Inequalities Problem
- Thread starter anaseer
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- Apr 12, 2005
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It requires a little exploration. Generally, at the very least, two things are required. Dive in and have no fear. It's an inequality. You won't break it.
All those increasing exponents. I'm thinking induction might be useful, but I don't even have a good conjecture until I play with it a little.
Maybe 99 is just too big to see. Let's try something a bit more tractable. Maybe 3?
1 + x + x^2 + x^3 = hmmmmm... I think I see it.
1(1+x^2) + x(1+x^2) = (1+x)(1+x^2) -- Since 1+x^2 is ALWAYS positive, the only way to get the product less than zero is 1+x<0
Let's try 5.
1 + x + x^2 + x^3 + x^4 + x^5 = 1(1 + x^2 + x^4) + x(1 + x^2 + x^4) = (1+x)1 + x^2 + x^4) -- yup. There it is again. The big one is greater than zero. Solve 1+x<0
Finally, go back to 99.
1 + x + x^2 + ... + x^99 = 1(1+x^2+x^4+...+x^98) + x(1+x^2+x^4+...+x^98) = (1+x)(1+x^2+x^4+...+x^98) -- i believe. How about you?
Does it work for final even exponents? Is there an add exponent for which it doesn't work?
See, exploration can be fun!!
All those increasing exponents. I'm thinking induction might be useful, but I don't even have a good conjecture until I play with it a little.
Maybe 99 is just too big to see. Let's try something a bit more tractable. Maybe 3?
1 + x + x^2 + x^3 = hmmmmm... I think I see it.
1(1+x^2) + x(1+x^2) = (1+x)(1+x^2) -- Since 1+x^2 is ALWAYS positive, the only way to get the product less than zero is 1+x<0
Let's try 5.
1 + x + x^2 + x^3 + x^4 + x^5 = 1(1 + x^2 + x^4) + x(1 + x^2 + x^4) = (1+x)1 + x^2 + x^4) -- yup. There it is again. The big one is greater than zero. Solve 1+x<0
Finally, go back to 99.
1 + x + x^2 + ... + x^99 = 1(1+x^2+x^4+...+x^98) + x(1+x^2+x^4+...+x^98) = (1+x)(1+x^2+x^4+...+x^98) -- i believe. How about you?
Does it work for final even exponents? Is there an add exponent for which it doesn't work?
See, exploration can be fun!!
Hello, anaseer!
\(\displaystyle \text{Note that if }x\text{ is positive, the left side greater than 1 ... and is positive.}\)
\(\displaystyle \text{Since the left side is negative, then }x\text{ must be negative.}\)
. . \(\displaystyle \text{Hence, }1-x\text{ is positive.}\)
\(\displaystyle \text{Multiply both sides by positive }(1-x):\)
. . \(\displaystyle (1-x)\left(1+x+x^2 + x^3 + \hdots + x^{99}\right) \;<\;(1-x)\cdot0\)
. . . . . . . . . . . . . . . . . . . . .\(\displaystyle 1 - x^{100} \;<\;0\)
. . . . . . . . . . . . . . . . . . . . . . \(\displaystyle -x^{100} \;<\;-1\)
. . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle x^{100} \;>\;1\)
. . . . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle |x| \;>\:1\)
. . \(\displaystyle \text{Hence: }\;x \:>\:1 \;\;\text{ or }\;\;x \:<\:-1\)
\(\displaystyle \text{Since }\text{ is negative: }\;x \:<\:-1\)
\(\displaystyle \text{Note that if }x\text{ is positive, the left side greater than 1 ... and is positive.}\)
\(\displaystyle \text{Since the left side is negative, then }x\text{ must be negative.}\)
. . \(\displaystyle \text{Hence, }1-x\text{ is positive.}\)
\(\displaystyle \text{Multiply both sides by positive }(1-x):\)
. . \(\displaystyle (1-x)\left(1+x+x^2 + x^3 + \hdots + x^{99}\right) \;<\;(1-x)\cdot0\)
. . . . . . . . . . . . . . . . . . . . .\(\displaystyle 1 - x^{100} \;<\;0\)
. . . . . . . . . . . . . . . . . . . . . . \(\displaystyle -x^{100} \;<\;-1\)
. . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle x^{100} \;>\;1\)
. . . . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle |x| \;>\:1\)
. . \(\displaystyle \text{Hence: }\;x \:>\:1 \;\;\text{ or }\;\;x \:<\:-1\)
\(\displaystyle \text{Since }\text{ is negative: }\;x \:<\:-1\)