How does this approximation work?

[Leo Liu]

Homework Statement

$$\frac{h}{h+x-y}\approx 1-\frac x h+\frac y h$$

Relevant Equations

.

My physics textbook does the approximation in the homework statement. Here, x and y are variables and are much smaller than h. I attempted to figure out why it is valid with $(1+x)^-1\sim 1-x$. However, after trying to convert the initial equation into 1+x form, I obtained $h(1-(h+x-y-1))$ which does not equal to $1-\frac x h+\frac y h$, the desired result. Could anyone tell me why?

 

[fresh_42]

What do you know about $(x-y)^2\,?$ And did you try to resolve the quotients?

 

[Doc Al]

Homework Statement:: $$\frac{h}{h+x-y}\approx 1-\frac x h+\frac y h$$
Relevant Equations:: .

My physics textbook does the approximation in the homework statement. Here, x and y are variables and are much smaller than h. I attempted to figure out why it is valid with $(1+x)^1\sim 1-x$. However, after trying to convert the initial equation into 1+x form, I obtained $h(1-(h+x-y-1))$ which does not equal to $1-\frac x h+\frac y h$, the desired result. Could anyone tell me why?

I assume you mean $(1+x)^{-1}\sim 1-x$.

Just divide the numerator and denominator of your initial expression by $h$ and you'll have a form you should recognize. (Not clear to me what you did.)

 

[Leo Liu]

Just divide the numerator and denominator of your initial expression by h and you'll have a form you should recognize.

$$=\frac{h/h}{(h+x-y)/h}=\frac{1}{1+(x/y-y/h)}=1-x/h+y/h$$
I get it. Thanks!

(Not clear to me what you did.)

I was trying to convert the denominator into $1-(h+x-y+1)$, where $h+x-y+1=u$, and therefore the result is $1-u$. The h in the numerator was separated from the fraction to make the numerator equal to 1.

 

[vela]

Homework Statement:: $$\frac{h}{h+x-y}\approx 1-\frac x h+\frac y h$$
Relevant Equations:: .

My physics textbook does the approximation in the homework statement. Here, x and y are variables and are much smaller than h. I attempted to figure out why it is valid with $(1+x)^-1\sim 1-x$. However, after trying to convert the initial equation into 1+x form, I obtained $h(1-(h+x-y-1))$ which does not equal to $1-\frac x h+\frac y h$, the desired result. Could anyone tell me why?

Let's assume for a moment that all the variables have units of length. The original fraction then is clearly unitless, but the expression you derived is all messed up unit-wise. That's usually a sign you made a mistake.

Generally, the trick is to factor out the big factor in the denominator. For example,
$$h + x = h\left(1+\frac x h \right).$$ And $x$ being small compared to $h$ means $x/h \ll 1$.

 

[vela]

I was trying to convert the denominator into 1−(h+x−y+1), where h+x−y+1=u, and therefore the result is 1−u.

After fixing the sign mistake, this would only work if $h$, $x$, and $y$ were unitless. Adding and subtracting 1 wouldn't make sense otherwise. Also, the approximation might not be very good because you can't infer that $u \ll 1$ from $x\ll h$ and $y\ll h$.

 

[Leo Liu]

That's usually a sign you made a mistake.

Yeah I've indeed made a dimension error, but I don't know where I got wrong. This is my reasoning:
$$\frac h{h+x-y}=h\cdot\frac 1{1+(h+x-y-1)}=h\cdot (1-(h+x-y-1))$$

 

[fresh_42]

I get that the approximation is equivalent to $x\approx y.$

 

[Charles Link]

It works whenever $|x-y|<<|h|$.
There are a couple of ways that this can happen.

 

[fresh_42]

It works whenever $|x-y|<<|h|$.
There are a couple of ways that this can happen.

The relation to $h$ is irrelevant. We only need that $h\neq 0$ which is implicitly given anyway. If $1\ll |x-y|$ then it is wrong, regardless what $h$ is. Simply multiply the entire thing with the common denominator and $h$ vanishes. If $h$ is very large such that $x-y$ doesn't matter, then we can as well assume $x\approx y$.

 

[Charles Link]

The relation to $h$ is irrelevant. We only need that $h\neq 0$ which is implicitly given anyway. If $1\ll |x-y|$ then it is wrong, regardless what $h$ is. Simply multiply the entire thing with the common denominator and $h$ vanishes. If $h$ is very large such that $x-y$ doesn't matter, then we can as well assume $x\approx y$.

Please look at it again. I don't agree with your statement. We all make mistakes, but I don't think I made one here. :)

 

[vela]

Yeah I've indeed made a dimension error, but I don't know where I got wrong.

I answered this in post 6, but you probably didn't see it before you posted. In any case, just to expand on this a bit more, say $x$ and $h$ have units of length. The expression $h+1$ doesn't make sense unit-wise because you're adding a number, which has no dimensions, to $h$ which has units of length. On the other hand, the quantity $x/h$ is dimensionless, so an expression like $1 + x/h$ does make sense.

Also, in general, when you expand in a series, the quantity you're expanding in powers of needs to be dimensionless.

 

[fresh_42]

$$\frac{h}{h+x-y}\approx 1-\frac x h+\frac y h$$

\begin{align*}
\Longleftrightarrow \;&h^2 \approx (h+x-y)(h-x+y)=h^2-(x-y)^2\\
\Longleftrightarrow \;&0\approx -(x-y)^2\\
\Longleftrightarrow \;&x\approx y
\end{align*}

 

[Charles Link]

\begin{align*}
\Longleftrightarrow \;&h^2 \approx (h+x-y)(h-x+y)=h^2-(x-y)^2\\
\Longleftrightarrow \;&0\approx -(x-y)^2\\
\Longleftrightarrow \;&x\approx y
\end{align*}

I think it might be a poor technique to subtract the same thing from both sides of an "approximately" expression. See also post 11.

 

[vela]

With $h=0.11$, $x=2.0$, and $y=2.1$ so that $x \approx y$, you get
$$\frac{h}{h+x-y} = 11$$ which isn't close to
$$1 - \frac xh + \frac y h \approx 1.9$$

 

[fresh_42]

I think it might be a poor technique to subtract the same thing from both sides of an "approximately" expression.

Since when? I subtracted the same amount on both sides! If $a$ is close to $b$ so is $a+c$ close to $b+c$.

\begin{align*}
\frac{h}{h+x-y}&\approx 1-\frac x h+\frac y h\\
\Longleftrightarrow \;&\frac{h}{h+x-y} =1-\frac x h+\frac y h +\varepsilon^2 \\
\Longleftrightarrow \;&h^2 = (h+x-y)(h-x+y)+\varepsilon^2 \cdot h \cdot ((h+x-y))\\
\Longleftrightarrow \;&h^2 =h^2-(x-y)^2+\varepsilon^2 (h^2+h(x+y))\\
\Longleftrightarrow \;&0= -(x-y)^2+\varepsilon^2 (h^2+h(x+y))\\
\Longleftrightarrow \;&|x-y| =\varepsilon\cdot \sqrt{h^2+h(x+y)}
\end{align*}
hence all what it takes is that $\varepsilon$ is close to zero, which is expressed by the original approximation.

 

[fresh_42]

With $h=0.11$, $x=2.0$, and $y=2.1$ so that $x \approx y$, you get
$$\frac{h}{h+x-y} = 11$$ which isn't close to
$$1 - \frac xh + \frac y h \approx 1.9$$

If $h\approx \varepsilon$, i.e. the error is as big as the quantities are, you are right.

 

[vela]

I think it might be a poor technique to subtract the same thing from both sides of an "approximately" expression.

I don't think that's the problem. @fresh_42 started with an approximation that assumes the condition you stated, namely $|x-y| \ll h$. In particular, since the approximation is only to first-order, that's equivalent to saying
$$ \left(\frac{x-y}{h}\right)^2 \approx 0,$$ which certainly holds if $x \approx y$ regardless of $h$. But you shouldn't really compare the quantity $x-y$ to 0 in this problem as these are dimensionful quantities. It only makes sense to compare $\frac{x-y}{h}$, which is dimensionless, to 0.

 

[fresh_42]

You are right, we need $\varepsilon \ll h$. So if we use the error margin then $h$ doesn't vanish (post #16). I assumed indeed that my error is close enough to zero so that multiplications wouldn't affect it. Sorry.

 

[Charles Link]

You can have $1000 \approx 1001$, but you do not necessarily have $0 \approx 1$. Most of the time it is ok to treat the approximately expression with the same algebraic rules as the equality expression, but this is one case where it didn't work.

 

[fresh_42]

I thought of $a\approx b$ as $a=b+(1/n)$ with large $n$.

 

[Charles Link]

I thought of $a\approx b$ as $a=b+(1/n)$ with large $n$.

Post 13 is where it is incorrect. It took me a minute or two to see why you can't necessarily subtract $h^2$ from both sides of the approximately expression. Instead you need to divide both sides by $h^2$.
Finally, when we have $1 \approx 1-(x-y)^2/h^2$, the next step is to say $(x-y)^2/h^2<< 1$.

 

[fresh_42]

Post 13 is where it is incorrect. It took me a minute or two to see why you can't necessarily subtract $h^2$ from both sides of the approximately expression. Instead you need to divide both sides by $h^2$.

Well, whether there is a difference between $1000\approx 1001$ and $0\approx 1$ is a matter of definition or context ;-) But as mentioned, too many small $\varepsilon$ in my life.