Questions on functions and supremums and infimums.

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In summary, in the first conversation, the first, third, and last statements are correct, while the others are incorrect. In the second conversation, it is shown that f(x)=cos(x)+sqrt(3)sin(x) is a bounded function with a supremum of 2, and a minimum of -2. The approach used is correct, but an alternative approach using the derivative can also be used.
  • #1
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1)let g:B->A f:A->B be functions, and gof is the identity function, prove/disprove the following:
g=f^-1;f is onto;f is 1-1;g is onto;g is 1-1.
iv'e showed that the first the third and the last are correct, and the others are wrong, am i right?

2)prove that f(x)=cos(x)+sqrt(3)sin(x) is a bounded function and find its supremum and infimum.

well to prove that it's a bounded function was easy, it's bounded by:
-1-sqrt(3)<=f(x)<=1+sqrt(3)
now to find the supremum and infimum, the supremum is 2, and i have proven it this way:
first we need to show that it's an upper bound, i.e cos(x)+sqrt(3)sinx<=2
we have 2-sqrt(3)sin(x)>=1 and we can suppose that cos(x) is positive, bacause the upper bound should be greater also from the least value of f(x) which can be obtained when cos(x)<0, so we can suppose that cos(x)>0 (is this assumption valid?) then we have sqrt(1-sin^2x)<=2-sqrt(3)sin(x) 1-sin^2x<=4+3sin^2x-4sqrt(3)sin(x)
4sin^2x-4sqrt(3)sin(x)+3>=0
2sin^2x-sqrt3)^2>=0 which is ofcourse correct.

now we need to show that for every c>=f(x) 2<c, (in order to show that's least upper bound), so we suppose that c<2, then c>=cos(x)+sqrt(3)sin(x)
again we assume cos(x)>0 so we have c^2-2csqrt(3)sin(x)+3sin^2(x)>=1-sin^2x 4sin^2x-2sqrt3sin(x)+c^2-1>=0 [/tex]\delta=4*3c-16c^2+16=-4c^2+16<=0[/tex] so we have (2-c)(2+c)<=0 c>=2 or c<=-2
but this clearly contradicts our assumptions, is my appraoch correct?


p.s
just something I am interested about, how do you prove that 0.123456789101112... (all the natural numbers) is irrational?
 
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  • #2
It is known that the decimal expansion of any rational number is eventually repeating.
 
  • #3
what about the other two questions?
 
  • #4
loop quantum gravity said:
1)let g:B->A f:A->B be functions, and gof is the identity function, prove/disprove the following:
g=f^-1;f is onto;f is 1-1;g is onto;g is 1-1.
iv'e showed that the first the third and the last are correct, and the others are wrong, am i right?
In order that a function HAVE an inverse, it must be both 1-1 and onto, and its inverse must be both 1-1 and onto. However, consider this example: A= {a, b}, B= {x,y,z} . f(a)= x, f(b)= y. g(x)= a, g(y)= b, g(z)= b. then gof(a)= g(x)= a, gof(b)= g(y)= b so gof is the identity function. Is g= f-1? What about the other questions?

2)prove that f(x)=cos(x)+sqrt(3)sin(x) is a bounded function and find its supremum and infimum.

well to prove that it's a bounded function was easy, it's bounded by:
-1-sqrt(3)<=f(x)<=1+sqrt(3)
now to find the supremum and infimum, the supremum is 2, and i have proven it this way:
first we need to show that it's an upper bound, i.e cos(x)+sqrt(3)sinx<=2
we have 2-sqrt(3)sin(x)>=1 and we can suppose that cos(x) is positive, bacause the upper bound should be greater also from the least value of f(x) which can be obtained when cos(x)<0, so we can suppose that cos(x)>0 (is this assumption valid?) then we have sqrt(1-sin^2x)<=2-sqrt(3)sin(x) 1-sin^2x<=4+3sin^2x-4sqrt(3)sin(x)
4sin^2x-4sqrt(3)sin(x)+3>=0
2sin^2x-sqrt3)^2>=0 which is ofcourse correct.

now we need to show that for every c>=f(x) 2<c, (in order to show that's least upper bound), so we suppose that c<2, then c>=cos(x)+sqrt(3)sin(x)
again we assume cos(x)>0 so we have c^2-2csqrt(3)sin(x)+3sin^2(x)>=1-sin^2x 4sin^2x-2sqrt3sin(x)+c^2-1>=0 [/tex]\delta=4*3c-16c^2+16=-4c^2+16<=0[/tex] so we have (2-c)(2+c)<=0 c>=2 or c<=-2
but this clearly contradicts our assumptions, is my appraoch correct?
Actually, in this case, the sup and inf are the maximum and minimum values so taking the derivative and setting it equal to 0 is a good start.
You can also note that the "root mean square" of the coefficients, 1 and [itex]\sqrt{3}[/itex] is [itex]\sqrt{1+ 3}= 2[/itex]. Factoring out a 2, you have [itex]f(x)= 2(\frac{1}{2}cos(x)+ \frac{\sqrt{3}}{2}sin(x))[/itex]. Now use the fact that sin(x+ a)= sin(a)cos(x)+ cos(a)sin(x). Here, we have [itex]sin(a)= \frac{1}{2}[/itex] and [itex]cos(a)= \frac{\sqrt{3}}{2}[/itex] so [itex]a= \frac{\pi}{6}[/itex].
[tex]f(x)= 2 sin(x+ \frac{\pi}{6})[/tex]
Aren't the max and min (sup and inf) of that obvious?
 
  • #5
In order that a function HAVE an inverse, it must be both 1-1 and onto, and its inverse must be both 1-1 and onto. However, consider this example: A= {a, b}, B= {x,y,z} . f(a)= x, f(b)= y. g(x)= a, g(y)= b, g(z)= b. then gof(a)= g(x)= a, gof(b)= g(y)= b so gof is the identity function. Is g= f-1? What about the other questions?
no bacuse we don't have a reverse image for f(g(z)) doesn't equal z.
so it means we don't have it so still i think that f isn't onto, but it's 1-1, cause:
if f(x1)=y1=y2=f(x2) in B, then x1=g(f(x1))=g(f(x2))=x2.
from the example you showed clearly g isn't one to one, but it's onto, so i guess i need to prove that g is onto.
then for every x in A there exists y in B such that g(y)=x when y=f(x).
thanks, halls.
 
  • #6
halls, am i right in what i wrote?
im having second thoughts about it. (-:
 

Related to Questions on functions and supremums and infimums.

1. What is a function?

A function is a mathematical relationship between two variables, where each input has a unique output. It can be represented by an equation, table, or graph.

2. What is a supremum?

A supremum, also known as a least upper bound, is the smallest number that is greater than or equal to all the elements in a set. It is denoted by sup(S).

3. What is an infimum?

An infimum, also known as a greatest lower bound, is the largest number that is less than or equal to all the elements in a set. It is denoted by inf(S).

4. How do you find the supremum and infimum of a set?

To find the supremum, you need to identify the largest element in the set. To find the infimum, you need to identify the smallest element in the set. If the set has no upper or lower bound, then the supremum and infimum do not exist.

5. What is the relationship between supremum and infimum?

The supremum and infimum are related as follows: if sup(S) = a and inf(S) = b, then for any element x in the set S, b ≤ x ≤ a. This means that the infimum is the lower bound and the supremum is the upper bound of the set S.

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