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Well done Rhema from Crown Woods School,
Simon from Elizabeth College, Guernsey,Trevor from Queen
Elizabeth's School, Barnet and Andrei from Tudor Vianu National
College, Bucharest, Romania, you all sent in good solutions. To
find the solution as a numerical approximation, Andrei used
interval halving and a graphical method, Trevor and Simon used the
Newton-Raphson method, Trevor used a spreadsheet for the
calculations and Simon wrote a program in C to do the
calculations.
Case (i)
If $a = b$ the equation $a^x + b^x = 1$ where $0< a, b < 1$
and $a + b < 1$ becomes $2a^x =1 $. Then $\log 2 + x \log a = 0$
so the solution can be given in three equivalent forms: $$x =
{-\log 2 \over \log a} = {\log 0.5 \over \log a } = {\log 2 \over
\log 1/a}.$$ Note we can use natural logarithms here or logarithms
to any base.
Case (ii)
Solve $a^x + b^x = 1$, where $a=1/2$ and $b=1/4$. The equation is:
$$(1/2)^x + (1/4)^x = 1,$$ Here you could substitute $y = (1/2)^x$
but Trevor multiplied by $4^x$ and used the substitution $y = 2^x$.
By Trevor's method: $$4^x/2^x + 1 = 4^x$$ So $$2^x + 1 = (2^x)^2.$$
and the equation becomes: $y + 1= y^2$ or $y^2- y- 1 = 0$. Using
the quadratic formula, the negative solution to the quadratic can
be ignored as $2^x$ is never negative, so the solution is: $$y={1+
\sqrt 5 \over 2}$$ (note this equals the golden ratio, $\phi $).
Therefore $2^x = \phi$ and so $x\ln 2 = \ln \phi$ and the solution
is: $$x= {\ln \phi \over \ln 2} ={ \ln (1 + \sqrt 5)/2 \over \ln
2}$$ giving $x = 0.69424$ to 5 significant figures
Case (iii)
This is Andrei's method using interval
halving:
Here I observe that I cannot solve the equation directly. First I
shall prove that for any $0 < a, b < 1$, with $a + b < 1$,
the equation $a^x + b^x = 1$ has a unique solution. Let $f: [0,1]
\to R, f(x) = a^x + b^x - 1$ then $$f(0) = 1 > 0, f(1) = a + b -
1 < 0$$ and $f$ is a continuous function, so $f(x) = 0$ has at
least one solution in its interval of definition. I shall now prove
it is unique. The derivative $$f'(x) = a^x \ln a + b^x \ln b <
0,$$ as $a$ and $b < 1$. So, $f$ is strictly decreasing and the
solution is unique.
For the case of the problem, $a =1/2$ and $b = 1/3$ I cannot find
the exact value of the root, but I can find a good approximation of
the solution. I shall use the interval halving method. Let $x_0$ be
the root. The first value that I choose is 1/2.
\begin{eqnarray} f(1/2) = 0.28 > 0 \Rightarrow
1/2 < x_0 < 1 & : & f(3/4) = 0.03 > 0\Rightarrow
3/4 < x_0 < 1 \\ f(7/8) = -0.07 < 0 \Rightarrow 3/4 <
x_0 < 7/8 & : & f(13/16) = -0.02 < 0 \Rightarrow 3/4
< x_0 < 13/16 \\ f(25/32) = 0.005 > 0 \Rightarrow 25/32
< x_0 < 13/16 & : & f(51/64) = - 0.007 < 0
\Rightarrow 25/32 < x_0 < 51/64\\ f(101/128) = -0.001 < 0
\Rightarrow 25/32 < x_0 < 101/128 & : & f(201/256) =
0.002 > 0 \Rightarrow 201/256 < x_0 < 101/128 \\
f(403/512) = 0.0006 > 0 \Rightarrow 403/512 < x_0 <
101/128 & : & f(707/1024) = -0.0001 < 0 \Rightarrow
403/512 < x_0 < 807/1024.\\ \end{eqnarray}
So, $0.7871 < x_0 < 0.7880$. Continuing the procedure, I
could find a better approximation for the root $x_0$.
|
Another method, though less precise, would have been a
graphical one. In this graph I have plotted f(x) and the solution
is given by the intersection with the x-axis. |
Simon used the Newton-Raphson method
and a computer to work out the answers to this part
Differentiating the function $f(x) = a^x + b^x - 1$ gives
$f'(x)=\ln a . a^x + \ln b . b^x.$ Using the Newton-Raphson
formula: $$x_{r+1} = x_r - {f(x_r)\over f'(x_r)}$$ gives $$x_{r+1}
= x_r - {a^{x_r} + b^{x_r} - 1 \over \ln a . a^{x_r} + \ln b .
b^{x_r}}.$$ As the notes allow us to use a programmable calculator,
which I interpreted to mean computer, as that is essentially what a
computer is, I coded a script in C to do this problem for me.
The approximate value for x: 0.784453067133075 (using $a=1/2,
\ b=1/3$.)
Here is Trevor's
spreadsheet for this
calculation.