Need help, URGENT maths question: (1 Viewer)

1008 · May 31, 2016

InteGrand said:
I think you forgot to use product rules or quotient rules etc. where needed. Like for the (∂/∂r)[∂ƒ/∂θ* (sin(θ)/r)] term in that second line, you should use product rule because the second function here (sin(θ)/r) is dependent on r (the sin(θ) part is a constant wrt r, but the r isn't of course, so we do need to use product rule).

Right thanks I'll get back to you on that question. Any thoughts on the other one?

InteGrand · May 31, 2016

1008 said:
Right thanks I'll get back to you on that question. Any thoughts on the other one?

That one is also fairly tedious I think. I might have a look at it later.

1008 · May 31, 2016

No worries

1008 · Jun 1, 2016

Had a few more questions regarding probability, if someone can help me (please):

For the first one: very that this function f is a probability density, that is, that:
f(x) is greater than or equal to zero and $\int_{-\infty}^{\infty}f(x)dx = 1$

Thanks

leehuan · Jun 1, 2016

Pretty sure I asked Q61 already.

For the Laplace distribution just rewrite the integral as

$\int_{-\infty}^0{\frac{1}{2}e^xdx}+\int_0^\infty{\frac{1}{2}e^{-x}dx}$

Pretty obvious as to why f(x)>0

InteGrand · Jun 1, 2016

1008 said:
Had a few more questions regarding probability, if someone can help me (please):

For the first one: very that this function f is a probability density, that is, that:
f(x) is greater than or equal to zero and $\int_{-\infty}^{\infty}f(x)dx = 1$

Thanks

$\noindent The last question here is basically just an integration exercise.$

$\noindent Let's assume the random variable is defined on the whole real line (so integrals we do will be from $-\infty$ to $+\infty$). The first step is to find the constant of proportionality $c$ such that $f_{X}\left(x\right) = c e^{-x^2 + x}$. This is found by recalling that the density must integrate to 1. So we need $\int _{-\infty} ^{\infty} c e^{-x^2 + x} \text{ d}x = 1$. You should end up getting something along the lines of $c=\frac{1}{e^{\frac{1}{4}}\sqrt{\pi}}$ if you work through the details $\Big{(}$use \emph{completing the square} in the exponential and recall the \textit{Gaussian integral} $\int _{-\infty} ^{\infty} e^{- \omega^2}\text{ d}\omega = \sqrt{\pi}$$\Big{)}$.$

$\noindent Now that we've worked out the density, the mean will be $\mu \equiv \mathbb{E}\left[X\right] = \int _{-\infty} ^{\infty}cx e^{-x^2 +x}\text{ d}x$. This can be done again using completing the square and a substitution. Basically you should end up having to integrate something like $xe^{-\left(x-\frac{1}{2}\right)^2}$ over the real line, at which point you can sub. $y=x-\frac{1}{2}$ and you'll be able to get it into an easy-to-integrate form (keeping in mind the Gaussian integral again).$

$\noindent Once we've found the mean, all we need is to find $\mathbb{E}\left[X^2\right]$ to be able to get the variance. To do this, we'll need to end up integrating something like $\left(y+\frac{1}{2}\right)^2 e^{-y^2}$ over the real line. If this thing is expanded, one term will end up being a Gaussian integral, one term will just be the mean we calculated already, and a final term will involve calculating something like $\int _{-\infty}^{\infty}y^2 e^{-y^2}\text{ d}y$. To do this last one, note that it is $2 \int _{0}^{\infty}y \cdot ye^{-y^2}\text{ d}y$, which we can use integration by parts on.$

leehuan · Jun 1, 2016

InteGrand said:
$\noindent The last question here is basically just an integration exercise.$

$\noindent Let's assume the random variable is defined on the whole real line (so integrals we do will be from $-\infty$ to $+\infty$). The first step is to find the constant of proportionality $c$ such that $f_{X}\left(x\right) = c e^{-x^2 + x}$. This is found by recalling that the density must integrate to 1. So we need $\int _{-\infty} ^{\infty} c e^{-x^2 + x} \text{ d}x = 1$. You should end up getting something along the lines of $c=\frac{1}{e^{\frac{1}{4}}\sqrt{\pi}}$ if you work through the details $\Big{(}$use \emph{completing the square} in the exponential and recall the \textit{Gaussian integral} $\int _{-\infty} ^{\infty} e^{- \omega^2}\text{ d}\omega = \sqrt{\pi}$$\Big{)}$.$

$\noindent Now that we've worked out the density, the mean will be $\mu \equiv \mathbb{E}\left[X\right] = \int _{-\infty} ^{\infty}cx e^{-x^2 +x}\text{ d}x$. This can be done again using completing the square and a substitution. Basically you should end up having to integrate something like $xe^{-\left(x-\frac{1}{2}\right)^2}$ over the real line, at which point you can sub. $y=x-\frac{1}{2}$ and you'll be able to get it into an easy-to-integrate form (keeping in mind the Gaussian integral again).$

$\noindent Once we've found the mean, all we need is to find $\mathbb{E}\left[X^2\right]$ to be able to get the variance. To do this, we'll need to end up integrating something like $\left(y+\frac{1}{2}\right)^2 e^{-y^2}$ over the real line. If this thing is expanded, one term will end up being a Gaussian integral, one term will just be the mean we calculated already, and a final term will involve calculating something like $\int _{-\infty}^{\infty}y^2 e^{-y^2}\text{ d}y$. To do this last one, note that it is $2 \int _{0}^{\infty}y \cdot ye^{-y^2}\text{ d}y$, which we can use integration by parts on.$

Oh right that's how to do it... I was like how do I integrate this because I left my answers at home but I forgot about completing the square entirely

Sent from my iPhone using Tapatalk

1008 · Jun 3, 2016

leehuan said:
Pretty sure I asked Q61 already.

For the Laplace distribution just rewrite the integral as

$\int_{-\infty}^0{\frac{1}{2}e^xdx}+\int_0^\infty{\frac{1}{2}e^{-x}dx}$

Pretty obvious as to why f(x)>0

Could you direct me to which thread you asked Q61 on?

1008 · Jun 3, 2016

InteGrand said:
$\noindent The last question here is basically just an integration exercise.$

$\noindent Let's assume the random variable is defined on the whole real line (so integrals we do will be from $-\infty$ to $+\infty$). The first step is to find the constant of proportionality $c$ such that $f_{X}\left(x\right) = c e^{-x^2 + x}$. This is found by recalling that the density must integrate to 1. So we need $\int _{-\infty} ^{\infty} c e^{-x^2 + x} \text{ d}x = 1$. You should end up getting something along the lines of $c=\frac{1}{e^{\frac{1}{4}}\sqrt{\pi}}$ if you work through the details $\Big{(}$use \emph{completing the square} in the exponential and recall the \textit{Gaussian integral} $\int _{-\infty} ^{\infty} e^{- \omega^2}\text{ d}\omega = \sqrt{\pi}$$\Big{)}$.$

$\noindent Now that we've worked out the density, the mean will be $\mu \equiv \mathbb{E}\left[X\right] = \int _{-\infty} ^{\infty}cx e^{-x^2 +x}\text{ d}x$. This can be done again using completing the square and a substitution. Basically you should end up having to integrate something like $xe^{-\left(x-\frac{1}{2}\right)^2}$ over the real line, at which point you can sub. $y=x-\frac{1}{2}$ and you'll be able to get it into an easy-to-integrate form (keeping in mind the Gaussian integral again).$

$\noindent Once we've found the mean, all we need is to find $\mathbb{E}\left[X^2\right]$ to be able to get the variance. To do this, we'll need to end up integrating something like $\left(y+\frac{1}{2}\right)^2 e^{-y^2}$ over the real line. If this thing is expanded, one term will end up being a Gaussian integral, one term will just be the mean we calculated already, and a final term will involve calculating something like $\int _{-\infty}^{\infty}y^2 e^{-y^2}\text{ d}y$. To do this last one, note that it is $2 \int _{0}^{\infty}y \cdot ye^{-y^2}\text{ d}y$, which we can use integration by parts on.$

Thanks

Since I have exams coming up in a week

please excuse pile of questions I am hurling towards you IG

$\\\text{(i) Regulatory authorities have developed a method for detecting illegal share trading.}\\\text{When a trade is illegal, the method detects it 99} \%\text{ of the time}\\\text{When a trade is legal, the mthod detects it as illegal 5}\%\text{ of the time}\\\text{Over a trial period the method detected 22}\%\text{ of the trades as illegal.}\\\text{a) Find the probability that a randomly chosen trade was illegal.}\\\text{b) If the method detected a trade as illegal, what was the probability it was actually a legal trade?}\\\\\text{(ii) The number of claims received by an insurance company follows a Poisson distribution.}\\\text{On average, one customer in 800 makes a claim in a year. If the company has 4000 such}\\\text{policies, estimate the probability the company receives more than 3 claims in one year.}\\\text{(iii) Let X be a random variable with probability density function }f(x).\\\text{Find the probability density function }g\text{ for }Y = e^X\text{ in terms of }f.$

InteGrand · Jun 3, 2016

1008 said:
Thanks

Since I have exams coming up in a week please excuse pile of questions I am hurling towards you IG

$\\\text{(i) Regulatory authorities have developed a method for detecting illegal share trading.}\\\text{When a trade is illegal, the method detects it 99} \%\text{ of the time}\\\text{When a trade is legal, the mthod detects it as illegal 5}\%\text{ of the time}\\\text{Over a trial period the method detected 22}\%\text{ of the trades as illegal.}\\\text{a) Find the probability that a randomly chosen trade was illegal.}\\\text{b) If the method detected a trade as illegal, what was the probability it was actually a legal trade?}\\\\\text{(ii) The number of claims received by an insurance company follows a Poisson distribution.}\\\text{On average, one customer in 800 makes a claim in a year. If the company has 4000 such}\\\text{policies, estimate the probability the company receives more than 3 claims in one year.}\\\text{(iii) Let X be a random variable with probability density function }f(x).\\\text{Find the probability density function }g\text{ for }Y = e^X\text{ in terms of }f.$

QUESTION (i)

(I said 'test' a lot below, I'm referring to the method for detecting illegal trades by that.)

• $\noindent $I$ be the event a trade is illegal (so $I^c$ is the event a trade is legal)$
• $\noindent $T$ be the event that the \textsl{test} says a randomly chosen trade is illegal (so $T^c$ is the event a randomly chosen trade is said to be legal by the test).$

$\noindent a) We are asked to find $\mathbb{P}\left(I\right)$. We are given: $\mathbb{P}\left(T\mid I\right) = 0.99$, $\mathbb{P}\left(T \mid I^c \right) = 0.05$, and $\mathbb{P}\left(T\right) = 0.22$. Keep in mind that $\mathbb{P}\left(A^c \mid B\right) = 1-\mathbb{P}\left(A \mid B\right)$. Now, let $x = \mathbb{P}\left(I\right)$, so $1-x = \mathbb{P}\left(I^c\right)$. So from the \textsl{Law of Total Probability}, we have$

$\begin{align*} \mathbb{P}\left(T\right) &=\mathbb{P}\left(T \mid I \right)x + \mathbb{P}\left(T \mid I^{c}\right) \left(1-x\right). \quad (1) \end{align*}$

$\noindent Now we know all the values for the probabilities above! (They're listed earlier, from the given data in the question.) This means we can solve the above equation to get $x$, which will be our answer (I'll let you do that since I don't have pen and paper or calculator on me and the numbers are rather tedious).$

$\noindent b) We are asked to find $\mathbb{P}\left(I^{c}\mid T\right)$. Note that from \textsl{Bayes' rule}, the answer will be$

$\begin{align*} \mathbb{P}\left(I^c \mid T\right) = \frac{\mathbb{P}\left(T \mid I\right)\mathbb{P}\left(I\right)}{\mathbb{P}\left(T\right)}. \end{align*}$

$\noindent The above gives us the answer because we know all the values of the quantities above: $\mathbb{P}\left(T \mid I\right) = 0.99$ (given), $\mathbb{P}\left(I\right)=x$ (value found in part (a)), and $\mathbb{P}\left(T\right) = 0.22$ (given).$

So for these Q's, the important things are to set out some notation at the start, write out the given data carefully in terms of our notation, figure out exactly what the Q's are asking in terms of our notation, and keep in mind the Law of Total Probability and Bayes' rule (as well as Pr(A^c| B) = 1 – Pr(A | B) and our other probability rules of course).

InteGrand · Jun 3, 2016

1008 said:
Thanks

Since I have exams coming up in a week please excuse pile of questions I am hurling towards you IG

$\\\text{(i) Regulatory authorities have developed a method for detecting illegal share trading.}\\\text{When a trade is illegal, the method detects it 99} \%\text{ of the time}\\\text{When a trade is legal, the mthod detects it as illegal 5}\%\text{ of the time}\\\text{Over a trial period the method detected 22}\%\text{ of the trades as illegal.}\\\text{a) Find the probability that a randomly chosen trade was illegal.}\\\text{b) If the method detected a trade as illegal, what was the probability it was actually a legal trade?}\\\\\text{(ii) The number of claims received by an insurance company follows a Poisson distribution.}\\\text{On average, one customer in 800 makes a claim in a year. If the company has 4000 such}\\\text{policies, estimate the probability the company receives more than 3 claims in one year.}\\\text{(iii) Let X be a random variable with probability density function }f(x).\\\text{Find the probability density function }g\text{ for }Y = e^X\text{ in terms of }f.$

QUESTION (ii)

$\noindent For questions like this, the important thing is to define a useful random variable(s), write down or work out its distribution, and work out what the question is asking of us in terms of our random variable.$

$$\noindent Let $X$ be the number of claims received by the company in a year (a random variable). Since there are 4000 policies in total, on average, $5$ claims are made per year. So $X \sim \text{Poisson}\left(\lambda\right)$, where $\lambda = 5$. We want to find $\mathbb{P}\left(X > 3\right)$, which is $1-\mathbb{P}\left(X \leq 2\right)$. Using the Poisson pmf, we have $1-\mathbb{P}\left(X \leq 2\right) = 1- \sum _{k = 0}^{2} e^{-\lambda }\frac{\lambda^k }{k!} = 1- \left(e^{-\lambda} + \lambda e^{-\lambda} + \frac{\lambda^2}{2!} e^{-\lambda}\right)$. Now sub. in $\lambda = 5$ to get the answer (factor out $e^{-\lambda}$ to reduce the amount of typing in a calculator), and check whether it is reasonable (always good to do).$

InteGrand · Jun 3, 2016

1008 said:
Thanks

Since I have exams coming up in a week please excuse pile of questions I am hurling towards you IG

$\\\text{(i) Regulatory authorities have developed a method for detecting illegal share trading.}\\\text{When a trade is illegal, the method detects it 99} \%\text{ of the time}\\\text{When a trade is legal, the mthod detects it as illegal 5}\%\text{ of the time}\\\text{Over a trial period the method detected 22}\%\text{ of the trades as illegal.}\\\text{a) Find the probability that a randomly chosen trade was illegal.}\\\text{b) If the method detected a trade as illegal, what was the probability it was actually a legal trade?}\\\\\text{(ii) The number of claims received by an insurance company follows a Poisson distribution.}\\\text{On average, one customer in 800 makes a claim in a year. If the company has 4000 such}\\\text{policies, estimate the probability the company receives more than 3 claims in one year.}\\\text{(iii) Let X be a random variable with probability density function }f(x).\\\text{Find the probability density function }g\text{ for }Y = e^X\text{ in terms of }f.$

QUESTION (iii)

$\noindent Let $y>0$, the cdf of $Y$ be $G$ and the cdf of $X$ be $F$. Then$

$\begin{align*}G(y) &= \mathbb{P}\left(Y \leq y\right) \\ &= \mathbb{P}\left(e^X \leq y \right) \\ &= \mathbb{P}\left(X \leq \ln y \right) \quad (\text{exponential function is increasing}) \\ &= F\left(\ln y\right).\end{align*}$

$\noindent Now, differentiating both sides and using the chain rule on the RHS gives us $g(y) = \frac{1}{y} f\left(\ln y\right)$, which is the answer.$

$\noindent \textbf{Remark.} Using the same method, we can show that in general if $Y = h\left(X\right)$, where $h$ is a function with an inverse $h^{-1}$ that is differentiable (so $h$ is monotone as it has an inverse) and $X$ and $Y$ are continuous random variables with pdf's $f_{X}$ and $f_{Y}$ respectively, then $f_{Y} \left(y\right) = f_{X}\left( h^{-1} \left(y\right)\right) \left |\frac{\mathrm{d}\left(h^{-1}\left(y\right)\right)}{\mathrm{d}y}\right |$.$

leehuan · Jun 3, 2016

Judging by what Q61 looks like it would be under continuous random variables

ALL of my pleas for help are under the extracurricular section for maths

http://community.boredofstudies.org...opics/350587/continuous-random-variables.html

1008 · Jun 3, 2016

InteGrand said:
QUESTION (iii)

$\noindent Let $y>0$, the cdf of $Y$ be $G$ and the cdf of $X$ be $F$. Then$

$\begin{align*}G(y) &= \mathbb{P}\left(Y \leq y\right) \\ &= \mathbb{P}\left(e^X \leq y \right) \\ &= \mathbb{P}\left(X \leq \ln y \right) \quad (\text{exponential function is increasing}) \\ &= F\left(\ln y\right).\end{align*}$

$\noindent Now, differentiating both sides and using the chain rule on the RHS gives us $g(y) = \frac{1}{y} f\left(\ln y\right)$, which is the answer.$

$\noindent \textbf{Remark.} Using the same method, we can show that in general if $Y = h\left(X\right)$, where $h$ is a function with an inverse $h^{-1}$ that is differentiable (so $h$ is monotone as it has an inverse) and $X$ and $Y$ are continuous random variables with pdf's $f_{X}$ and $f_{Y}$ respectively, then $f_{Y} \left(y\right) = f_{X}\left( h^{-1} \left(y\right)\right) \left |\frac{\mathrm{d}\left(h^{-1}\left(y\right)\right)}{\mathrm{d}y}\right |$.$

Wow thanks!

Here are some more, I have no idea what to do:
$\\\text{(i) A function }f(x,y)\text{ is said to be homogeneous of degree }n\text{ if }f(tx,ty)=t^nf(x,y)\text{ for all }t>0.\text{ Show that }\text{ such a function satisfies the equation }\\x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} =nf\\\\\text{(ii) (a) Let }a\text{ and }b\text{ be real numbers. Consider the partial differentation equation }\\a \frac{\partial u }{\partial x} +b \frac{\partial u}{\partial y}=0\text{ } (1)\\\text{Use the chain rule to show that for a suitable choice of the constants }\alpha\text{ and }\beta\text{, every function}\\\text{ of the form }u(x,y)=f(\alpha x+\beta y),\text{ where }f\text{ is differentiable, satisfies (1)} \\\text{(b) Generalise this remark to find solutions }u:\mathbb{R}^n \longrightarrow \mathbb{R}\text{ of }\\a_1 \frac{\partial u }{\partial x_1} +a_2 \frac{\partial u}{\partial x_2} + ... +a_n \frac{\partial u }{\partial x_n}=0$

EDIT: what is a homogeneous equation anyway?

InteGrand · Jun 3, 2016

1008 said:
Wow thanks!

Here are some more, I have no idea what to do:
$\\\text{(i) A function }f(x,y)\text{ is said to be homogeneous of degree }n\text{ if }f(tx,ty)=t^nf(x,y)\text{ for all }t>0.\text{ Show that }\text{ such a function satisfies the equation }\\x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} =nf\\\\\text{(ii) (a) Let }a\text{ and }b\text{ be real numbers. Consider the partial differentation equation }\\a \frac{\partial u }{\partial x} +b \frac{\partial u}{\partial y}=0\text{ } (1)\\\text{Use the chain rule to show that for a suitable choice of the constants }\alpha\text{ and }\beta\text{, every function}\\\text{ of the form }u(x,y)=f(\alpha x+\beta y),\text{ where }f\text{ is differentiable, satisfies (1)} \\\text{(b) Generalise this remark to find solutions }u:\mathbb{R}^n \longrightarrow \mathbb{R}\text{ of }\\a_1 \frac{\partial u }{\partial x_1} +a_2 \frac{\partial u}{\partial x_2} + ... +a_n \frac{\partial u }{\partial x_n}=0$

For the first one, partially differentiate the homogeneity equation wrt t (that's how you get n to pop out) and then sub. in t = 1 after that. To differentiate the LHS of that wrt t,

$\noindent Let $u=xt$, $v=yt$ and $z = f(xt,yt) = f(u,v)$. Then $u_t = x$ and $v_t = y$ (subscripts denote partial derivatives). So via chain rule, $z_t = f_1 (u,v) u_t + f_2 (u,v) v_t = x f_1(tx,ty) + yf_2 (tx,ty) = nt^{n-1} f(x,y)$. The reason it equals $nt^{n-1} f(x,y)$ is that this is the partial derivative of the RHS of homogeneity equation wrt $t$). Now set $t=1$ to find that $x f_1 (x,y) + yf_2 (x,y) = nf(x,y)$, as required.$

InteGrand · Jun 3, 2016

1008 said:
Wow thanks!

Here are some more, I have no idea what to do:
$\\\text{(i) A function }f(x,y)\text{ is said to be homogeneous of degree }n\text{ if }f(tx,ty)=t^nf(x,y)\text{ for all }t>0.\text{ Show that }\text{ such a function satisfies the equation }\\x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} =nf\\\\\text{(ii) (a) Let }a\text{ and }b\text{ be real numbers. Consider the partial differentation equation }\\a \frac{\partial u }{\partial x} +b \frac{\partial u}{\partial y}=0\text{ } (1)\\\text{Use the chain rule to show that for a suitable choice of the constants }\alpha\text{ and }\beta\text{, every function}\\\text{ of the form }u(x,y)=f(\alpha x+\beta y),\text{ where }f\text{ is differentiable, satisfies (1)} \\\text{(b) Generalise this remark to find solutions }u:\mathbb{R}^n \longrightarrow \mathbb{R}\text{ of }\\a_1 \frac{\partial u }{\partial x_1} +a_2 \frac{\partial u}{\partial x_2} + ... +a_n \frac{\partial u }{\partial x_n}=0$

Second one is also an application of chain rule.

$Let $ u(x,y) = f(\alpha x + \beta y)$. Then via chain rule,$

$\noindent $u_x= f^{\prime} (\alpha x + \beta y) \cdot \alpha$ and $u_y = f^\prime (\alpha x + \beta y) \cdot \beta$ (using subscript notation for partial derivatives). If we sub. these into the given partial differential equation ($au_x + bu_y = 0$), we see we want$

$\noindent $af^{\prime} (\alpha x + \beta y) \cdot \alpha + bf^{\prime} (\alpha x + \beta y) \cdot \beta=0$, i.e. $f^\prime (\alpha x +\beta y)\cdot \left(a\alpha + b\beta \right) = 0$. Clearly this'll be true if we choose $\alpha,\beta$ such that $a\alpha + b\beta =0$. So any function $u(x,y) = f(\alpha x+\beta y)$ with $a\alpha + b\beta = 0$ will satisfy the given partial differential equation.$

$For the generalisation, any function of the form $u(x,y) = f\left(\alpha_1 x_1 + \cdots + \alpha_n x_n\right)$ with $f$ being a scalar-valued differentiable function and the $\alpha_i$'s chosen so that $a_1 \alpha_1 +\cdots + a_n \alpha_n = 0$ will satisfy the PDE.$

leehuan · Jun 3, 2016

That same question involving f(tx,ty) appeared in a past paper in a different form.

$$a) Find $\frac{d}{dt}f(tx,ty)$ in terms of $\frac{\partial f}{\partial x}(tx,ty)$ and $\frac{\partial f}{\partial y}(tx,ty)$

And I don't get the answers.

$$Let $g(x,y,t)=f(tx,ty)=t^n f(x,y)$. Then $\frac{\partial g}{\partial t}=nt^{n-1}f(x,y)$

$$Also let $g(t,x,y)=f(u,v)$ with $u=tx, v=ty$. Then by the chain rule:$\\ \frac{\partial g}{\partial t} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial t}=x\frac{\partial f}{\partial u}+y \frac{\partial f}{\partial u}$

Sure ok... No idea why they changed the order of the variables of g but still ok...

${\color{red}\text{Hence, for fixed }x,y}\\ \frac{d}{dt}f(tx,ty)= {\color{red} x\frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} }$

What. How?

InteGrand · Jun 3, 2016

leehuan said:
That same question involving f(tx,ty) appeared in a past paper in a different form.

$$a) Find $\frac{d}{dt}f(tx,ty)$ in terms of $\frac{\partial f}{\partial x}(tx,ty)$ and $\frac{\partial f}{\partial y}(tx,ty)$

And I don't get the answers.

$$Let $g(x,y,t)=f(tx,ty)=t^n f(x,y)$. Then $\frac{\partial g}{\partial t}=nt^{n-1}f(x,y)$

$$Also let $g(t,x,y)=f(u,v)$ with $u=tx, v=ty$. Then by the chain rule:$\\ \frac{\partial g}{\partial t} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial t}=x\frac{\partial f}{\partial u}+y \frac{\partial f}{\partial u}$

Sure ok... No idea why they changed the order of the variables of g but still ok...

${\color{red}\text{Hence, for fixed }x,y}\\ \frac{d}{dt}f(tx,ty)= {\color{red} x\frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} }$

What. How?

$\noindent If $x$ and $y$ are fixed, then $f\left(tx,ty\right)$ is just a function of $t$, and so then we can write $\frac{\mathrm{d}}{\mathrm{d}t}$. Further, $f\left(tx,ty\right)=g\left(x,y,t\right)$, so the derivative of $f$ wrt $t$ is just that of $g$ wrt t (to help, you might want to write $z=g\left(x,y,t\right)=f\left(tx,ty\right)$, and then note they are talking about $\frac{\mathrm{d}z}{\mathrm{d}t}$ with $x,y$ fixed). Also, here the $\frac{\partial f}{\partial x}$ is referring to $f_{1}$, just as the earlier line's $\frac{\partial f}{\partial u}$ was. This is why it might be a good idea to use numerical subscript notation for the derivative here to reduce confusion, since the $x$ and $y$'s are both being considered as fixed numbers and also the designated variables in the red part.$

1008 · Jun 3, 2016

leehuan said:
That same question involving f(tx,ty) appeared in a past paper in a different form.

$$a) Find $\frac{d}{dt}f(tx,ty)$ in terms of $\frac{\partial f}{\partial x}(tx,ty)$ and $\frac{\partial f}{\partial y}(tx,ty)$

And I don't get the answers.

$$Let $g(x,y,t)=f(tx,ty)=t^n f(x,y)$. Then $\frac{\partial g}{\partial t}=nt^{n-1}f(x,y)$

$$Also let $g(t,x,y)=f(u,v)$ with $u=tx, v=ty$. Then by the chain rule:$\\ \frac{\partial g}{\partial t} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial t}=x\frac{\partial f}{\partial u}+y \frac{\partial f}{\partial u}$

Sure ok... No idea why they changed the order of the variables of g but still ok...

${\color{red}\text{Hence, for fixed }x,y}\\ \frac{d}{dt}f(tx,ty)= {\color{red} x\frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} }$

What. How?

Wtf?

1008 · Jun 3, 2016

InteGrand said:
$\noindent If $x$ and $y$ are fixed, then $f\left(tx,ty\right)$ is just a function of $t$, and so then we can write $\frac{\mathrm{d}}{\mathrm{d}t}$. Further, $f\left(tx,ty\right)=g\left(x,y,t\right)$, so the derivative of $f$ wrt $t$ is just that of $g$ wrt t (to help, you might want to write $z=g\left(x,y,t\right)=f\left(tx,ty\right)$, and then note they are talking about $\frac{\mathrm{d}z}{\mathrm{d}t}$ with $x,y$ fixed). Also, here the $\frac{\partial f}{\partial x}$ is referring to $f_{1}$, just as the earlier line's $\frac{\partial f}{\partial u}$ was. This is why it might be a good idea to use numerical subscript notation for the derivative here to reduce confusion, since the $x$ and $y$'s are both being considered as fixed numbers and also the designated variables in the red part.$

Can't you just do it like this?:
$\\\frac{\mathrm{d} }{\mathrm{d} t}f(tx,ty) = \frac{\partial f}{\partial x} \frac{\partial x }{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}\\\text{and so }\\ \frac{\mathrm{d} f}{\mathrm{d} t} = x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y}$

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