The temperature T(x) at each point x on the surface of Mars (a sphere) is a continuous function. Show that there is a point x on the surface such that T(x)=T(−x)

Baladdaa9 2022-07-23 Answered
The temperature T ( x ) at each point x on the surface of Mars (a sphere) is a continuous function. Show that there is a point x on the surface such that T ( x ) = T ( x )
(Hint: Represent the surface of Mars as { x R 3 : | | x | | = 1 }.)
Consider the function f ( x ) = T ( x ) T ( x )
So.....
I consider an unit sphere is locating at the origin of a x y z-plane.
As | | x | | = 1, I can say with r a d i u s = 1 = x 2 + y 2 + z 2
To find there is a point T ( x ) = T ( x ), we use the formula f ( x ) = T ( x ) T ( x ) and show somehow f ( x ) will equal to 0 ???
It will be a point in the upper hemisphere and another point with the exactly opposite vector (if using i j k plane) on the lower hemisphere
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

tykoyz
Answered 2022-07-24 Author has 17 answers
Hint: Pick some point x. If you are lucky, f ( x ) = 0, but probably this does not occur. In the other case, what can you say about the relation between f ( x ) and f ( x )? Now connect x and x using a path along the sphere...

We have step-by-step solutions for your answer!

Bernard Boyer
Answered 2022-07-25 Author has 5 answers
If f ( x ) = T ( x ) T ( x ) is zero, we are done. Assume there does not exist an x that makes f ( x ) zero. Then f ( x ) > 0 or < 0 for all x. If say f ( x ) > 0 is never satisfied, then replacing x by x we get a contradiction. Similarly if f ( x ) < 0 never holds. So there must be points for which f ( x ) > 0 and points for which f ( x ) < 0. The intermediate value theorem then implies there must exist a point y at which f ( y ) = 0 (contradiction).

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-07-18
My textbook Elementary Classical Analysis claims that by Darboux's theorem (the intermediate value theorem for derivatives), if a function f : R R has a nonzero derivative on R , then is f strictly monotonic (i.e., either f ( x ) > 0 on R or f ( x ) < 0 on R ).
The claim is definitely true if f s domain were instead a closed interval, but since R is open, I don't understand why Marsden's claim should be true.
asked 2022-09-21
Let h : [ 0 , 1 ] R be continuous. Prove that there exists w [ 0 , 1 ] such that
h ( w ) = w + 1 2 h ( 0 ) + 2 w + 2 9 h ( 1 2 ) + w + 1 12 h ( 1 ) .
I tried several things with this problem. I first tried a new function
g ( x ) = h ( x ) x + 1 2 h ( 0 ) + 2 x + 2 9 h ( 1 2 ) + x + 1 12 h ( 1 ) = h ( x ) ( x + 1 ) ( 1 2 h ( 0 ) + 2 9 h ( 1 2 ) + 1 12 h ( 1 ) ) .
Evaluating g ( 0 ) and g ( 1 ) didn't really work.

Since I want to find a point x 0 where g ( x 0 ) > 0 and another x 1 where g ( x 1 ) < 0, I started thinking how I can make g ( x 0 ) > 0,or
h ( x 0 ) ( x 0 + 1 ) ( 1 2 h ( 0 ) + 2 9 h ( 1 2 ) + 1 12 h ( 1 ) ) > 0 ,
or
h ( x 0 ) x 0 + 1 > 1 2 h ( 0 ) + 2 9 h ( 1 2 ) + 1 12 h ( 1 ) .
Since the LHS is the gradient from the point ( 1 , 0 ) to ( x 0 , h ( x 0 ) ), I need to somehow prove that that
1 2 h ( 0 ) + 2 9 h ( 1 2 ) + 1 12 h ( 1 )
is always less than the steepest gradient from ( 1 , 0 ) to ( x 0 , h ( x 0 ) ). However, I am stuck here. Any help?
asked 2022-07-20
Let f : [ 0 , 1 ] R be continuous with f ( 1 ) = f ( 0 ). Prove that if h ( 0 , 1 2 ) is not of the form 1 n , then there does not necessarily exist | x y | = h satisfying f ( x ) = f ( y ). Provide an example that illustrates this using h = 2 5 .
So I was given the hint that I can use a modified s i n function, however I'm not really sure how I would go about that. Preferably, an example not using that would be great.
My thoughts so far are that I use a proof by contradiction saying that x , y [ 0 , 1 ] such that f ( x ) = f ( 0 ) and | x y | = h. And I need to get to the point where h = 1 n , which would create the contradiction (I assume that's the endpoint?). However, how would I use h = 2 5 to prove that? Is it trivial in that h = 1 n is false? Also, it isn't given that n N , so maybe I shouldn't assume that?
asked 2022-05-21
I'm aware that R is not the only set that satisfies the least upper bound property, the p-adics do also. Does the intermediate value theorem also hold in the p-adics then?

What are the spaces where the intermediate value theorem hold?
asked 2022-08-19
Let X be a compact metric space (feel free to impose more conditions as long as they're also satisfied by spheres) and F : X × [ 0 , 1 ] R a continuous function such that

1. F ( x , 0 ) > 0 for all x X
2. F ( x , 1 ) < 0 for all x X

Then, for each x X, let t 0 ( x ) [ 0 , 1 ] be the smallest such that F ( x , t 0 ( x ) ) = 0. Is t 0 : X [ 0 , 1 ] a continuous function?

A friend suggested that I applied the Maximum theorem, but to show that the relevant correspondence is lower semicontinuous I need to prove the following statement:

If x n x and F ( x , t ) = 0, there is a subsequence x n k and a sequence t k such that F ( x n k , t k ) = 0 and t k t. This doesn't seem very obvious or even true, but I'm not sure.

Any suggestions?
asked 2022-07-11
Given continuous function g : [ a , b ] n R n R . By Weistress g has a max and a min.

Can I also conclude its image contains all values in-between this maximum and minimum?

I need this result to complete a proof but cannot seem to find a generalisation of the Intermediate Value Theorem to R n .
asked 2022-09-23
I am currently doing a problem that asks me to use the intermediate value theorem to show that x 1 / 3 = 1 x lies between 0 and 1. I want to start by evaluating the function at 0 and 1, but it seems the function is undefined because if you plug in 1, you get 1 = 0. This doesn't seem right. Is my interpretation correct?

New questions