Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the number of positive solutions to

$$ (x^{1000} + 1)(1 + x^2 + x^4 + \cdots + x^{998}) = 1000x^{999}? $$

I tried to solve it. First I used by using sum of Geometric Progression. Then the equation becomes too complicated and is in the power of 1998. How can I get the number of positive solutions with that equation?

Thanks in advance.

share|improve this question
    
Its x to the power 1000 , and x to the power 998 , and x to the power 999. i am sorry i tried but it was not coming in proper format. –  vikiiii Mar 30 '12 at 3:26
    
When exponents have multiple digits, you need to put curly braces (i.e., { } ) around them. –  Matthew Conroy Mar 30 '12 at 3:30
    
Since $(1-x^2)(1 + x^2 + x^4 + … + x^{998})= 1-x^{1000}$ we have $1 + x^2 + x^4 + … + x^{998} = \frac{1-x^{1000}}{1-x^2}$ so $$(x^{1000}+1)(1 + x^2 + x^4 + … + x^{998})=\frac{(1+x^{1000})(1-x^{1000})}{1-x^2}=\frac{1-x^{2000}}{1-x^2}$$ hence we are looking for solutions to $\frac{1-x^{2000}}{1-x^2}=1000x^{999}$. Equivalently, we want $$0=(1-x^{2000})-(1000x^{999})(1-x^{2000})=1000x^{2999}-x^{2000}-1000x^{999}+1.$‌​$ –  Alex Becker Mar 30 '12 at 3:52
1  
It should rather be $(1 - x^{2000} - 1000x^{999} + 1000x^{1001}) = 0.$ –  Suresh Mar 30 '12 at 4:08

3 Answers 3

For a cleverness free solution:

The number of positive roots counting multiplicity is $2$: the root $1$ is repeated twice.

You can use Descartes' rule of signs, which is a general method which can be useful sometimes.

Your equation (after multiplying by $x^2-1$) is

$$ P(x) = x^{2000} - 1000x^{1001} + 1000x^{999} - 1 = 0$$

which has $3$ sign changes.

So the number of positive roots is either $1$ or $3$ (counting multiple roots multiple times).

The derivative is $$P'(x) = 2000 x^{1999} - 1000\times 1001 x^{1000} + 1000 \times 999 x^{998}$$

Notice that $P(1) = 0$, $P'(1) = 0$.

Since $1$ is a root, and also of the derivative, the number of positive roots is $3$, counting $1$ at least two times.

But since we introduced an extraneous positive root by multiplying by $x^2 -1$, the number of positive roots of your original equation, counting multiplicity is $2$.

The second derivative

$$P''(x) = 2000\times 1999 x^{1998} - 1000\times1000\times1000 x^{999} + 1000 \times 999 \times 999$$

and $P''(1) = 0$. Thus the root $1$ is of multiplicity $2$.

This also applies to the equation $x^{2n} - n(x^{n+1} - x^{n-1}) - 1 = 0$ for $n \gt 1$.

See Also: Sturm's theorem.

share|improve this answer
    
how it can be solved as A.M. >= G.M. as you wrote in the below comment? –  vikiiii Mar 30 '12 at 19:18
    
@vikiiii: Tao's answer has it: "since $1 + x^2 + \dots + x^{2m-2} \ge m x^{m-1}$ (algeraic average is great than blah...)". –  Aryabhata Mar 30 '12 at 19:30

Consider$$(x^{m} + 1)(1 + x^2 + x^4 + \cdots + x^{m-2}) = mx^{m-1}?$$ by multiplying $x^2-1$ in tow sides of equation,we have $$(x^{m} + 1)(x^m-1)=mx^{m-1}(x^2-1)$$, Last we have $$x^{2m} - m(x^{m + 1} - x^{m - 1}) - 1 = 0 \tag{#}$$ Factorization $$\begin{align*} &x^{2m} - m(x^{m + 1} - x^{m - 1}) - 1\\ &=(x^{2m}-1)-mx^{m-1}(x^2-1)\\ &=(x^2-1)(1+x^2+x^4+\cdots x^{2m-2})-mx^{m-1}(x^2-1)\\ &=(x^2-1)(1+x^2+x^4+\cdots x^{2m-2}-mx^{m-1})\\ \end{align*}$$ since $1+x^2+x^4+\cdots x^{2m-2}\geq mx^{m-1}$(algebraic average is great than geometry average $\forall x>0$)if and only if $1=x^2= \cdots$ ,then$$1+x^2+x^4+\cdots x^{2m-2}= mx^{m-1}$$ so the root of equation(#)is $1$ and -$1$,but sine we first multiply with$(x^2-1)$, so maybe there are some extraneous roots in it,substitute $1$ and$-1$ for original equation,we conclude that the root of original equation is $1$.

share|improve this answer
1  
Just simply multiplying out $(1+x^{1000})(1 + x^2 + \dots + x^{998})$ gives us $1 + x^2 + x^4 + \dots + x^{998} + x^{1000} + \dots + x^{1998}$. Now we can apply AM $\ge$ GM. There is no need to bring $x^2 -1$ in. –  Aryabhata Mar 30 '12 at 5:22
    
@Aryabhata you are right. –  noname1014 Mar 30 '12 at 5:28
1  
There is no need to even multiply...You can apply the AM-GM inequality directly to $1+x^{1000}$ and $1 + x^2 + \dots + x^{998}$... –  N. S. Mar 30 '12 at 5:36

I think a family of problems to consider would be $x^{2n} - n(x^{n + 1} - x^{n - 1}) - 1 = 0.$ Substituting $ n = 1000$ we get your case.

share|improve this answer
    
This is more suited to be a comment. –  Aryabhata Mar 30 '12 at 5:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.