Need help, URGENT maths question: (4 Viewers)

1008 · May 15, 2016

InteGrand said:
I believe this Q. was asked here before, and answered by Sy123: http://community.boredofstudies.org...1437/intermediate-value-theorem-question.html .

Thanks for that InteGrand.

I have another question. What would be the quickest method to do this?

$\\\text{Find the point of intersection of the line }\\x = \begin{pmatrix}3\\2\\2\end{pmatrix} + \lambda_{1} \begin{pmatrix}-7\\-1\\-5\end{pmatrix}\\\text{And the plane }\\\omega = \begin{pmatrix}-13\\1\\-23\end{pmatrix} +\xi \begin{pmatrix}-4\\-3\\5\end{pmatrix}+\mu \begin{pmatrix}-5\\-1\\-5\end{pmatrix}$

The method I know involves converting the plane to a Cartesian equation and then substituting the x1, x2 and x3 variables of the cartesian equation with the line. This took very long for this question, and I got it wrong...

InteGrand · May 15, 2016

1008 said:
Thanks for that InteGrand.

I have another question. What would be the quickest method to do this?

$\\\text{Find the point of intersection of the line }\\x = \begin{pmatrix}3\\2\\2\end{pmatrix} + \lambda_{1} \begin{pmatrix}-7\\-1\\-5\end{pmatrix}\\\text{And the plane }\\\omega = \begin{pmatrix}-13\\1\\-23\end{pmatrix} +\xi \begin{pmatrix}-4\\-3\\5\end{pmatrix}+\mu \begin{pmatrix}-5\\-1\\-5\end{pmatrix}$

The method I know involves converting the plane to a Cartesian equation and then substituting the x1, x2 and x3 variables of the cartesian equation with the line. This took very long for this question, and I got it wrong...

The method you used isn't too long, and will lead to the right answer if done correctly. You'll basically end up with the plane equation of the form ax + by + cz = d . Then sub. in the line coordinates to this and solve for lambda (which is easy to solve for since it's just a linear equation in lambda), and then sub. this solution back into the parametric equation for the line.

Another way to do it is to equate the RHS's of both plane and line equations and try solve for lambda (and mu and xi, though you'll only need lambda, but you might want to get mu and xi too for checking purposes). This will involve using Gaussian Elimination.

1008 · May 15, 2016

InteGrand said:
The method you used isn't too long, and will lead to the right answer if done correctly. You'll basically end up with the plane equation of the form ax + by + cz = d . Then sub. in the line coordinates to this and solve for lambda (which is easy to solve for since it's just a linear equation in lambda), and then sub. this solution back into the parametric equation for the line.

Another way to do it is to equate the RHS's of both plane and line equations and try solve for lambda (and mu and xi, though you'll only need lambda, but you might want to get mu and xi too for checking purposes). This will involve using Gaussian Elimination.

Thanks for the reply InteGrand. I was thinking, isn't there a method to do this using dot products and/or cross product? The answer to this question was the point (3,2,2), which just happens to be the point in the line equation...

InteGrand · May 15, 2016

1008 said:
Thanks for the reply InteGrand. I was thinking, isn't there a method to do this using dot products and/or cross product? The answer to this question was the point (3,2,2), which just happens to be the point in the line equation...

Well it's basically a coincidence it happened to the point they give in the line's parametric equation (because they could have used a whole host of other points for the exact same line's equation (i.e. any other point on that line)).

The cross product would be used to find a normal to the plane, which would then allow us to find the Cartesian equation of the plane.

1008 · May 15, 2016

InteGrand said:
Well it's basically a coincidence it happened to the point they give in the line's parametric equation (because they could have used a whole host of other points for the exact same line's equation (i.e. any other point on that line)).

The cross product would be used to find a normal to the plane, which would then allow us to find the Cartesian equation of the plane.

Could you please demonstrate how one would go about doing this?

InteGrand · May 15, 2016

1008 said:
Could you please demonstrate how one would go about doing this?

$\noindent In general, say we're given a plane equation in $\mathbb{R}^{3}$: $\Pi: \bold{x} = \bold{a} + \eta \bold{v}_1 + \xi \bold{v}_2$ (parametric vector equation). Then a normal to the plane is $\bold{n} := \bold{v}_1\times \bold{v}_2$ (in other words, cross the two direction vectors and that gives us a normal). Now, the equation for the plane can be written as: $\bold{n}\cdot \left(\bold{x}-\bold{a}\right)=0 \Longleftrightarrow \bold{n}\cdot \bold {x} = \bold{n} \cdot \bold{a}$. Since we are given $\bold{a}$ and we found $\bold{n}$, we can write this out as a Cartesian equation.$

$\noindent E.g. say we found $\bold{n}$ to be $\begin{bmatrix}2 \\ -1 \\ 5\end{bmatrix}$ and our given point on the plane was $\bold{a} = \begin{bmatrix}-3 \\ 2 \\ 6\end{bmatrix}$ (\textsl{just making up random numbers for the sake of example}). Then if $\bold{x}\equiv \begin{bmatrix}x \\ y \\ z\end{bmatrix}$, note that $\bold{n}\cdot \bold{x} = 2x-y+5z$ (i.e. the coefficients are just the normal vector's elements), and $\bold{n}\cdot \bold{a} = -6 -2 + 30 = 22$. So the plane's equation is $2x-y+5z=22$.$

1008 · May 16, 2016

InteGrand said:
$\noindent In general, say we're given a plane equation in $\mathbb{R}^{3}$: $\Pi: \bold{x} = \bold{a} + eta \bold{v}_1 + \xi \bold{v}_2$ (parametric vector equation). Then a normal to the plane is $\bold{n} := \bold{v}_1\times \bold{v}_2$ (in other words, cross the two direction vectors and that gives us a normal). Now, the equation for the plane can be written as: $\bold{n}\cdot \left(\bold{x}-\bold{a}\right)=0 \Longleftrightarrow \bold{n}\cdot \bold {x} = \bold{n} \cdot \bold{a}$. Since we are given $\bold{a}$ and we found $\bold{n}$, we can write this out as a Cartesian equation.$

$\noindent E.g. say we found $\bold{n}$ to be $\begin{bmatrix}2 \\ -1 \\ 5\end{bmatrix}$ and our given point on the plane was $\bold{a} = \begin{bmatrix}-3 \\ 2 \\ 6\end{bmatrix}$ (\textsl{just making up random numbers for the sake of example}). Then if $\bold{x}\equiv \begin{bmatrix}x \\ y \\ z\end{bmatrix}$, note that $\bold{n}\cdot \bold{x} = 2x-y+5z$ (i.e. the coefficients are just the normal vector's elements), and $\bold{n}\cdot \bold{a} = -6 -2 + 30 = 22$. So the plane's equation is $2x-y+5z=22$.$

Wow, thanks!

I had another question:
$\\\text{Let X be a discrete random variable with Poisson distribution with parameter }\lambda\\\text{a) Show that }\\ E(X^{2})=\sum_{k=1}^{\infty}\frac{ke^{-\lambda}\lambda^{k}}{(k-1)!}\\\text{b) By changing the summation variable, show that }E(X^{2})=\lambda E(X)+\lambda \\\text{c) Using (b), show that Var(X)= }\lambda$

Just wanted to know if this is a valid method: (next post)

1008 · May 16, 2016

$\\\text{a) }E(X) = \sum_{k=1}^{\infty} kp_{k}\\\therefore E(X^{2}) = \sum_{k=1}^{\infty} k^{2}p_{k}\\=\sum_{k=1}^{\infty} \frac{k^{2}e^{-\lambda}\lambda^{k}}{k!}\\=\sum_{k=1}^{\infty} \frac{ke^{-\lambda}\lambda ^{k}}{(k-1)!}\\\text{b) Let }l = k-1\\\therefore E(X^{2})= \sum_{l=0}^{\infty} \frac{(l+1)e^{-\lambda}\lambda^{l+1}}{l!}\\=\lambda (\sum_{l=0}^{\infty} \frac{e^{-\lambda }\lambda^{l}}{(l-1)!}+\sum_{l=0}^{\infty}\frac{e^{-\lambda}\lambda^{l}}{l!})\\=\lambda(\sum_{l=1}^{\infty}\frac{e^{-\lambda}\lambda^{l}}{(l-1)!}+e^{-\lambda}e^{\lambda})\text{ (As l=0 isn't defined in the summation and by the definition of the exponential)}\\=\lambda E(X)+\lambda$

I dunno how to do C though...

InteGrand · May 16, 2016

1008 said:
$\\\text{a) }E(X) = \sum_{k=1}^{\infty} kp_{k}\\\therefore E(X^{2}) = \sum_{k=1}^{\infty} k^{2}p_{k}\\=\sum_{k=1}^{\infty} \frac{k^{2}e^{-\lambda}\lambda^{k}}{k!}\\=\sum_{k=1}^{\infty} \frac{ke^{-\lambda}\lambda ^{k}}{(k-1)!}\\\text{b) Let }l = k+1\\\therefore \sum_{l=0}^{\infty} \frac{(l+1)e^{-\lambda}\lambda^{l+1}}{l!}\\=\lambda (\sum_{l=0}^{\infty} \frac{e^{-\lambda }\lambda^{l}}{(l-1)!}+\sum_{l=0}^{\infty}\frac{e^{-\lambda}\lambda^{l}}{l!})\\=\lambda(\sum_{l=1}^{\infty}\frac{e^{-\lambda}\lambda^{l}}{(l-1)!}+e^{-\lambda}e^{\lambda})\text{ (As l=0 isn't defined in the summation)}\\=\lambda E(X)+\lambda$

I dunno how to do C though...

$\noindent Part (c) is the easy part if the other two are done, assuming you know or have shown that $\mathbb{E}\left[X\right] = \lambda$. In that case, $\mathrm{Var}\left[X\right] =\mathbb{E}\left[X^2\right]-\big{(}\mathbb{E}\left[X\right] \right\big{)}^2 = \lambda^2 + \lambda - \lambda^2 = \lambda $ (using $\mathbb{E}\left[X\right] = \lambda$ in the part (b) formula).$

InteGrand · May 16, 2016

1008 said:
$\\\text{a) }E(X) = \sum_{k=1}^{\infty} kp_{k}\\\therefore E(X^{2}) = \sum_{k=1}^{\infty} k^{2}p_{k}\\=\sum_{k=1}^{\infty} \frac{k^{2}e^{-\lambda}\lambda^{k}}{k!}\\=\sum_{k=1}^{\infty} \frac{ke^{-\lambda}\lambda ^{k}}{(k-1)!}\\\text{b) Let }l = k-1\\\therefore \sum_{l=0}^{\infty} \frac{(l+1)e^{-\lambda}\lambda^{l+1}}{l!}\\=\lambda (\sum_{l=0}^{\infty} \frac{e^{-\lambda }\lambda^{l}}{(l-1)!}+\sum_{l=0}^{\infty}\frac{e^{-\lambda}\lambda^{l}}{l!})\\=\lambda(\sum_{l=1}^{\infty}\frac{e^{-\lambda}\lambda^{l}}{(l-1)!}+e^{-\lambda}e^{\lambda})\text{ (As l=0 isn't defined in the summation)}\\=\lambda E(X)+\lambda$

I dunno how to do C though...

To do part b), the change of variable should be l = k-1, rather than +1 (it's (the change of variables and limits of summation) done basically like how we'd do a change of variables in integration).

1008 · May 16, 2016

InteGrand said:
$\noindent Part (c) is the easy part if the other two are done, assuming you know or have shown that $\mathbb{E}\left[X\right] = \lambda$. In that case, $\mathrm{Var}\left[X\right] =\mathbb{E}\left[X^2\right]-\big{(}\mathbb{E}\left[X\right] \right\big{)}^2 = \lambda^2 + \lambda - \lambda^2 = \lambda $ (using $\mathbb{E}\left[X\right] = \lambda$ in the part (b) formula).$

Thanks

InteGrand said:
To do part b), the change of variable should be l = k-1, rather than +1 (it's (the change of variables and limits of summation) done basically like how we'd do a change of variables in integration).

Sorry just fixed that up, it was a typo lol!

InteGrand · May 16, 2016

1008 said:
Wow, thanks!

I had another question:
$\\\text{Let X be a discrete random variable with Poisson distribution with parameter }\lambda\\\text{a) Show that }\\ E(X^{2})=\sum_{k=1}^{\infty}\frac{ke^{-\lambda}\lambda^{k}}{(k-1)!}\\\text{b) By changing the summation variable, show that }E(X^{2})=\lambda E(X)+\lambda \\\text{c) Using (b), show that Var(X)= }\lambda$

Just wanted to know if this is a valid method: (next post)

$\noindent Here's how to do part (b).$

$Starting from $\mathbb{E} \left[X^2 \right]= \sum_{k=1}^{\infty} \frac{k\lambda^k \mathrm{e}^{-\lambda}}{\left(k-1\right)!}$, we use the change of variable $j=k-1\Longleftrightarrow k = j+1$. When $k=1$, $j=0$, and as $k\to \infty$, $j\to \infty$, so we have$

$\begin{align*} \mathbb{E} \left[X^2 \right] &= \sum_{j=0}^{\infty} \frac{\left(j+1\right)\lambda^{j+1} \mathrm{e}^{-\lambda}}{j!} \\ &= \sum_{j=0}^{\infty} \left( \frac{j\lambda^{j+1}\mathrm{e}^{-\lambda}}{j!}+\frac{\lambda^{j+1}\mathrm{e}^{-\lambda}}{j!} \right) \\ &= \sum_{j=0}^{\infty} \frac{j \lambda^{j+1}\mathrm{e}^{-\lambda}}{j!} + \sum_{j=0}^{\infty} \frac{\lambda^{j+1}\mathrm{e}^{-\lambda}}{j!} \quad (\text{splitting up the sum as both sums here are convergent}) \\ &= \lambda \left(\underbrace{\sum_{j=0}^{\infty} \frac{j \lambda^{j}\mathrm{e}^{-\lambda}}{j!}}_{\text{this is just }\mathbb{E}\left[X\right] \text{ by definition}}\right) + \lambda \left(\underbrace{\sum_{j=0}^{\infty} \frac{\lambda^{j}\mathrm{e}^{-\lambda}}{j!}}_{\text{this is the sum of the p.m.f. over all possible values, so equals 1}}\right) \quad (\text{pulling out a }\lambda \text{ from each sum}) \\ &= \lambda \mathbb{E}\left[X\right] + \lambda.\end{align*}$

Zoinked · May 16, 2016

Still not convinced that InteGrand isnt a super computer

.

leehuan · May 16, 2016

mattstaker said:
Still not convinced that InteGrand isnt a super computer .

Replies are not generated within a matter of 10 seconds of algorithms. They're generated out of niceness in helping others

Zoinked · May 16, 2016

leehuan said:
Replies are not generated within a matter of 10 seconds of algorithms. They're generated out of niceness in helping others

I know I know. Guys a beast. Wanna know what he studies/does for a living

1008 · May 17, 2016

InteGrand said:
$\noindent Here's how to do part (b).$

$Starting from $\mathbb{E} \left[X^2 \right]= \sum_{k=1}^{\infty} \frac{k\lambda^k \mathrm{e}^{-\lambda}}{\left(k-1\right)!}$, we use the change of variable $j=k-1\Longleftrightarrow k = j+1$. When $k=1$, $j=0$, and as $k\to \infty$, $j\to \infty$, so we have$

$\begin{align*} \mathbb{E} \left[X^2 \right] &= \sum_{j=0}^{\infty} \frac{\left(j+1\right)\lambda^{j+1} \mathrm{e}^{-\lambda}}{j!} \\ &= \sum_{j=0}^{\infty} \left( \frac{j\lambda^{j+1}\mathrm{e}^{-\lambda}}{j!}+\frac{\lambda^{j+1}\mathrm{e}^{-\lambda}}{j!} \right) \\ &= \sum_{j=0}^{\infty} \frac{j \lambda^{j+1}\mathrm{e}^{-\lambda}}{j!} + \sum_{j=0}^{\infty} \frac{\lambda^{j+1}\mathrm{e}^{-\lambda}}{j!} \quad (\text{splitting up the sum as both sums here are convergent}) \\ &= \lambda \left(\underbrace{\sum_{j=0}^{\infty} \frac{j \lambda^{j}\mathrm{e}^{-\lambda}}{j!}}_{\text{this is just }\mathbb{E}\left[X\right] \text{ by definition}}\right) + \lambda \left(\underbrace{\sum_{j=0}^{\infty} \frac{\lambda^{j}\mathrm{e}^{-\lambda}}{j!}}_{\text{this is the sum of the p.m.f. over all possible values, so equals 1}}\right) \quad (\text{pulling out a }\lambda \text{ from each sum}) \\ &= \lambda \mathbb{E}\left[X\right] + \lambda.\end{align*}$

Wow, thanks!

1008 · May 17, 2016

mattstaker said:
I know I know. Guys a beast. Wanna know what he studies/does for a living

I agree. And @leehuan, yes InteGrand is a great guy!

1008 · May 17, 2016

I had another question:

$\noindent A brewery brews one type of beer which is marketed under three different brands. In a survey of 150 first year students, 58 drink at least Brand A, 49 drink at least Brand B and 57 drink at least Brand C. 14 drink brand A and brand C, 13 drink brand A and brand B and 17 drink both brand B and brand C. 4 students drink all three brands. How many students drink none of these three brands?$

InteGrand · May 17, 2016

1008 said:
I had another question:

$\noindent A brewery brews one type of beer which is marketed under three different brands. In a survey of 150 first year students, 58 drink at least Brand A, 49 drink at least Brand B and 57 drink at least Brand C. 14 drink brand A and brand C, 13 drink brand A and brand B and 17 drink both brand B and brand C. 4 students drink all three brands. How many students drink none of these three brands?$

$\noindent Let $A$ be the set of all students who drink Brand A, and similarly define sets $B$ and $C$ (by `drink Brand A', we mean `at least' Brand A, to use the question's phrasing.). Let $U$ be the universal set. We are asked to find $|A^c \cap B^c \cap C^c|$. We are given: $|U| = 150, |A| = 58, |B|=49, |C| = 57, |A\cap C|=14, |A\cap B|=13, |B\cap C|=17$ and $|A\cap B \cap C| = 4$. Now, using various set laws, we have$

$\begin{align*}|A^c \cap B^c \cap C^c| &= |U| - |A\cup B \cup C| \\ &= |U|- \left(|A|+|B|+|C| - |A\cap B| - |B\cap C| - |A\cap C| + |A\cap B \cap C| \right) \quad (\text{inclusion/exclusion principle}) \\ &= 150- \left(58+49 + 57 - 13 - 17 - 14 + 4\right) \\ &= 26.\end{align*}$

$\noindent Hence the answer is 26.$

1008 · May 17, 2016

InteGrand said:
$\noindent Let $A$ be the set of all students who drink Brand A, and similarly define sets $B$ and $C$ (by `drink Brand A', we mean `at least' Brand A, to use the question's phrasing.). Let $U$ be the universal set. We are asked to find $|A^c \cap B^c \cap C^c|$. We are given: $|U| = 150, |A| = 58, |B|=49, |C| = 57, |A\cap C|=14, |A\cap B|=13, |B\cap C|=17$ and $|A\cap B \cap C| = 4$. Now, using various set laws, we have$

$\begin{align*}|A^c \cap B^c \cap C^c| &= |U| - |A\cup B \cup C| \\ &= |U|- \left(|A|+|B|+|C| - |A\cap B| - |B\cap C| - |A\cap C| + |A\cap B \cap C| \right) \quad (\text{inclusion/exclusion principle}) \\ &= 150- \left(58+49 + 57 - 13 - 17 - 14 + 4\right) \\ &= 26.\end{align*}$

$\noindent Hence the answer is 26.$

Thanks InteGrand. I've another question:

$$\noindent 1. Urn 1 contains 2 red balls and 3 black balls. Urn 2 contains 4 red balls and 5 black balls.$\\$a) Suppose a ball is drawn at random from urn 1 and placed into Urn 2. If a ball is then drawn at random from the 10 balls in urn 2, what is the probability that it is red? (I get this part) $\\$b) In the previous part, given that the ball drawn from urn 2 is red, what is the proba- bility that the ball transferred from urn 1 was black?$\\$2. Establish, using mathematical induction, Bonferoni's inequality: $\\P(A_{1}\cap A_{2}\cap ...\cap A_{n}) \geq 1-[P(A_{1}^{c})+P(A_{2}^{c}+...+P(A_{n}^{c}))$

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