Archive for June, 2017

I am unworthy of a Miele washing machine

June 14, 2017

So.

I wanted a washing machine delivered on a Wednesday, when I am at home.

I rang up Miele on Saturday 3 June an arranged for one to be delivered on 14 June.

They sent me many texts saying it would be delivered between 1130 and 1330 on 14 June.

They sent me an email saying it would be delivered between 1130 and 1330 on 14 June.

On 14 June they rang me to arrange delivery. I said they were due to deliver between 1130 and 1330. They said they had not been able to find one or had entered two orders or something. Maybe I could have one by special courier if I waited in a couple of days.

I said it would have been better to let me know beforehand.  I cancelled the order.

Their washing machines may be perfectly adequate. It looks like I will never know now!

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Joseph Henry Blackburne lived here

June 4, 2017

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50 Sandrock Road

J. H. Blackburne dominated British chess during the second half of the 19th century, and at one point he was the world’s second most successful player.

He is perhaps best known for losing heavily to Wilhelm Steinitz and for taking it badly, but according to the biography by Tim Harding he was living in 9 Whitbread Road, Brockley at the time of the 1901 census, later moving to 45 Sandrock Road and then number 50 in the same road, where he died on 1 September 1924.

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45 Sandrock Road

So number 45 has changed over the years more radically than number 50, but not as radically as the place in 9 Whitbread Road.

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Presumed site of 9 Whitbread Road

Now then, it is known that Blackburne was bombed-out during a German raid in the First World War, but the dates are such that it’s unlikely the view above came into being that way.

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J. H. Blackburne (1841-1924)

Now then Steinitz apparently lived in Shoreditch, which only adds to my suspicions that he was really Karl Marx on his day off…

How popular is Russia in the UK?

June 4, 2017

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We are sometimes asked how popular Russia is among people in Britain.

In 2015, Chatham House reported results from surveys in 2012 and 2014 asking respondents which countries they felt especially favourable and unfavourable towards.   Above, we derive an overall favourability score from (favourable – unfacourable) , take the average of  this 2012 and 2014 and plot it against the change between the two dates.

We see that Russia is out on its own in terms of being unpopular and becoming more so, followed at a respectful distance by Israel.  This is presumably due to a series of notable events in this period: BP, Pussy Riot, Greenpeace, Ukraine, none of which were well received.

Interestingly enough, a BBC poll of about the same period of attitudes to different states in a sample of 22 countries showed Russia as among the least popular but without the same deterioration.

Stationary points of f(x)=x+√x

June 4, 2017
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The curves in the pic are f(x) and f'(x).

Question: The function is f(x)=x+√x. Can you find its min or max value? If you calculate f'(x) and put it to 0, you get x=1/4. But draw the curve and there is no such max/min there.

One suggestion:   Clearly f'(x) has two asymptotes at x=0 and y=1 and so can never be 0 for any real value of x. But why does differentiating lead to the result x=1/4? I agree that f(x) = x-√x has a clear minimum there, but it’s a different function. My answer was that since there’s an infinity at x=0 f'(x) is undefined, but I’m not convinced that’s the answer.

OK, so √x is defined as the positive value of x^(1/2).

Let’s see if we can differentiate it:

x = √x√x  where x, √x >= 0.

D(x)  = 1 = 2√x*D(x) => D(x) = 1/(2√x)

So, f'(x) = 1 + 1/(2√x)

if f'(x) =0, then √x = -1/2.  But √x is >= 0 by definition, so there is no solution.

Let’s try to be a little more systematic while staying within the bounds of high school maths.

We consider the function  g(x,c) = x + x^(1/2)

where x^(1/2) = (√x)*(-1)^c

and we choose c to be either 0 or 1.

So g'(x) = 1 + (1/2√x)*(-1)^c

Setting g'(x) = 0 =>

√x = (-1/2)*(-1)^c

But since  √x is positive, c must be 1 rather than 0 for this to have a solution.

So by accepting a solution here we force the original function to be g(x) = x – √x.