How Can Sequences and Series Prove a Real Solution to the Equation x^11+2x^5=2?

[pablito21]

A={xεR:X^11+2X^5<2} let a=supA By choosing a suitable sequence of elements of belonging to A and which tends to a as n->inf, or otherwise, show that a^11+2a^5=<2.Choose another sequence this time of all real numbers not belonging to A to show that a^11+2a^5>=2 and hence show that a^11+2a^5=2,so the equation x^11+2x^5=2 has a real solution

any help would be really appreciated!how can i solve it

 

[Office_Shredder]

For the first sequence, use a-1/n. a-1/n<a for all n, so is in A (since the function is increasing, all elements less than a are in A). When you plug a-1/n into the function, you get something like:

a^11 + 2a^5 +1/n*(bunch of stuff) < 2

Similiarly for taking a+1/n for the second part gives

a^11 + 2a^5 + 1/n*(bunch of stuff) >=2

Try to work it from there

 

[tacman]

To solve this problem, we first need to understand what a sequence and series are. A sequence is a list of numbers in a specific order, while a series is the sum of the terms in a sequence. In this problem, we are given a set A, which is defined as all real numbers x such that x^11+2x^5<2. We are also given a=supA, which means that a is the least upper bound of the set A.

To show that a^11+2a^5=<2, we can choose a sequence of elements from A that tends to a as n->inf. One such sequence could be {a-1/n}, where n is a positive integer. As n increases, the terms in this sequence get closer and closer to a, and as a result, the terms in the sequence also satisfy the condition x^11+2x^5<2. Therefore, we can say that a^11+2a^5=<2.

On the other hand, to show that a^11+2a^5>=2, we can choose a sequence of all real numbers not belonging to A. One such sequence could be {a+1/n}, where n is a positive integer. As n increases, the terms in this sequence get closer and closer to a, but they do not satisfy the condition x^11+2x^5<2. Therefore, we can say that a^11+2a^5>=2.

Combining these two results, we can conclude that a^11+2a^5=2. This means that the equation x^11+2x^5=2 has a real solution, which is a. This solution satisfies the condition x^11+2x^5<2, and it is also the least upper bound of the set A. Therefore, we can say that a is the unique solution to the equation x^11+2x^5=2.

In conclusion, by choosing suitable sequences of elements from A and not belonging to A, we were able to show that a^11+2a^5=<2 and a^11+2a^5>=2, respectively. This helped us to prove that a^11+2a^5=2 and that the equation x^11+2x^5=2 has a real solution, which is a.