# How does this approximation work?

• Leo Liu
That something is usually the base. So in this case, ##x/h## would need to be expanded into ##x^h##, which is what you get when you take the derivative of the original expression.Please look at it again.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 ##xf

#### 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?

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

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.

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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##.

• Leo Liu
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:
$$\frac h{h+x-y}=h\cdot\frac 1{1+(h+x-y-1)}=h\cdot (1-(h+x-y-1))$$

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

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

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##.

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. :)

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.

• Leo Liu
$$\frac{h}{h+x-y}\approx 1-\frac x h+\frac y h$$
\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*}

• Leo Liu
\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.

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
$$1 - \frac xh + \frac y h \approx 1.9$$

• 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.

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.

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.

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.

• 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##.

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 ##.

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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.

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