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Suppose that $$ b_1 = const .f_1(x) \left( b_2 + \frac{a} {f_2(x)} \right) $$ becomes $$ b_1 = const. b_2 $$ because $f_1(x) \rightarrow 1$ and $f_2(x) \rightarrow \infty$ when $x \rightarrow \infty$. Do you agree that it allows us to write $$ const.b_2 = const .f_1(x) \left( b_2 + \frac{a} {f_2(x)} \right) ?$$

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no, because what if $x$ was $5$ (or anything else) then $f_2(x)$ wouldn't be $\infty$ and $f_1(x)$ wouldn't be $1$ and the equality wouldn't hold, but you could write "const.$b_2 = \lim_{x\to\infty}$ const.$f_1(x)(b_2 + \frac{a}{f_2(x)})$" –  Deven Ware Sep 23 '11 at 4:16
    
If $b_1$, $b_2$, $a$ are constants (independent of $x$), and $f_1$ and $f_2$ are continuous, and the given equation is supposed to hold for all $x$, then it's okay. –  Ted Sep 23 '11 at 4:36
    
Thanks a bunch, Deven. This appeared to be something akin to the boundary value problems in differential equations. How would you comment the seeming similarity of the above incorrect substitution to the boundary value problems? –  ganzewoort Sep 23 '11 at 4:39
    
Ted, but the first equation shows that $b_1$ depends on $x$, doesn't it? Therefore, Deven's reply holds. –  ganzewoort Sep 23 '11 at 4:42
    
@ganzewoort I'm sorry but I don't know exactly what a "boundary value problem" is since I've never studied differential equations but hopefully someone else can help you on that :) –  Deven Ware Sep 23 '11 at 4:56
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1 Answer 1

The substitution is not directly valid because:

b1=Constant.b2

is not an absolute equality. It is a constrained equality by the constraint x --> infinity, so you can't just use it without that constraint.

You can see an example if you use this specific case: f1(x) = (1+1/x) and f2(x)=x

Hope this helps.

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Thanks, Emmad to you too. –  ganzewoort Sep 23 '11 at 5:03
    
Emmad, I quess your reply also answers my question about the seeming similarity with boundary value problems in differential equations. In the boundary value problems the equalities given as boundary conditions are absolute, unlike the case at hand, right? –  ganzewoort Sep 23 '11 at 5:08
    
Well, I guess the case is not the same. To my limited knowledge, in boundary problems of Diff. Equations, the result is a function that obeys the boundary conditions (constraints). That result is not the generalized result, however, it obeys the original condition under the given constraints only. A good example may be the one in en.wikipedia.org/wiki/Boundary_value_problem in this case, y(x)=2 Sin(x) is not the general result but a special one satisfying y(0)=0 and y(pi/2)=2 –  Emmad Kareem Sep 23 '11 at 5:58
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