)
X ≈ 1.674092319
Substituting X into a(X) to get:
A ≈ 0.6491313602
See you next time,
Eddie
All original content copyright, © 2011-2019. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
Introduction
Considering the following problem:
We have to equations:
F = f(X)
A = a(X)
Assume that X is close to zero (0) as possible. Therefore, we are only considering one root.
We are given the value of F and we are tasked to find A. In order to solve for A, we must solve for F first.
Example 1:
f(X) = 2 *X
a(X) = 1/(X^3 + 3)
F = -0.04
Solving for X:
-0.04 = 2 * X
-0.02 = X
Calculating A:
A = 1/((-0.02)^3 + 3)
A = 0.3333342222
Example 2:
f(X) = e^(X^2 - 0.5)
a(X) = ln(X + 1)/(ln X + 1)
F = 10
10 = e^(X^2 - 0.5)
ln 10 = X^2 -0.5
X^2 = ln 10 + 0.5
(we'll only use the principal root in this example)
X = √(ln 10 + 0.5)
X ≈ 1.674092319
Substituting X into a(X) to get:
A ≈ 0.6491313602
See you next time,
Eddie
All original content copyright, © 2011-2019. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
