- #1

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help with the following taylor series:

(y+dy)^0.5

Thanks

(y+dy)^0.5

Thanks

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- Thread starter chenrim
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- #1

- 17

- 0

help with the following taylor series:

(y+dy)^0.5

Thanks

(y+dy)^0.5

Thanks

- #2

BruceW

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- #3

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and by definition : fy(y)dy= P{y<Y<=y+dy}

i can substitute Y with (X)^2 and then i take the square root from both sides and get :

P{ sqrt(y)<X<= sqrt(y+dy) }}

now i want to show the PDF of X by the definition so i want to develop the right side of the inquality into a taylor series.

sqrt(y+dy) This is the term i want to develop into a taylor series . y and dy both are numbers. i tried to use Taylor's formula but couldn't get it right,

Hope you understand my question

thanks

- #4

BruceW

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- #5

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Yes you right X also should be taken negative.

i didnt understand the step between line 2 and 3

https://drive.google.com/file/d/0B4wgc0vIE7CCUmpXLXJ0TEtIeUU/view?usp=sharing [Broken]

i didnt understand the step between line 2 and 3

https://drive.google.com/file/d/0B4wgc0vIE7CCUmpXLXJ0TEtIeUU/view?usp=sharing [Broken]

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- #6

Stephen Tashi

Science Advisor

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i didnt understand the step between line 2 and 3

https://drive.google.com/file/d/0B4wgc0vIE7CCUmpXLXJ0TEtIeUU/view?usp=sharing [Broken]

It looks like an argument using differentials. If [itex] f_X [/itex] is the pdf of the random variable [itex] X [/itex] then the probability of the event [itex] \{x: a < x \le a + h\} \approx f_X(a) h [/itex]. This is applied when [itex] a = \sqrt{y} [/itex] and [itex] h = \frac{\triangle y} {2 \sqrt{y} } [/itex] and again when [itex] a = -\sqrt{y} [/itex].

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