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Find the number of solutions to the equation $$2011^x+2012^x+2013^x-2014^x=0$$

The answer seems to be zero, but I have no idea why. Please avoid considering complex solutions and other scary things.

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A solution is $x\approx1226.7$ according to Wolfram Alpha. –  Eul Can Aug 4 at 9:46
    
When X=0, the result is 2, no? –  JoeTaxpayer Aug 4 at 15:45

4 Answers 4

up vote 8 down vote accepted

Starting equation is equivalent to this one:

$$ \left(\frac{2011}{2014}\right)^x + \left(\frac{2012}{2014}\right)^x + \left(\frac{2013}{2014}\right)^x =1. $$

Denote $a=\frac{2011}{2014}, b=\frac{2012}{2014}, c=\frac{2013}{2014}$.

Function $f(x)=a^x+b^x+c^x-1$ is continuous and decreasing (since $\ln a$, $\ln b$,$\ln c$ are negative):

$$f'(x) = a^x\ln a + b^x \ln b+c^x\ln c<0.$$

$f(0)=2$, $f(+\infty)=-1$, so there is only $1$ solution.

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The answer can't be zero solutions.

Let $f(x)=2011^x+2012^x+2013^x-2014^x$

Obviously $f(x)$ is continuous (sum of continuous functions), and $f(0)=2$. But what about the behavior as $x\to\infty$?

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The $2014^X$ seems to dominate the other three, implying the l.h.s. becomes negative, sunsequently implying that there is one solution for positive reals. What about the negative reals? –  Shubham Aug 4 at 9:56
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Compare just $2011^x$ and $2014^x$ for negative real $x$. Then use the fact that $2012^x$ and $2013^x$ are both $>0$ for all real $x$. –  Silynn Aug 4 at 10:03

By Laguerre's extension of Descartes' rule of sign,

The number of real roots of Dirichlet polynomials of the form $$\alpha_1 \beta_1^x + \alpha_2 \beta_2^x + \cdots + \alpha_n \beta_n^x \quad\text{ subject to }\quad \begin{cases}\alpha_1, \ldots, \alpha_n \ne 0\\\beta_1 > \beta_2 > \cdots \beta_n > 0\end{cases} $$ is no more than number of sign changes in the finite sequence $(\alpha_1, \ldots, \alpha_n)$.

In addition, counting multiplicity, the difference between the number of real roots and number of sign changes is an even number.

If we apply this to the Dirichlet polynomial $$2014^x - 2013^x - 2012^x - 2011^x,$$ the number of sign changes is $1$ and hence the number of real roots is $1$.

For more info, you can either look up the original paper (in French)

  • E Laguerre, Sur la théorie des équations numériques, J. Math. Pures et Appl. 9 (1883)

or a modern introduction of same subject

  • G.J.O Jameson, Counting zeros of generalized polynomials: Descartes' rule of signs and Laguerre's extensions, (Math. Gazette 90, no. 518 (2006), 223-234).
    An online copy can be found here.
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This kind of equation does not show solutions which could be expressed analytically and only numerical methods, such as Newton, should be used.

But before, you could notice that this function varies extremely fast and, in opinion, it should be better to rewrite the equation as $$\log(2011^x+2012^x+2013^x)=\log(2014^x)$$ If there were only one term, it will be the intersection of two straight lines.

Starting with a "reasonable" guess $x_0$, it will be updated according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, let us write $$f(x)=\log(2011^x+2012^x+2013^x)-\log(2014^x)$$ As said, the problem is to start with a reasonable value. So, let us try with $3 \times 2012^x-2014^x=0$ which gives $x_0=\frac{\log (3)}{\log \left(\frac{1007}{1006}\right)} \simeq 1105.75$ and now apply Newton.

The successive iterates are then : $1225.43$, $1226.69$ which is the solution for six significant figures.

Using this transform makes the convergence very fast. Let us try with $x_0=0$; the iterates of Newton method are :$1105.66$, $1225.43$, $1226.69$.

As you can see, all the trick is the switch to logarithms.

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Unfortunately, I am not aware of Newton's numerical methods. This question turned up in our review test and we were required to just analytically tell how many solutions were possible. –  Shubham Aug 4 at 10:27
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In such a case plot what I called $f(x)$ and you will see that it is almost a straight line with negative slope since $2014^x$ varies much faster than any other term. There is only one possible solution. –  Claude Leibovici Aug 4 at 10:29

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