some problems from previous post

Forum for the GRE subject test in mathematics.
Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

some problems from previous post

Postby Nameless » Tue Oct 14, 2008 9:00 am

Dear friends,

The following was posted very long time ago, I am posting it here. Let kill them :


Deal All, I have some more problems which need your kind help:

1) f(x)=3x^2 when x is irrational but -5x^2 when x is rational, ask f(0) is continuous? differentiable?
I think it is both continuous and differentiable at 0, but not sure the reasons

2) A similar one as above, g(x) = 1 when x is rational and e^x when x is irrational, the answer is g(x) is continuous only at 0. Why?


3) f’(0)=f’’(0)=1, g=f(x^10), ask the 11-th differentiation of g at 0.

4) When x>=0, x-x^3<=f(x)<=x, f(0) is differentiable?

5) Ring R has the following properties but R*R (pointwise) doesn’t have:
a. R is field
b. R is finite
c. R is communicative

6) find out the number of connected component of e^z where |z|=1

7) f: X->Y is continuous bijection, which are correct?
a. if X is compact , so is Y
b. if X is Hausdorff space, so is Y
c. if X is compact and Y is Hausdorff space , then inverse of f exist

Cool f’’(0) < 0 and f’(0) = 0, let T=f(0)+2f(2)+2f(4)+2f(6); I=integration of f(x) from 0 to 6; R=2f(2)+2f(4)+2f(6); sort T, I and R

9) Fair coins are tossed and when either four consecutive heads of tails appear the process stops. What is the probability of two consecutive head or tail or any one of them in one row?



#2 was solved already
I start with # 5 ( the easiest one) :D

5) Ring R has the following properties but R*R (pointwise) doesn’t have:
a. R is field
b. R is finite
c. R is communicative


I have no idea with this question :D

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm

Postby CoCoA » Tue Oct 14, 2008 1:07 pm

1) both functions approach zero as x->0, so any sequence f(x_n)->0 as x->0, therefore continuous at 0; I think since same thing happens to individual derivative of each function, it is differentiable also.

4) What is f(x) when x<0? You can do limits and algebra with the supplied information, but don't you need at open interval around a point for differentiability at the point? That is, there is a one-sided limit, but is there a one-sided derivative?

5) Does this mean R*R={a^2:a in R}? Then F_2*F_2 = F_2 so it has all these properties.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Tue Oct 14, 2008 1:16 pm

Hi CoCoA,
I just dug up a previous post, and think that it will be helpful if we can solve them together . So I'm not sure about the correctness of given information in each question.

And we do have one-side derivative but it is not common.

For #5 . I guest : R*R=RxR - the Castesian product of R- i.e RxR={(a,b) : a, b in R}. Hence, if we understand that RxR is the Castersian product of R then a ) is not correct for all R, even R is a field since we can choose (0,1) in R such that (0,1) does not have its inverse


Let kill other questions :D

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby moo5003 » Tue Oct 14, 2008 2:20 pm

#2 Is only continuous at f(0) = 1 since for any other value x in the domain you can find a sequence namely a sequence of irrationals that will not have 1 as the limit and therefore will not agree with other sequences namely a rational one. Obviously if you want to make this a proof of rigor you can use epsilon/delta language but the general idea is above.

#3 After writing out the chain rule for the first 3 derivatives I believe the answer to be f''(0)*10! = 10!

g(x) = f(x^10)
g'(x) = f'(x^10)*(10x^9)
g''(x) = f''(x^10)(10x^9)^2 + f'(x^10)*[10*9*x^8]
g^(3)(x) = f'''(x^10)(10x^9)^3 + f''(x^10)(2)(10x^9)[10*9x^8] + f''(x^10)(10x^9)[10*9x^8] + f'(x^10)(10*9*8x^7)

Edit: After working out the derivative again, Cocoa is correct this should be zero since every term has far to high power of x to be eliminated except for the last term which has to few.

Question
For number 5 how sure are we that it means the Reals? Because RxR would not be a field nor would it be finite.
Last edited by moo5003 on Tue Oct 14, 2008 5:10 pm, edited 3 times in total.

User avatar
lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

Postby lime » Tue Oct 14, 2008 2:49 pm

g''(x) = f''(x^10)(10x^9) + f'(x^10)*[10*9*10x^8]

Your second derivative is incorrect.

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby moo5003 » Tue Oct 14, 2008 2:50 pm

Fixed, I just wrote it down wrong.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Tue Oct 14, 2008 4:05 pm

Question
For number 5 how sure are we that it means the Reals? Because RxR would not be a field nor would it be finite.



Problem #5 depends on what ring we are working on.
But the question is "what properties the ring RxR does not have for any choice of R- ( R is no need to be real number )

For example :
let R=Z_2 then R is finite and commutative so is RxR , hence b) and c) are true in this case BUT RxR is NOT a field Since (0,1) is in RxR has no inverse so a) in is correct

let kill #6, 7

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm

Postby CoCoA » Tue Oct 14, 2008 4:48 pm

#3
g(x) = f(x^10)
g'(x) = f'(x^10)*(10x^9)
g''(x) = f''(x^10)(10x^9) + f'(x^10)*[10*9*x^8]
g^(3)(x) = f'''(x^10)(10x^9) + f''[10*9x^8] + f''(x^10)(10x^9)[10*9x^8] + f'(x^10)(10*9*8x^7)

Careful -
g''(x)= (f'(x^10))'*(10x^9) + f'(x^10)*(10x^9)' =[f''(x^10)*10x^9]*10x^9 + f'(x^10)*(10*9*x^8)
= f''(x^10)*10^2*x^18 + f'(x^10)*10*9x^8;
have to remember to use the chain rule on [(f^(n)(x^10)]'. In particular, you see that all but the last term in the sum will have too many powers of x to go away in the higher derivs, and evaluated at zero will be zero. Thus we should get
g^(10)(x)=(terms with powers of x that are (much?) greater than 2) + f^(10)(x^10)*10!. The eleventh deriv is just a chain rule again, no product rule, so it is (terms with powers of x) + 10!*f^(11)(x^10)*10x^9.

Evaluated at zero, this is: zero.

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby moo5003 » Tue Oct 14, 2008 5:11 pm

Went through it again, you are correct.

CoCoA
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Joined: Wed Sep 03, 2008 5:39 pm

Postby CoCoA » Tue Oct 14, 2008 8:24 pm

#6: 1 I think. Let A={z:|z|=1}. A is connected, and f(z)=e^z is a continuous function, so f(A) is a connected set.

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm

Postby CoCoA » Tue Oct 14, 2008 8:33 pm

#7 - we are given f is a bijection, so I think part C should ask if inverse of f is continuous

ieoi
Posts: 4
Joined: Wed Oct 15, 2008 10:12 am

Postby ieoi » Sat Nov 01, 2008 6:22 am

I agree with CoCoA about #6, #7.

especially, if #7 is asking inverse of f is continuous, then it is true.

In order to check that inverse of f is continuous, we require that f maps open set to open set, equivalently f maps closed set to closed set. (f is bijective)

I will use several Theorems.
1) if f is continuous, compact set maps to compact set.
2) closed subset of compact space is compact
3) In Hausdorff space, compact subset is closed.

Since X is compact, A, closed subset of X, is compact. Then, f(A) is compact. Moreover, Y is Hausdorff space, so compact subset f(A) of Y is closed. In conclusion, inverse of f is continuous.




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