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Differential Calculus

Blazing goIITian

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14 Aug 2009 11:45:16 IST

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651

Rigorous Question.........

None

Well these qs are given with the sole purpose of seeing the method and not answer:

A:)

f(x) = 0 ; x is irrational

f(x)=1 ; x is rational

comment on the continuity of F(x)

B:)

Proove by counter-example that if

F : I equals and belongs to R

F(x) is a continous and differentiable function and that f'(x) exists for all x belonging to I

then f'(x) may not be continuous.

C:)

X(n) is a sequence and if

X(n+1) = (X(n)³ + 2)/7

and 0<x(1) < 1

then proove that X(n) converges and find its limit and proove that it is equal to a root of x³ -7x + 2 = 0

D:)

find the limit of the sequence

X(n) = aⁿ / n

under various value of a where a is a postive real number.

E:)

Proove that A.B = {a_x b_x , a_yb_y , a_z b_z}

is a scalar quantity

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Comments (6)

bladeX -rise of a new sun

Blazing goIITian

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15 Aug 2009 09:58:44 IST

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Hint for 1st one use e-d definition

Anirudh

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16 Aug 2009 09:14:56 IST

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Attempted answer for part (a) : Let f(x) be continuous for all rational x. Then the limit of f(x) has to be 1,by the definition of continuity.This means |f(x)-1|

Anirudh

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16 Aug 2009 09:15:41 IST

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Let f(x) be continuous for all rational x. Then the limit of f(x) has to be 1,by the definition of continuity. This means |f(x)-1|

Anirudh

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16 Aug 2009 09:16:13 IST

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Attempted answer for part(a).Let f(x) be continuous for all rational x. Then the limit of f(x) has to be 1,by the definition of continuity.

This means |f(x)-1|<e for all x1 (not equal to x) such that |x1-x|<d.

But here |f(x)-1|<e for any positive value of e for all positive rational values of x1.

this means d and e are not related, thus the limit of the function is not one .

We have arrived at a contradiction ,thus the function is discontinuous.

Let f(x) be continuous for all irrational x. we arrive at a similar contradiction, thus proving f(x) is discontinuous for irrational x as well. This means f(x) is discontinuous for all real x.

Anirudh

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16 Aug 2009 09:20:55 IST

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Sorry i tried to post the proof before and it didn't come,that's the reason for so many messages.

bladeX -rise of a new sun

Blazing goIITian

Joined: 12 Dec 2007

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16 Aug 2009 10:53:34 IST

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First I will post the answer to the 1st qs. Anirudh thanks for trying and you were very near to the answer.

Let x₀ be a rational and if the function F(x) : which is a Drichlet Function is continuous here then

| x - x0 | < d implies |F(x)- F(x0)|<e

F(x0) = 1

|x-xo|<d implies |F(x)-1|<e

e can be anything so let e <1

now between (x0 - d , x0 + d) we have a irrational point due to density of rational point and as d is not zero.

so let that point be Xc.

|F(Xc) - 1 | < e < 1

F(Xc) = 0

|1| < e < 1

which is a contradiction

similarly we can proove for the other case when x0 is irrational

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