Find Help w/ Taylor Series: (y+dy)^0.5

• chenrim
In summary, you are trying to develop a Taylor series for the pdf of X from first principles, and you need to take into account that X can be negative. After this, I think it is easiest to use the cumulative distribution, and differentiate to get the probability density functions.
chenrim
help with the following taylor series:

(y+dy)^0.5

Thanks

I'm guessing that y is a constant and dy is your variable? and you are trying to expand around dy=0 ? well, you should try to use the definition of the Taylor series, and see what you get :)

The thing is that i have an r.v (random variable) of the form Y=X^2

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

yes, I think I understand. you want to show how the pdf of X (from first principles) is related to the pdf of Y. Generally, you also need to take into account that X can be negative. (unless you want to specifically say that the random variable X cannot be negative). After this, I think it is easiest to use the cumulative distribution, and differentiate to get the probability density functions.

Yes you right X also should be taken negative.
i didnt understand the step between line 2 and 3

Last edited by a moderator:
chenrim said:
i didnt understand the step between line 2 and 3

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

Last edited by a moderator:

1. What is the purpose of finding help with Taylor Series for (y+dy)^0.5?

The purpose of finding help with Taylor Series for (y+dy)^0.5 is to approximate the value of a square root function, which is often used in mathematical and scientific calculations.

2. How is the Taylor Series used to find the square root of (y+dy)^0.5?

The Taylor Series is a mathematical tool that allows us to approximate the value of a function by using its derivatives at a specific point. By using the Taylor Series for (y+dy)^0.5, we can calculate an approximation for the square root of any value of (y+dy)^0.5.

3. What are the steps involved in finding the Taylor Series for (y+dy)^0.5?

The steps involved in finding the Taylor Series for (y+dy)^0.5 are:

1. Find the derivatives of (y+dy)^0.5 at a specific point.
2. Substitute the values of the derivatives into the Taylor Series formula.
3. Simplify the series by grouping like terms.
4. Apply the values of (y+dy)^0.5 and its derivatives to the simplified series.
5. The resulting series is the Taylor Series for (y+dy)^0.5.

4. What is the significance of finding the Taylor Series for (y+dy)^0.5?

The Taylor Series for (y+dy)^0.5 is significant because it allows us to approximate the value of a square root function, which is a commonly used mathematical operation. This approximation can then be used in various calculations and models in science and engineering.

5. Are there any limitations to using the Taylor Series for (y+dy)^0.5?

Yes, there are limitations to using the Taylor Series for (y+dy)^0.5. The accuracy of the approximation depends on the number of terms used in the series, and as the number of terms increases, so does the complexity of the calculations. Additionally, the Taylor Series can only approximate the square root function within a certain range, and the accuracy decreases as the input value moves further away from the starting point.

• Calculus
Replies
2
Views
1K
• Calculus
Replies
5
Views
12K
• Calculus
Replies
3
Views
967
• Calculus
Replies
2
Views
1K
• Calculus
Replies
17
Views
3K
• Calculus
Replies
3
Views
1K
• Calculus
Replies
9
Views
821
• Calculus
Replies
3
Views
1K
• Calculus
Replies
9
Views
1K
• Calculus
Replies
1
Views
1K