Wednesday, April 11, 2012

The two things you need to know about Homeopathy


Since this is World Homeopathy Awareness Week, I thought I would point out the two things you should know about homeopathy.

The first is the story of its origin. Homeopathy owes its existence to a Doctor Samuel Hahnemann, who in 1789 was interested in the effect of cinchona bark, from a South American tree, known to be effective in the treatment of malaria. Hahnemann ingested the substance, and found that it gave him nausea, and other symptoms that he felt were consistent with those experienced by those infected with malaria. It occurred to him, I presume, that there was some kind of poetic validity in the idea that a substance could produce symptoms in a healthy person similar to those it could treat in a sick person; in any case, he made that 'principle of similia' the core principle of homeopathy. This one observation lead to the entire business of homeopathy, with all manner of substances being used to treat all kinds of conditions, following this principle. It also became necessary to dilute the substances beyond all measurement, but that’s another story. The point here is that the whole operation started from an observation of the effects of a known, effective, existing malaria treatment, which was extrapolated to cover virtually any substance and any condition. The work of Hahnemann is still considered extremely important, and ground-breaking, by homeopaths today.

The second thing you need to know about homeopathy: Today -- more than two hundred years after the invention of homeopathy, inspired by Hahnemann’s work with a known, effective, existing malaria treatment -- homeopathy cannot cure malaria, it cannot treat malaria, it cannot prevent malaria to any measurable degree. 

And the only conclusion one can reasonably come to, after considering these two points? It's pretty obvious, but it does not seem to bother homeopaths at all.

There’s more to know about homeopathy than that, of course, but if you just want to know if it has any value as a system of medicine, it seems to me that those two facts should suffice.

Cinchona bark is still effective as a malaria treatment, since it is a natural source of quinine, a substance of which Hahnemann was not aware. There are now better treatments.

Given malaria’s direct role in the origins of Homeopathy, it is unsurprising that homeopaths very much want to believe that it really works for malaria. In 2006, Dr Simon Singh conducted an undercover 'sting' - in which a woman approached ten practising homeopaths, telling them she was planning to travel to Central and Southern Africa, and that the anti-malarial drugs her doctor had provided made her queasy. All of them offered homeopathic remedies as 'alternative' protection. You can read about it here.



In this video, Melanie Oxley, representing the British Society of Homeopaths, in a debate discussing the situation, demonstrates the lack of responsibility and accountability in the homeopathic community. But I want to draw your attention to her comments at 3:35 in the video, as the debate is coming to a close.

She manages to get these in right at the end, so there's no time for Dr. Singh to answer. But, after a long, defensive posture of making excuses for homeopaths who tell customers their 'remedies' can prevent malaria, and making even worse excuses for why they won't be held accountable in any way -- she attempts to justify their actions. By pointing out that
"Indeed the first remedy ever proved -- as we call it -- discovered -- was a remedy for malaria. You have to remember that traditional medicines have effects."
An utterly jaw-dropping but very revealing utterance, in which she is undoubtedly referring to Hahnemann's work with cinchona. A remedy which is efficacious for treatment (complete with side effects), in its natural form, but has no effect, prophylactic or therapeutic, (beyond placebo) when prepared by homeopaths. A remedy which was presumed to work via similia -- 'like cures like', but which actually works through a non-magical biological mechanism unknown to Hahnemann. And through the twisted, backward, superstitious, self-deluding logic which pervades almost all of their thinking, homeopaths have somehow managed to believe its beneficial effect as due to homeopathic principles, whereas in fact, the preposterous homeopathic core principle of 'similia' is originally based on a false conjecture of the mechanism of this particular substance. The cart is squarely before the horse. And thus, to the homeopathic view, the effect of the cinchona bark is presumed to imply, without need for evidence, the effect of the diluted-to-nothing homeopathic preparation of the bark.

(the subject of homeopathic 'proving' I would like to leave for another time; it's more of the same kind of logic.)

So, Oxley appears to be telling us this:  because this horrible fallacy is the keystone of the origin story of homeopathy, that's enough to justify believing that homeopathy works on malaria, and enough to justify selling it to credulous clients to protect them for this serious disease. It doesn't seem to make any difference that it doesn't work, that people will come back from Africa suffering from multiple organ failures. To recognize that would to be to put the entire foundation of homeopathy in question, wouldn't it now?

By making this statement just before running out of time, she also manages to avoid answering the one question she had been brought in to answer, the question of what, if anything, the Society was planning to do about homeopaths who endanger their clients in this way. Also, she refers to 'traditional medicines', another fallacy - this reference brings to mind various plant-derived remedies, which can indeed be effective, in an attempt to claim that goodwill for homeopathy. I fail to see what's 'traditional' or 'natural' about an extreme dilution of Berlin Wall or Mobile Phone, etc.

So, have a good World Homeopathy Awareness Week. 

Further reading:

HPA issues winter malaria warning
http://www.nursingtimes.net/home/clinical-specialisms/infection-control/hpa-issues-winter-malaria-warning/5039467.article

"It has also stressed the importance of having proper certified anti-malaria medication, highlighting the fact there is no evidence that homeopathic remedies are effective at either treating or preventing the disease."


Homeopathy: what does the "best" evidence tell us?
http://www.ncbi.nlm.nih.gov/pubmed/20402610
“The findings of currently available Cochrane reviews of studies of homeopathy do not show that homeopathic medicines have effects beyond placebo.”

Homeopathy for Malaria

http://theness.com/neurologicablog/index.php/homeopathy-for-malaria/
"We can say with confidence that homeopathic products are worthless in general, and specifically for malaria prevention or treatment."

"Homeopathic Resistant Malaria"
http://www.ncbi.nlm.nih.gov/pubmed/9815426
"We therefore urge the readers to stand up against the dangerous use of homeopathic drugs and instead motivate travelers to use protective malaria prophylaxis."

Alarm as homeopathic treatments are free to make health claims without trials
http://www.dailymail.co.uk/news/article-412818/Alarm-homeopathic-treatments-free-make-health-claims-trials.html
"The Society of Homeopaths said its members are bound by a code of ethics designed to protect patients."

A Condensed History Of Homeopathy

Tuesday, April 10, 2012

Proof By Induction

This is, I believe, a first year algebra topic; I find it rather interesting since it's a way of performing a mathematical proof which appears, at first glance, to rely on circular reasoning, but in fact does not. When this topic is first presented in an algebra course, extra care needs to be taken to show what's going on, and why it's not circular.

Apologies for the bad 'text-based math'.

Let's use the usual example, which is the expression S(n), being the sum of all the integers 1 through n, with n being an integer ≥ 1.

Generally, for an inductive proof you need to know beforehand exactly what you intend to prove.
So first, we'll find S(n) with an informal discussion:
Make a table with n rows and 2 columns.
The first column has the values 1 through n.
The second column has the same values in reverse order (from n down to 1).
So, each row adds to n+1; there are n rows; the total of all the numbers in the table is thus n(n+1). The table contains two copies of the list, so
S(n) = n(n+1)/2

Now, we'll prove that by induction. To avoid making an error of deduction, we're going to define these two functions:
S(n) = the sum of the numbers from 1 to n
H(n) = n(n+1)/2
H(n) being a 'hypothesis' function; and we will forget that we already know that they are the same. I.e. we won't rely on that knowledge in the proof.

So, to prove by induction that S(n) = H(n) for all integer n ≥ 1:

  1. first prove that this is true for n=1
  2. then, prove that if it's true for n=j, then it must also be true for n=j+1.
If we manage to prove both of these, we will have proven it's true for n=2, since we can set j=1, and then the second step says "if S(1)=H(1), then also S(2)=H(2). And, thus, by setting j=2, it's proven for S(3)=H(3). And inductively -- not circularly -- we prove S(n)=H(n) for all higher integers.

The first part is easy:
S(1) = sum of { 1 } = 1
H(1) = n(n+1)/2 for n = 1, which is (1)(2)/2 = 1

So, we have proven it for n=1. We will step carefully through the second part of the proof, to avoid making any circular inferences.

Let 'j' be any value of n for which we know that S(n) = H(j).

So:
S(j) = H(j) [ given ]
but:
S(j+1) = S(j) + (j+1) [ adding the next integer to the sequence]
S(j+1) =H(j) + j+1 [ since S(j) = H(j) ]
S(j+1) = j(j+1)/2 + j + 1 [ substitute H(j) ]
S(j+1) = ( j*j + j + 2*j + 2)/2 [expand and collect terms over 2 ]
S(j+1) = ( j*j + 3*j + 2 ) /2 [collect terms]
S(j+1) = (j+1)(j+2)/2 [ factor the quadratic ]
S(k) = k(k+1)/2 [ substitute k = j+1 ]
S(k) = H(k) [ substitute H(k) ]
So, we are back where we started with S(j)=H(j)? No, because j is not just a parameter, it's the specific value for which we (by premise) know that S and H are the same. Since k=j+1, we've proven something we didn't have before. Let's eliminate k to make this clearer:

S(j+1) = H(j+1)

So we've shown, using only specific knowledge of S(n)'s properties, and the definition of H(n), that if S(n)=H(n) for n=j, it must also hold for n=j+1

So, we've proven this for all integer n ≥ 1.

If you don't like having to do the factoring to 'come up with' H(j+1) -- in a sense it seems like this requires you to anticipate the result -- you can finish like this instead:
S(j+1) = ( j*j + 3*j + 2 ) /2 [as before]
.. and note that
H(j+1) = (j+1)(j+1+1)/2 [ sub n=j+1 into H(n) ]
.. thus the difference between these is
H(j+1)-S(j+1) = ((j*j + 3*j+2) - (j*j + 3*j+2))/2 [expand, subtract, collect over 2]
H(j+1)-S(j+1) = 0 [ since all terms cancel]

Thus, again, H(j+1) = S(j+1). Somewhat less satisfying, but more amenable to 'brute force' algebra.

If these seem too 'magic' or self-fulfilling - "Can't you just do that with any function?" a good way to assure yourself that it really works is to try 'proving' something else, something which isn't right (because no, you can't just do that with any function). I'm going to replace H(n) with this 'wrong' hypothesis function
W(n) = n(n-1)+1

So, as before, we start by trying to prove W(1) = S(1):
W(1) = (1)(0) + 1 = 1 = S(1)

Hey, that works! In fact it works for n=2 as well: W(2)=S(2) = 3, I've deliberately chosen a good 'wrong' function to make it more interesting.

So, on to the second part of our (hopefully doomed-to-failure) proof. We want to show that if W(j)=S(j), then W(j+1) = S(j+1).

S(j) = W(j) [ given ]
but:
S(j+1) = S(j) + (j+1) [ adding the next integer to the list]
S(j+1) =W(j) + j+1 [ since S(j) = W(j) ]
S(j+1) = j(j-1)+1 + j + 1 [ substitute W(j) ]
S(j+1) = (j*j - j +1) + j + 1 [expand ]
S(j+1) = ( j*j +2) [collect terms]

Note, there's no error in this; it's perfectly valid; if S(j)=W(j) then S(j+1)= (j*j+2), and we know that the premise is true for j=1 and j=2 at least. But how to work this towards S(j+1)=W(j+1)? I'll use the second approach above and subtract:

W(j+1) = (j+1)(j+1-1) + 1 [substitute]
W(j+1) = j*j + j + 1 [ expand and collect terms]
... thus the difference is
W(j+1) - S(j+1) = (j*j+ j + 1) - (j*j+2 ) [difference]
W(j+1) - S(j+1) = j - 1 [simplify]

Again, this is all correct, no errors, but we have simply failed to show that W(j+1)=S(j+1) for all j. In fact, we've shown that is only true for j=1. So from this, and from W(1)=S(1), we can indeed conclude that W(2)=S(2), but from there it goes no further, the inductive proof has failed.

This amounts to testing our method of testing things; if I had been able to prove the W(n) function using a methodology - and we know W(n) is wrong - I would have no choice but to conclude that my methodology of proof was flawed.

To a mathematician, this exercise with W(n) is utterly superfluous, and you are thus unlikely to find it any textbook. Because the 'proof-by-induction' is simply a strategy for applying the general method of mathematical proofs - any particular use of it will stand (or fail) on that basis -- and of course, showing that a particular methodology fails to prove one non-true hypothesis doesn't mean it will fail to prove all others. But I'm primarily an engineer, we sometimes like to deliberately test faulty things just to be sure that the test will in fact find them to be faulty. And often the way in which things fail gives you insight into how they are supposed to work.

Etymology
I don't know exactly why this is called 'by induction'; the word seems to generally refer to something causing something else ("How can fire induce stone?"- Wormtongue), electrical e1ngineers talk about magnetic fields inducing current in conductors. in the mathematical sense, each proof 'induces' the proof of its neighbor, but there's definitely an implication that this process carries on to an infinite -- or at least arbitrarily long -- extent, and this element is not present in other uses of the word.