# Homework Help: Approximation: The right way to do it

1. Jun 3, 2016

### Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

ΔE = hf

3. The attempt at a solution

f = ΔE/h = $k\left [ \frac{1}{(n-1)^2} - \frac{1}{n^2} \right ]$

= $\frac{k(2n-1)}{n^2(n-1)^2}$

= $\frac{k}{n^4}(2n-1)(1-\frac{1}{n})^{-2}$

= $\frac{2kn}{n^4}(1-\frac{1}{2n})(1+\frac{2}{n} +$ higher powers $)$

≈ $\frac{2k}{n^3}(1-\frac{3}{2n})$ . If I neglect the second term in the bracket , then I get option 3) , but I doubt whether I can do that because I need to retain atleast first order terms .

how should I proceed ?

Any help is appreciated .

Thanks

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2. Jun 3, 2016

### haruspex

Why do you think you need to retain the -3/2n term (which, incidentally, is wrong)?

3. Jun 3, 2016

### Vibhor

It should be ≈ $\frac{2k}{n^3}(1+\frac{3}{2n})$ . Don't we retain the first order terms while making approximations ?

4. Jun 3, 2016

### haruspex

It depends on the approximation. You keep as many terms as necessary to get an answer that is sufficiently accurate for purpose.
In many cases, e.g. in developing differential calculus, the zeroth order terms cancel, so it is essential to keep first order terms. In the present problem, the zeroth order term is enough.

5. Jun 3, 2016

### Vibhor

Ok .
Could you please explain with examples , at least two

6. Jun 3, 2016

### haruspex

Limit problems illustrate it. What is limit of (1-cos(x))x-2 as x tends to 0? When you expand the cos(x) as a power series, you have to keep the x2 term, but no further. But if you wanted the limit of (1-cos(x))x-1 you would only need to keep the zeroth and first order terms. Sometimes you find you did not keep enough and have to start again.

In a thread I recently got in an argument over, people who should have known better (including the problem setter) were effectively keeping some second order terms and discarding others. That's a no-no. It can lead you into thinking you have a valid solution when you do not.

7. Jun 3, 2016

### Vibhor

I think in both the cases we need to write cos(x) ≈ 1 - x2/2 .

What do you mean by "you would only need to keep the zeroth and first order terms " ?

8. Jun 3, 2016

### Ray Vickson

For large $n>0$ we have
$$\frac{1}{(n-1)^2} = \frac{1}{n^2} \left(1 - \frac{1}{n} \right)^{-2} = \frac{1}{n^2} \left( 1 + \frac{2}{n} + \cdots \right) \doteq \frac{1}{n^2} + \frac{2}{n^3}$$

9. Jun 3, 2016

### haruspex

If we keep only the zeroth and first order terms of the cos(x) expansion in (1-cos(x))x-1 we get (1-1+(0)x+O(x2))x-1=0+O(x), which tends to 0.

10. Jun 3, 2016

### Vibhor

We have to multiply this by $\frac{k(2n-1)}{n^2}$

$≈ k \left ( \frac{2}{n^3} + \frac{3}{n^4} - \frac{2}{n^5}\right )$

Now , since $\frac{2}{n^3}$ is the biggest and the most dominant term , hence we neglect the other terms and keep only the $\frac{2}{n^3}$ .

Is this what you are trying to convey ?

11. Jun 3, 2016

### Ray Vickson

The way you are doing it requires the factor $k(2n-1)/n^2$, but not the way I am doing it. Look again.

12. Jun 4, 2016

### haruspex

The sign is wrong. Vibhor corrected it in post #3.

13. Jun 4, 2016

### haruspex

Ray, I see no real difference between your method and Vibhor's. Isn't it just a matter of the order of the steps? If you take one more term in your expansion you will get Vibhor's original result (with the sign correction).

14. Jun 4, 2016

### Ray Vickson

Well, I was hoping to get the OP to look at things more simply. It is a lot easier to get
$$f/k = \frac{1}{(n-1)^2} - \frac{1}{n^2}$$
directly as
$$f/k = \left( \frac{1}{n^2} + \frac{2}{n^3} + \frac{3}{n^4} + \cdots \right) - \frac{1}{n^2} \\ = \frac{2}{n^3} + \frac{3}{n^4} + \cdots$$
than by writing
$$f/k = \frac{2n-1}{n^2} \left( \frac{1}{n^2} + \frac{2}{n^3} + \frac{3}{n^4} + \cdots \right)$$
and then doing more algebra plus collecting terms, etc. The OP did ask for the right way to do it.

15. Jun 4, 2016

### nrqed

Oops. Gotcha. I was so focused on checking the 3/2 that I did not pay attention to the sign. Sorry.

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