text:
### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2005

MCQ (Single Correct Answer)
If radius of the $\matrix{ {27} \cr {13} \cr }$ $Al$ nucleus is estimated to be $3.6$ fermi then the radius of $\matrix{ {125} \cr {52} \cr } \,Te$ nucleus is estimated to be nearly
A
$8$ fermi
B
$6$ fermi
C
$5$ fermi
D
$4$ fermi

## Explanation

KEY CONCEPT : $R = {R_0}{\left( A \right)^{1/3}}$

Here A = Mass number

$\therefore$ ${{{R_1}} \over {{R_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^{1/3}}$

$= {\left( {{{27} \over {125}}} \right)^{1/3}} = {3 \over 5}$

${R_2} = {5 \over 3} \times 3.6 = 6$ fermi
2

### AIEEE 2004

MCQ (Single Correct Answer)
A nucleus disintegrated into two nuclear parts which have their velocities in the ratio of $2:1.$ The ratio of their nuclear sizes will be
A
${3^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}:1$
B
$1:{2^{1/3}}$
C
${2^{1/3}}:1$
D
$1:{3^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$

## Explanation

From conservation of momentum ${m_1}{v_1} = {m_2}{v_2}$

$\Rightarrow \left( {{{{m_1}} \over {{m_2}}}} \right) = \left( {{{{v_2}} \over {{v_1}}}} \right)\,\,$ given $\,\,{{{v_1}} \over {v{}_2}} = 2$

$\Rightarrow {{{m_1}} \over {{m_2}}} = {1 \over 2}$

$\Rightarrow {{r_1^3} \over {r_2^3}} = {1 \over 2}$

$\Rightarrow \left( {{{{r_1}} \over {{r_2}}}} \right) = {\left( {{1 \over 2}} \right)^{1/3}}$
3

### AIEEE 2004

MCQ (Single Correct Answer)
The binding energy per nucleon of deuteron $\left( {{}_1^2\,H} \right)$ and helium nucleus $\left( {{}_2^4\,He} \right)$ is $1.1$ $MeV$ and $7$ $MeV$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
A
$23.6\,\,MeV$
B
$26.9\,\,MeV$
C
$13.9\,\,MeV$
D
$19.2\,\,MeV$

## Explanation

The nuclear reaction of process is $2_1^2H \to {4 \over 2}$ He

Energy released $= 4 \times \left( 7 \right) - 4\left( {1.1} \right) = 23.6\,MeV$
4

### AIEEE 2004

MCQ (Single Correct Answer)
An $\alpha$-particle of energy $5$ $MeV$ is scattered through ${180^ \circ }$ by a fixed uranium nucleus. The distance of closest approach is of the order of
A
${10^{ - 12}}\,cm$
B
${10^{ - 10}}\,cm$
C
$1A$
D
${10^{ - 15a}}\,cm$

## Explanation

KEY NOTE :

Distance of closest approach

${r_0} = {{Ze\left( {2e} \right)} \over {4\pi {\varepsilon _0}E}}$

Energy, $E = 5 \times {10^6} \times1.6 \times {10^{ - 19}}J$

$\therefore$ ${r_0} = {{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)} \over {5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}$

$\Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm$

### Joint Entrance Examination

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Class 12


text:
Remove/estimate correlation of parameters (or get to know if correlation is real)

For the R I used one of the institutional clusters. I installed it using

install.packages("rstan", dependencies = TRUE)
install.packages("hBayesDM", dependencies=TRUE)

(I followed the instructions from STAN), I remember having problems with it and we’re solving it for quite some time but unfortunately, I do not remember the details.

For Pystan, I’m using Ubuntu 18.04 LTS or Win10. I have anaconda on both together with python 3.7. In Windows I mostly followed this protocol https://pystan.readthedocs.io/en/latest/windows.html#windows :

conda update conda
conda install libpython m2w64-toolchain -c msys2
conda install numpy cython -c conda-forge
pip install pystan

On Ubuntu I also don’t remember, I guess only install pystan worked. But I never had any problem before this one on any of the devices with any code.
Is it working on yours?

I’m on Ubuntu 19.10 running R 3.6.1 and RStan 2.19.3, and the model compiles fine for me. I do get an error when I attempt to sample:

Error in new_CppObject_xp(fields$.module, fields$.pointer, ...) :
Exception: mismatch in number dimensions declared and found in context; processing stage=data initialization; variable name=Tsubj; dims declared=(1); dims found=()  (in 'model4bc36b91a558_temp' at line 4)

failed to create the sampler; sampling not done

which is due to you attempting to subset the data down to one participant, so when the Tsubj vector is created in R it actually gets turned into a single value while Stan is expecting a vector. I vaguely recall there was a way to handle this scenario but can’t remember precisely. Can you post a data set with multiple subjects?

here

@mike-lawrence, @ahartikainen, my apologies, I have not realised I have given pickled data and did not mention it. Also, @mike-lawrence, it has been prepared for pystan so it is in a format of python dictionary. I can resave it in another way if it would help you (as I said, I’m not familiar with R so I don’t know what would be the best).

So the code works as follows:

import pickle

data_path = '/home/user/LossAversion/'

with open(os.path.join(data_path, 'data_LA_exp1_STAN_correlation.txt'), "rb") as fp:   # Unpickling

Edit:
The data in not pickled way (but still python dict) would be:
csv – data_LA_exp1_STAN_correlation_unpickle.csv (72.3 KB)
txt – data_LA_exp1_STAN_correlation_unpickle.txt (101.5 KB)

Hey!

I don’t know if you saw my message. My question would be, why there are some gamble which are -1, since these would cause trouble for bernoulli. Also instead of Phi_approx, I’d just use inv_logit.

I think I’ve got soe working code, but it runs slowly, and depending on what your answer is wrt tothe -1 issue, it could likely be sped up significantly.

Cheers!
Max

Hi Max,

thanks for the message, I got it. So I double-checked the -1 in the data. When I prepare my data, I initialise it with -1 and then replace where I have valid trials. Of course, the max trials for a subject, Tsubj' is then smaller by the number of invalid trials. So they should not cause any problems since they’re not accounted for by STAN.

I also double checked that the prospect.stan file:

works. It compiles and runs without any problems.

Why would you use inv_logit instead of Phi_approx?

And thanks again for your reply and reaching out, I very much appreciate.
J

I have looked into the console what stan prints out (haven’t thought about it before), gives me series of failures to sample (which is also happening with prospect but with decreasing rate of rejection until it starts sampling) and followed/ended by this:

Rejecting initial value:
Log probability evaluates to log(0), i.e. negative infinity.
Stan can't start sampling from this initial value.

Initialization between (-2, 2) failed after 100 attempts.
Try specifying initial values, reducing ranges of constrained values, or reparameterizing the model.
Rejecting initial value:
Log probability evaluates to log(0), i.e. negative infinity.
Stan can't start sampling from this initial value.

Ah, got that now, thanks!

I reckon its a bit more stable, as the approx Phi is just a scaled inverse logit (see here Normal_lcdf underflows much faster than pnorm(log=TRUE)).

Here’s my go at it:

This model works fine for me (tested with cmdstanr):

data {
int<lower=1> N;
int<lower=1> T;
int<lower=1, upper=T> Tsubj[N];
real<lower=0> gain[N, T];
real<lower=0> loss[N, T];
real cert[N, T];
int<lower=-1, upper=1> gamble[N, T];
}
parameters {
vector[3] mu_pr;
vector<lower=0>[3] sigma;
cholesky_factor_corr[3] cors;
vector[3] rho_lambda_tau_tilde[N];
}
transformed parameters {
vector<lower=0, upper=2>[N]  rho;
vector<lower=0, upper=5>[N]  lambda;
vector<lower=0, upper=30>[N] tau;
cholesky_factor_cov[3] L_Sigma = diag_pre_multiply(sigma, cors);

for (i in 1:N) {
vector[3] logit_rho_lambda_tau_pr = inv_logit(mu_pr + L_Sigma*rho_lambda_tau_tilde[i]);
rho[i]    = logit_rho_lambda_tau_pr[1] * 2;
lambda[i] = logit_rho_lambda_tau_pr[2] * 5;
tau[i]    = logit_rho_lambda_tau_pr[3] * 30;
}
}
model {
mu_pr ~ std_normal();
sigma ~ normal(0, 0.2);
cors ~ lkj_corr_cholesky(3);

for( i in 1:N)
rho_lambda_tau_tilde[i] ~ std_normal();

for (i in 1:N) {
for (t in 1:Tsubj[i]) {
real evSafe;    // evSafe, evGamble, pGamble can be a scalar to save memory and increase speed.
real evGamble;  // they are left as arrays as an example for RL models.

// loss[i, t]=absolute amount of loss (pre-converted in R)
evSafe   = pow(cert[i, t], rho[i]);
evGamble = 0.5 * (pow(gain[i, t], rho[i]) - lambda[i] * pow(loss[i, t], rho[i]));
gamble[i, t] ~ bernoulli_logit(tau[i] * (evGamble - evSafe));
}
}
}

This version is a bit faster (better “vectorization” of the bernoulli_logit):

data {
int<lower=1> N;
int<lower=1> T;
int<lower=1, upper=T> Tsubj[N];
real<lower=0> gain[N, T];
real<lower=0> loss[N, T];
real cert[N, T];
int<lower=-1, upper=1> gamble[N, T];
}
parameters {
vector[3] mu_pr;
vector<lower=0>[3] sigma;
cholesky_factor_corr[3] cors;
vector[3] rho_lambda_tau_tilde[N];
}
transformed parameters {
vector<lower=0, upper=2>[N]  rho;
vector<lower=0, upper=5>[N]  lambda;
vector<lower=0, upper=30>[N] tau;
cholesky_factor_cov[3] L_Sigma = diag_pre_multiply(sigma, cors);

for (i in 1:N) {
vector[3] logit_rho_lambda_tau_pr = inv_logit(mu_pr + L_Sigma*rho_lambda_tau_tilde[i]);
rho[i]    = logit_rho_lambda_tau_pr[1] * 2;
lambda[i] = logit_rho_lambda_tau_pr[2] * 5;
tau[i]    = logit_rho_lambda_tau_pr[3] * 30;
}

}
model {

mu_pr ~ std_normal();
sigma ~ normal(0, 0.2);
cors ~ lkj_corr_cholesky(3);

for (i in 1:N){

vector[Tsubj[i]] evSafe;
vector[Tsubj[i]] evGamble;

for (t in 1:Tsubj[i]){
evSafe[t] = pow(cert[i, t], rho[i]);
evGamble[t] = 0.5 * (pow(gain[i, t], rho[i]) - lambda[i] * pow(loss[i, t], rho[i]));
}

head(gamble[i], Tsubj[i]) ~ bernoulli_logit(tau[i] * (evGamble - evSafe));
rho_lambda_tau_tilde[i] ~ std_normal();

}

}

You might have to fix the generated quantities block, which I have ignored here.

Results

3 chains, 600 warm-up each, 600 post-warmup iterations each

   variable    mean median     sd    mad      q5     q95   rhat ess_bulk ess_tail
<chr>      <dbl>  <dbl>  <dbl>  <dbl>   <dbl>   <dbl>  <dbl>    <dbl>    <dbl>
1 mu_pr[1]  -0.150 -0.158 0.113  0.109  -0.328   0.0412  1.00      690.     904.
2 mu_pr[2]  -1.66  -1.65  0.0716 0.0698 -1.77   -1.54    0.999     991.    1079.
3 mu_pr[3]  -3.60  -3.60  0.237  0.243  -4.00   -3.22    1.00      846.    1057.
4 sigma[1]   0.444  0.441 0.0581 0.0592  0.353   0.545   1.00     1236.    1397.
5 sigma[2]   0.221  0.218 0.0542 0.0518  0.136   0.311   1.00      906.     702.
6 sigma[3]   0.902  0.901 0.0977 0.0996  0.742   1.06    1.01     1035.    1318.
7 cors[1,1]  1      1     0      0       1       1      NA          NA       NA
8 cors[2,1]  0.283  0.298 0.177  0.177  -0.0285  0.550   1.00     1317.    1249.
9 cors[3,1] -0.703 -0.714 0.0943 0.0903 -0.834  -0.537   1.00     1177.    1317.
10 cors[1,2]  0      0     0      0       0       0      NA          NA       NA

Cheers!
Max

1 Like

Thank you very much Max! The code works.
I have added the generated parameters, although I’m not sure about a few steps:

1. what is the difference between the quantities generated now with your code and quantities which I’d generated in the last block using:
  Omega_corr = multiply_lower_tri_self_transpose(cors);
// corr_matrix[3] Omega = cors * cors';

? Or in general between what is generated in this block and sampled, eg. mu_pr and corresponding generated quantities or the covariance/correlation matrices?

1. Is there any difference between:
  Omega_corr = multiply_lower_tri_self_transpose(cors);

Again, thank you very much!
and

  corr_matrix[3] Omega = cors * cors';

?

1. Would you/Should I use the inv_logit instead of Phi_approx also in the generated quantities block?

Bellow my stan code (working):

Stan code
data {
int<lower=1> N;            // num of subjects
int<lower=1> T;            // num of trials per subject
int<lower=1, upper=T> Tsubj[N];  // maximum possible num of trials
real<lower=0> gain[N, T];
real<lower=0> loss[N, T];  // absolute loss amount
real cert[N, T];
int<lower=-1, upper=1> gamble[N, T];
}
parameters {
vector[3] mu_pr;
vector<lower=0>[3] sigma;
cholesky_factor_corr[3] cors;
vector[3] rho_lambda_tau_tilde[N];
}
transformed parameters {
vector<lower=0, upper=2>[N]  rho;
vector<lower=0, upper=5>[N]  lambda;
vector<lower=0, upper=30>[N] tau;
cholesky_factor_cov[3] L_Sigma = diag_pre_multiply(sigma, cors);

for (i in 1:N) {
vector[3] logit_rho_lambda_tau_pr = inv_logit(mu_pr + L_Sigma*rho_lambda_tau_tilde[i]);
rho[i]    = logit_rho_lambda_tau_pr[1] * 2;
lambda[i] = logit_rho_lambda_tau_pr[2] * 5;
tau[i]    = logit_rho_lambda_tau_pr[3] * 30;
}
}
model {
mu_pr ~ std_normal();
sigma ~ normal(0, 0.2);
cors ~ lkj_corr_cholesky(3);

for( i in 1:N)
rho_lambda_tau_tilde[i] ~ std_normal();

for (i in 1:N) {
for (t in 1:Tsubj[i]) {
real evSafe;    // evSafe, evGamble, pGamble can be a scalar to save memory and increase speed.
real evGamble;  // they are left as arrays as an example for RL models.

// loss[i, t]=absolute amount of loss (pre-converted in R)
evSafe   = pow(cert[i, t], rho[i]);
evGamble = 0.5 * (pow(gain[i, t], rho[i]) - lambda[i] * pow(loss[i, t], rho[i]));
gamble[i, t] ~ bernoulli_logit(tau[i] * (evGamble - evSafe));
}
}
}

generated quantities {
real<lower=0, upper=2>  mu_rho;
real<lower=0, upper=5>  mu_lambda;
real<lower=0, upper=30> mu_tau;

matrix[3,3] Omega_corr;
matrix[3,3] Sigma;

real log_lik[N];

// For posterior predictive check
real y_pred[N, T];

// Set all posterior predictions to 0 (avoids NULL values)
for (i in 1:N) {
for (t in 1:T) {
y_pred[i, t] = -1;
}
}

mu_rho    = Phi_approx(mu_pr[1]) * 2;
mu_lambda = Phi_approx(mu_pr[2]) * 5;
mu_tau    = Phi_approx(mu_pr[3]) * 30;
//vector[3] logit_rho_lambda_tau_pr = inv_logit(mu_pr + L_Sigma*rho_lambda_tau_tilde[i]);
//rho[i]    = logit_rho_lambda_tau_pr[1] * 2;
//lambda[i] = logit_rho_lambda_tau_pr[2] * 5;
//tau[i]    = logit_rho_lambda_tau_pr[3] * 30;

Omega_corr = multiply_lower_tri_self_transpose(cors);
// corr_matrix[3] Omega_corr = cors * cors';

{ // local section, this saves time and space
for (i in 1:N) {
log_lik[i] = 0;
for (t in 1:Tsubj[i]) {
real evSafe;    // evSafe, evGamble, pGamble can be a scalar to save memory and increase speed.
real evGamble;  // they are left as arrays as an example for RL models.
real pGamble;

// loss[i, t]=absolute amount of loss (pre-converted in R)
evSafe   = pow(cert[i, t], rho[i]);
//evGamble = 0.5 * (pow(gain[i, t], rho[i]) - lambda[i] * pow(loss[i, t], rho[i]));
evGamble = 0.5 * (pow(gain[i, t], rho[i]) - lambda[i] * pow(fabs(loss[i, t]), rho[i]));
pGamble    = inv_logit(tau[i] * (evGamble - evSafe));
log_lik[i] += bernoulli_lpmf(gamble[i, t] | pGamble);

// generate posterior prediction for current trial
y_pred[i, t] = bernoulli_rng(pGamble);
}
}
}
}

A few clarifications:
0) why rho[i] = logit_rho_lambda_tau_pr[1] * 2; and not rho[i] = logit_rho_lambda_tau_pr[i][1] * 2;?
Is it because the vector[3] logit_rho_lambda_tau_pr = inv_logit(mu_pr + L_Sigma*rho_lambda_tau_tilde[i]); is actually a matrix?

1. the tilda multivariate parameter should be the one sampled and without any correlation. That works but why is that? Could you please explain a bit more? There is a bit weird relation between rho_lambda_tau_tilde[1] and rho_lambda_tau_tilde[2], looks like a parabola.

2. the \tau, \lambda, \rho parameters will be correlated but we know their correlation (Omega) or covariance (Sigma) since we have also estimated it, those are the cors and L_sigma respectively. Meaning if I then test a correlation between eg. rho[1], lambda[1], tau[1], I should find the same correlation as from cors, right? Or does this apply only to the mu_pr[1-3] distributions?

What would the correlation be for the means, ie if I extract means for each parameter (mean of rho[1], rho[2],...lambda[51]) and plot a correlation between them?

Because weirdly enough to me, if I do it, there is some exponential correlation between rho and tau and some weird (looks like parabolic) correlation between rho and lambda.

Cool! :)

In the model code it just builds the Cholesky facor of the covariance matrix. In the generated quantities it build the (proper) covariance matrix.

I don’t understand your question here, sorry.

No, not really. The first function is a specialized version and is more efficient for lower triangular matrices, which cors is. It’s a tiny matrix, so the difference is probably not noticeable…

Yes.

In the transformed parameters logit_rho_lambda_tau_pr is a local variable, a (column) vector of length 3. It is newly created for every i in N. Therefore, we don’t need the indexing by i.

Hm, not sure if I understood correctly. rho_lambda_tau_tilde is an array of size N (for every subject) of vectors of length 3 (for \tilde\rho, \tilde\lambda, \tilde\tau). All of these are uncorrelated standard normal random variables. Their correlation is then induced by the Cholesky factor of the covariance matrix, L_Sigma. This is the so called Matt trick, or more commonly now non-centered parameterization of the multivariate normal distribution. In the univariate case z = \mu_z + \sigma_z \tilde z, \tilde z \sim N(0,1) \rightarrow z \sim N(\mu_z, \sigma_z), and the multivariate case replaces the \sigma with a Cholesky factor of the covariance matrix and everything else with vectors.

I’m not sure what you’re indexing here and I don’t think looking at the tilde variables is that insightful here.

Well, technically the model captures the (linear) correlations of the \rho, \lambda, \tau on the logit scale. The multivariate prior on these parameters is effectively a (scaled) multivariate Logit-Normal distribution.The induced correlation of the actual (loosely speaking) parameters will surely be non-linear. So this:

… makes sense to me. I thought from the model code that you started off with, that this is what you were aiming at?

The good thing about modeling the correlations on the logit scale is, that the parameters are unconstrained in that space. We model them (hierarchically) there and then transform them to the constrained space.Thus, we gain flexibility at the cost of loosing the interpretability of the correlations as parameters of linear dependence.

Hope this clears thing up? :)

Cheers!
Max

2 Likes

Thank you very much for your detailed answer, I very much appreciate it, you answered most of my queries :). The few left are bellow:

The question was what is the difference between the mu_pr which I get from stan automatically and the mu_tau, mu_lambda, mu_rho which I get from the generated quantities block. Normally, this should do the reverse Matt trick – going back from centred to non-centred distribution if I understood correctly. But I got a bit lost with the multivariate distribution. (You know – I tried to write down the code and it did not work so I went one step back and realised that I’m not sure what’s going on :).)

Thank you :).

And thank you for this, yes, this is what I was asking about. I did not realise where the Cholesky factor jumps into play.

Sorry, I should have been clearer. By rho_lambda_tau_tilde[1] I mean that I take and plot the distribution of means for all subjects for the first dimension=parameter, ie. \tilde\rho. Similarly, rho_lambda_tau_tilde[2] corresponds to a distribution of N(number of subjects) means for \tilde\lambda and rho_lambda_tau_tilde[3] for \tilde\tau.

Yes, I aimed to estimate the correlation which exists between the parameters and which is imposed by the model/sampling and then interpret the result, to figure out what to do with it. The motivation is that I think that if I have a model in which two parameters are correlated, then I could be able to rewrite the model in a way that simplifies the model, eg. remove one of the parameters, because I should be able to recover the information given by the correlated parameter without this parameter. Correct me if I’m wrong.
So having established that there is a correlation and how the correlation looks like (using Omega_corr), the question would be if I can simplify/rewrite my model by removing some parameters. Or how do I actually use the knowledge of the parameters’ correlation.

Thanks again :).
Cheers,
Jan

Hey!

Yes these should be the same. They’re are merely different ways to arrive at the same result.

Yeah, got it – I think I glossed over the plot too quickly the first time. Hm, my answer still stands I guess. The relationship between the parameters are (modeled) non-linearly, so the plot does not surprise me that much.

Not sure if that’s valid reasoning, but I might be missing something. You could just fit the two parameter version an then compare posterior predictive distributions of the two models and see what the simpler model is missing, or where they differ. If I understood you correctly, then you want to recover one parameter from the other two in the simpler model? This strikes me as a bit unusual, but, again, I could be missing something. Usually, if parameters are correlated, you’d want to estimate them jointly – and if they are not correlated you might get away with estimating them marginally (Think of a hurdle model where you could estimate the two parts separately when you assume they’re not correlated.)

Why do you want a simpler model? To reduce fitting time? For interpretability? The former could be dealt with by re-working the model. The latter maybe by clever presentation of the posterior predictive draws. (not saying there could be other good reasons to fit the simpler model.) ;)

Cheers!

Hi Max,

thanks again :).

What do you mean by this? WAIC/LOOIC or something else?

For the second reason. Usually, I believe the best way to build up a model is based on some theory/expectations and with minimum possible parameters. Both because of its interpretability, clarity and simplicity. Having two parameters where one can be expressed using the other is a thing I want to avoid and correlation is hinting that this could be the case (not always true). And what do you mean by the

? :)

Thanks again :).

For example, yes. Or simple visual diagnostics even (looking at predicted probabilities etc.).

Well, actually the same as above: Finding interesting configurations of gain, loss, and cert and plotting what the models have to say about these (three parameter model vs. two parameter model).

Sorry, I’m not able to give any clear cut answers here. I think this goes into subject matter and what points you want to convey and which results are interesting (to you or your audience).

Cheers! :)

Thank you very much for the help and for you opinion :).

1 Like

format: plain


text:
## University Calculus: Early Transcendentals (3rd Edition)

$$f^{-1}(x)=1-\sqrt{x+1}$$ - Domain: $[-1,\infty)$. - Range: $(-\infty,1]$
$$y=f(x)=x^2-2x\hspace{1cm}x\le1$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=x^2-2x$$ $$y=(x^2-2x+1)-1$$ $$y=(x-1)^2-1$$ $$(x-1)^2=y+1$$ - Since $x\le1$, it follows that $(x-1)\le0$, so we will go with the negative value of $(x-1)$, which means: $$x-1=-\sqrt{y+1}$$ $$x=1-\sqrt{y+1}$$ 2) Interchange $x$ and $y$: $$y=1-\sqrt{x+1}$$ Therefore, $$f^{-1}(x)=1-\sqrt{x+1}$$ - Domain: $x$ is defined where $(x+1)\ge0$, or $x\ge-1$. So the domain of $f^{-1}$ is $[-1,\infty)$. - Range: $\sqrt{x+1}\ge0$ for $x\in[-1,\infty)$ Thus $-\sqrt{x+1}\le0$, and $1-\sqrt{x+1}\le1$ That means the range of $f^{-1}$ is $(-\infty,1]$ *Check: $$f(f^{-1}(x))=(1-\sqrt{x+1})^2-2(1-\sqrt{x+1})=[1+(x+1)-2\sqrt{x+1}]-2+2\sqrt{x+1}=x+2-2=x$$ $$f^{-1}(f(x))=1-\sqrt{x^2-2x+1}=1-\sqrt{(x-1)^2}=1-|x-1|$$ And since $x\le1$, we have $x-1\le0$, so $|x-1|=1-x$ $$f^{-1}(f(x))=1-(1-x)=1-1+x=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$


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# 12.13: Applications Using Rational Equations

Difficulty Level: At Grade Created by: CK-12
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Practice Applications Using Rational Equations
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Consider a flying airplane. If it takes the airplane the same amount of time to travel 3000 miles in calm air as to travel 2000 miles against a 100 mph wind, could you determine the speed (in mph) of the plane in calm air? After completing this Concept, you'll be able to use rational equations to solve real-world applications like this.

### Watch This

CK-12 Foundation: 1213S Solve Applications Using Rational Equations

### Guidance

A motion problem with no acceleration is described by the formula distance=speed×time\begin{align*}distance = speed \times time\end{align*}. Since speed is a measure of rate, the formula may also appear as distance=rate×time\begin{align*}distance = rate \times time\end{align*}. These problems can involve the addition and subtraction of rational expressions.

#### Example A

Last weekend Nadia went canoeing on the Snake River. The current of the river is three miles per hour. It took Nadia the same amount of time to travel 12 miles downstream as it did to travel 3 miles upstream. Determine how fast Nadia’s canoe would travel in still water.

Solution

Define variables:

Let s=\begin{align*}s =\end{align*} speed (rate of travel) of the canoe in still water

Then, s+3=\begin{align*}s + 3 =\end{align*} the speed of the canoe traveling downstream

s3=\begin{align*}s - 3 =\end{align*} the speed of the canoe traveling upstream

Construct a table:

Direction Distance (miles) Rate Time
Downstream 12 s+3\begin{align*}s + 3\end{align*} t\begin{align*}t\end{align*}
Upstream 3 s3\begin{align*}s - 3\end{align*} t\begin{align*}t\end{align*}

Write an equation:

Since distance=rate×time\begin{align*}distance = rate \times time\end{align*}, we can say that time=distancerate\begin{align*}time=\frac{distance}{rate}\end{align*}.

The time to go downstream is:t=12s+3The time to go upstream is:t=3s3Since the time it takes to go upstream and downstream are the same, we have:3s3=12s+3

Solve the equation:

Cross-multiply:3(s+3)=12(s3)Simplify:3s+9=12s36Solve:s=5 mi/h

Check: Upstream: t=128=112 hour\begin{align*}t=\frac{12}{8}=1 \frac{1}{2} \ hour\end{align*}; downstream: t=32=112 hour\begin{align*}t=\frac{3}{2}=1 \frac{1}{2} \ hour\end{align*}. The answer checks out.

#### Example B

Peter rides his bicycle. When he pedals uphill he averages a speed of eight miles per hour, when he pedals downhill he averages 14 miles per hour. If the total distance he travels is 40 miles and the total time he rides is four hours, how long did he ride at each speed?

Solution

Define variables:

Let d=\begin{align*}d =\end{align*} distance Peter bikes uphill at 8 miles per hour.

Construct a table:

Direction Distance (miles) Rate (mph) Time (hours)
Uphill d\begin{align*}d\end{align*} 8 t1\begin{align*}t_1\end{align*}
Downhill 40d\begin{align*}40 - d\end{align*} 14 t2\begin{align*}t_2\end{align*}

Write an equation:

We know that time=distancerate\begin{align*}time=\frac{distance}{rate}\end{align*}.

The time to go uphill is:The time to go downhill is:We also know that the total time is 4 hours:t1=d8t2=40d14d8+40d14=4

Solve the equation:

Find the lowest common denominator:Multiply all terms by the common denominator:Solve: LCD=567d+1604d=224d=21.3 mi

Check: Uphill: t=21.38=2.67 hours\begin{align*}t=\frac{21.3}{8}=2.67 \ hours\end{align*}; downhill: t=4021.314=1.33 hours\begin{align*}t=\frac{40-21.3}{14}=1.33 \ hours\end{align*}. The answer checks out.

#### Example C

A group of friends decided to combine their money and buy a birthday gift that cost $200. Later 12 of the friends decided not to participate any more. This meant that each person paid$15 more than their original share. How many people were in the group to begin with?

Solution

Define variables:

Let x=\begin{align*}x =\end{align*} the number of friends in the original group.

Make a table:

Number of people Gift price Share amount
Original group x\begin{align*}x\end{align*} 200 200x\begin{align*}\frac{200}{x}\end{align*}
Later group x12\begin{align*}x - 12\end{align*} 200 200x12\begin{align*}\frac{200}{x-12}\end{align*}

Write an equation:

Since each person’s share went up by 15 after 12 people refused to pay, we write the equation 200x12=200x+15\begin{align*}\frac{200}{x-12}=\frac{200}{x}+15\end{align*} Solve the equation: Find the lowest common denominator:Multiply all terms by the LCD:Cancel common factors and simplify:Eliminate parentheses:Get all terms on one side of the equation:Divide all terms by 15:Factor:Solve:LCD=x(x12)x(x12)200x12=x(x12)200x+x(x12)15200x=200(x12)+15x(x12)200x=200x2400+15x2180x0=15x2=180x24000=x212x1600=(x20)(x+8)x=20,x=8 The answer that makes sense is \begin{align*}x = \mathbf{20}\end{align*} people. Check: Originally200 shared among 20 people is $10 each. After 12 people leave,$200 shared among 8 people is $25 each. So each person pays$15 more. The answer checks out.

Watch this video for help with the Examples above.

CK-12 Foundation: Solving Applications Using Rational Equations

### Guided Practice

Carrie is a runner. When she runs uphill she averages a speed of 2 miles per hour, when she runs downhill she averages 5 miles per hour. If she runs through the hilly streets of San Francisco for 4.5 hours and the total distance she travels is 13.5 miles, how long did she run uphill and how long did she run downhill?

Solution

Define variables:

Let \begin{align*}t_1 =\end{align*} time Carrie runs uphill.

Let \begin{align*}t_2 =\end{align*} time Carrie runs downhill.

Let \begin{align*}d =\end{align*} the distance Carrie runs uphill.

Construct a table:

Direction Distance (miles) Rate (mph) Time (hours)
Uphill \begin{align*}d\end{align*} 2 \begin{align*}t_1\end{align*}
Downhill \begin{align*}13.5 - d\end{align*} 5 \begin{align*}t_2\end{align*}

Write an equation:

We know that \begin{align*}time=\frac{distance}{rate}\end{align*}.

Solve the equation:

Since \begin{align*}d=6\end{align*} is the distance Carrie to ran uphill, then \begin{align*}13.5-d=13.5-6=7.5\end{align*} is the distance Carrie ran downhill.

Check: Uphill: \begin{align*}t=\frac{6}{2}=3 \ hours\end{align*}; downhill: \begin{align*}t=\frac{13.5-6}{5}=1.5 \ hours\end{align*}. The answer checks out.

### Explore More

For 1-4, solve for \begin{align*}x\end{align*}.

1. \begin{align*}\frac{3x^2+2x-1}{x^2-1}=-2\end{align*}
2. \begin{align*}x+\frac{1}{x}=2\end{align*}
3. \begin{align*}3+\frac{1}{x+1}=\frac{2}{x}\end{align*}
4. \begin{align*}\frac{1}{x}-\frac{x}{x-2}=2\end{align*}

For 5-10, solve the following applications.

1. Juan jogs a certain distance and then walks a certain distance. When he jogs he averages 7 miles/hour and when he walks he averages 3.5 miles per hour. If he walks and jogs a total of 6 miles in a total of 1.2 hours, how far does he jog and how far does he walk?
2. A boat travels 60 miles downstream in the same time as it takes it to travel 40 miles upstream. The boat’s speed in still water is 20 miles per hour. Find the speed of the water.
3. Paul leaves San Diego driving at 50 miles per hour. Two hours later, his mother realizes that he forgot something and drives in the same direction at 70 miles per hour. How long does it take her to catch up to Paul?
4. On a trip, an airplane flies at a steady speed against the wind and on the return trip the airplane flies with the wind. The airplane takes the same amount of time to fly 300 miles against the wind as it takes to fly 420 miles with the wind. The wind is blowing at 30 miles per hour. What is the speed of the airplane when there is no wind?
5. A debt of $420 is shared equally by a group of friends. When five of the friends decide not to pay, the share of the other friends goes up by$25. How many friends were in the group originally?
6. A non-profit organization collected $2250 in equal donations from their members to share the cost of improving a park. If there were thirty more members, then each member could contribute$20 less. How many members does this organization have?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 12.13.

### Vocabulary Language: English

Rational Expression

Rational Expression

A rational expression is a fraction with polynomials in the numerator and the denominator.

## Date Created:

Oct 01, 2012

Jan 31, 2016
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Calderón-type theorems for operators with non-standard endpoint behavior on Lorentz spaces
Linköping University, Department of Mathematics, Mathematics and Applied Mathematics. Linköping University, The Institute of Technology. (Department of Mathematical Analysis, Faculty of Mathematics and Physics, Charles University, Praha, Czech Republic)
2012 (English)In: Mathematische Nachrichten, ISSN 0025-584X, E-ISSN 1522-2616, Vol. 285, no 11-12, 1450-1465 p.Article in journal (Refereed) Published
##### Abstract [en]

The Calderón theorem states that every quasilinear operator, which is bounded both from $L^{p_1,1}$ to $L^{q_1,\infty}$, and from $L^{p_2,1}$ to $L^{q_2,\infty}$ for properly ordered values of $p_1$$p_2$$q_1$$q_2$, is bounded on some rearrangement-invariant space if and only if the so-called Calderón operator is bounded on the corresponding representation space. We will establish Calderón-type theorems for non-standard endpoint behavior, where Lorentz Λ and M spaces will be the endpoints of the interpolation segment. Two distinctive types of non-standard behavior are to be discussed; we’ll explore the operators bounded both from Λ(X1) to Λ(Y1), and from Λ(X2) to M(Y2) using duality arguments, thus, we need to study the operators bounded both from Λ(X1) to M(Y1), and from M(X2) to M(Y2) first. For that purpose, we evaluate Peetre's K-functional for varied pairs of Lorentz spaces.

##### Place, publisher, year, edition, pages
Berlin, Germany: Wiley-VCH Verlagsgesellschaft, 2012. Vol. 285, no 11-12, 1450-1465 p.
##### Keyword [en]
Calderón theorem; Calderón operator; Interpolation of quasilinear operators; Lorentz spaces; msc (2010) Primary: 46B70; Secondary: 26D10
Mathematics
##### Identifiers
ISI: 000307008700012OAI: oai:DiVA.org:liu-80786DiVA: diva2:548266
##### Note

Funding Agencies||SVV-2010-261316|

Available from: 2012-08-30 Created: 2012-08-30 Last updated: 2012-09-05Bibliographically approved

#### Open Access in DiVA

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Malý, Lukáš
##### By organisation
Mathematics and Applied MathematicsThe Institute of Technology
##### In the same journal
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Why isn't neptunium used in nuclear reactors in nuclear power plants?

Why isn't neptunium used in nuclear reactors in nuclear power plants? Uranium is, and plutonium is. But neptunium isn't and it is in the middle of them. Is it like it is too hard to make it do fission? Can someone please kindly explain it to me?

• Fission is not an issue. It's hard to make neptunium, period. – Ivan Neretin Mar 20 '18 at 11:37
• Actually, neptunium (Np-237) is regularly used in nuclear reactors; however, not as nuclear fuel but as so-called fission dosimeter in irradiation capsules. (I have just received a delivery of six such capsules that will later be installed in a pressurized water reactor.) – Loong Mar 20 '18 at 11:47

There are several reasons that neptunium may not be used.

The first is the abundance of neptunium. From Wikipedia, it states the best source is from spent fuel rods. Basically, someone would need to handle very radioactive material to obtain a small amount of neptunium.

Next would be the likelihood of fission when struck with a neutron. If it is the same or less than uranium, I would go with uranium. It would have to be substantially more than uranium to offset the cost of refinement.

Also, there is the energy output per gram. I don't think it would be much greater than uranium, probably less.

The isotopes of neptunium that have long half-lives are $\ce{{}^{236}Np}$ and $\ce{{}^{237}Np}$. These are not produced in great abundance from $\ce{{}^{235}U}$ or $\ce{{}^{238}U}$.

To make plutonium from uranium, the uranium must absorb neutrons without fission. This is much more likely to happen then to make $\ce{{}^{236}Np}$ or $\ce{{}^{237}Np}$

• Technically, neptunium is an intermediate when uranium takes up neutrons. Starting with U-238, we go to U-239, which decays by electron beta emission to Np-239 and then Pu-239. The plutonium isotope, with a longer half-life than the neptunium intermediate, is primarily what we see. – Oscar Lanzi Apr 25 '18 at 22:54

Production rate is low for reactors using low-enrichment uranium. You get about one atom of 237Np per thousand atoms of 239Pu, so its easier to get the Pu out and use it rather than running and running your production reactor to get the 237Np.

Neptunium is not a good fuel, firstly there is the activation problem

Neptunium-237 can adsorb neutrons and either undergo fission or it can capture neutrons to form neptunium-238. This will beta decay to form plutonium-238 which is a horrible isotope of plutonium to work with in bulk. Plutonium-238 has a half life slightly shorter than 100 years, it has a very high specific activity and it does lots of horrible things that plutonium-239 does not do. A small trace of plutonium-238 will greatly increase the alpha activity of the plutonium produced by the reactor. This will make reprocessing and the spent fuel handling harder.

Then there is the problem of production.

The normal route to neptunium-237 is that a uranium-235 nucleous adsorbs a neutron, it forms an excited state of uranium-236. Normally we hope that the excited state of uranium-236 undergoes nuclear fission. But sometimes it manages to relax into its ground state.

The uranium-236 can adsorb another neutron to form uranium-237 which then beta decays to neptunium-237. An alternative route is to subject uranium-238 to fast neutrons such as fusion neutrons, this causes the uranium-238 to undergo the n.2n reaction. These very fast neutrons will not appear commonly in a normal nuclear reactor.


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# The R-Squared Statistic

This tutorial explains the R-Squared fit statistic through its “manual” calculation on R. The best way to engage this lesson is to program with these commands while you read. Tinker with the values. Be hands on. Packages used: ggplot2

## What is the R-Squared?

#### Defined

The R-Squared is a fit statistic, a metric that we use to measure how well our regression model fits the observed data.

Conventionally, people interpret the R-Squared as the percentage of observed variance captured by a model. More strictly, the statistic expresses the ratio of (1) the variance in the dependent variable that we cannot attribute to model error, the discrepancy between the model’s predicted and observed values and (2) the variance observed within the model’s dependent variable values.

I see the metric as a widely-known, rough metric of model fit whose greatest benefit is more related to ease of communicating a point to audiences than to it being a powerful metric to help guide your analysis.

#### Calculating

The metric is measured is the ratio of the regression sum of squares ($SS_{reg}$) to total sum of squares ($SS_{tot}\$):

$R^{2} = SS_{reg} / SS_{tot}$

Where:

• $SS_{reg}$ is the regression sum of squares, a metric that is widely used to describe the proportion of a variable’s variance that is captured by our regression model, and
• $SS_{tot}$ is the total sum of squares, a metric that captures overall variance in the regression’s dependent variable.

## Total Sum of Squares

#### Explanation

The total sum of squares ($SS_{tot}$) is a measure of a variable’s overall variance (i.e., how much its scores vary). When this variance score is lower, then a variables scores mostly cluster around the mean. Where it is higher, scores are more spread out.

So, for example, the variance of people’s internal body temperatures are lower than the weather outside, because people’s body temperatures will only differ by one or two degrees Celsius, whereas the weather outside can vary by tens of degrees Celsius.

#### Calculating

The equation below shows how to calculate the total sum of squares. It is calculated by (1) subtracting each observed score by the group’s overall mean, (2) squaring each of the resulting terms, and then (2) adding all these individual terms together.

Let $y_{mean}$ be the mean value of variable $y$, and let $y_{i}$ be the \$i^{th}$ observation of $y$.

$SS_{tot} = \sum(y_{i} - y_{mean})^2$

#### Illustrated

The total sum of squares is one way to measure a variable’s variability. The higher the score, the higher the observed variability.

Consider two temperature measures: We measure the body temperatures of 10 randomly-chosen people, and the outdoor temperatures of 10 randomly-chosen places.

# Generate 10 randomly-genereated body temperature scores.
# Assume mean of 36.8 degrees, and a standard deviation of 0.4
body.temps <- rnorm(10, 36.8, 0.4)
body.temps <- round(body.temps, 1)  #Rounding the number to make handout more legible
body.temps

##  [1] 36.6 36.3 36.8 36.6 36.6 36.9 36.8 36.5 37.1 37.1

# Generate 10 hypothetical outdoor temperatures
# Assume mean of 17 degrees, and a standard deviation of 8
outdoor.temps <- rnorm(10, 17, 8)
outdoor.temps <- round(outdoor.temps, 1) #Rounding for legibility
outdoor.temps

##  [1] 22.2 18.9  8.6 11.7 26.1 22.5 14.7 10.2 13.5 10.3

The outdoor temperatures vary more widely than the body temperatures, and so the outdoor temperatures register a higher total sum of squares, our metric for measuring variance:

sum((body.temps - mean(body.temps))^2)

## [1] 0.601

sum((outdoor.temps - mean(outdoor.temps))^2)

## [1] 338.261

These figures do not have such an easy, direct interpretation. They are useful in this larger excercise of assessing model fit, however.

##Calculation Deconstructed
Let’s begin by randomly generating a data set with 20 people’s ages, with a mean age of 45 and a standard deviation of 9.

ages <- rnorm(10, 45, 9)  #generate 20 random numbers, normal distribution,
#mean 45 and standard deviation of 9

ages <- round(ages, 0)    #Rounding variable to zero decimal places, so we
#don't have to look at decimals when I list them below.
ages

##  [1] 26 40 50 43 48 69 52 45 45 44

The mean age in this set is:

mean(ages)

## [1] 46.2

Calculate the difference between each observed variable and it mean. That will get you $y_{i} - y_{mean}$ for each individual value:

ages - mean(ages)

##  [1] -20.2  -6.2   3.8  -3.2   1.8  22.8   5.8  -1.2  -1.2  -2.2

Then square each term:

(ages - mean(ages))^2

##  [1] 408.04  38.44  14.44  10.24   3.24 519.84  33.64   1.44   1.44   4.84

Then sum the terms together. This will give you the total sum of squares score:

sum((ages - mean(ages))^2)  #This is the total sum of squares

## [1] 1035.6

A side note: We square the terms to ensure that variance is always expressed as a positive number, so that when we sum the individual scores together the negatives do not cancel the positive ones out. If we did not square the terms, then our variance estimates would be virtually zero. Try it out. Run the script a few times to try it with different randomly generated numbers:

ages - mean(ages)

##  [1] -20.2  -6.2   3.8  -3.2   1.8  22.8   5.8  -1.2  -1.2  -2.2

sum (ages - mean(ages))

## [1] -0.00000000000002842171

## Residual Sum of Squares

#### Our Example

Above, we focused on the fact that our dependent variable scores vary. We see different scores on some metric (e.g., temperature), and we might run a regression to figure out how to predict that metric score using other variables (say latitude or altitude).

For this part of the lesson, let’s focus on a new example: The owner of Paradise Ice Cream asks our friend Bob for help predicting ice cream sales. Better predictions will allow her to better estimate how much ice cream to make the night before. We are going to use temperature forecasts to predict ice cream sales.

Generate Data. Let’s begin by generating 60 hypothetical days, in which the daily forecasted temperature was Normally distributed with a mean of 25 degrees Celcius (SD = 4 degrees):

temps <- rnorm(60, 25, 4)
temps <- round(temps) #round for legibility
temps

##  [1] 24 25 25 31 32 32 29 25 26 24 20 23 20 31 23 23 24 28 28 21 21 32 20
## [24] 20 23 22 23 21 27 21 17 23 29 19 28 19 19 24 20 20 24 27 27 29 33 24
## [47] 24 29 24 30 22 25 24 27 21 26 24 17 24 29

Bob doesn’t know it yet, but in reality there is a relationship between temperature and Paradise Ice Cream sales. The true model is:

$SALES = 300 + 90 \times TEMP + \epsilon$

Where $SALES$ is ice cream shop’s total dollar sales, and $TEMP$ is the foretasted temperature in degrees in Celsius. So the model predicts that, for each degree of higher temperature, Paradise Ice Cream sells $90 more in ice cream. The $$\epsilon$$ term captures our model’s residual error – the number of dollars by which ice cream sales are off from our predictions. Let’s generate some hypothetical numbers. Let’s imagine that, on average, our model is typically off by plus/minus$500, and that the errors are Normally distributed.

error.1 <- rnorm(60, 0, 500)   #Generating 60 days' of error terms
error.1 <- round(error.1)

sales <- 300 + 90 * temps + error.1
sales <- round(sales, 0)  #rounding for legibility
sales

##  [1] 2356 2997 3269 2729 3180 3313 2092 1965 2179 2500 2367 2744 2673 3309
## [15] 2924 3015 2042 3841 3434 2660 2287 3532 2059 1667 2123 2022 1293 2339
## [29] 2942 2340 1249 2101 2679 1948 2447 1645 1951 2206 2050 2903 2354 2346
## [43] 2649 3149 3260 2460 2269 2473 2019 2724 1729 2758 2678 2776 2723 2709
## [57] 2642 2350 2638 2366

Let’s compile our sales, temps, and error vectors into a data frame, and then run a regression. The relationship is highly significant as I wrote it above. You can mess with the regression results by messing with my formulations of the “error” and “sales” vectors above:

#Compile into data frame
icecream <- data.frame(cbind(sales, temps))
head(icecream)  #Take a look at the first few rows of the data frame

##   sales temps
## 1  2356    24
## 2  2997    25
## 3  3269    25
## 4  2729    31
## 5  3180    32
## 6  3313    32

#The relationship visualized:
ggplot(icecream, aes(x = temps, y = sales)) + geom_point() + geom_smooth(method="lm", se= F)

#Run the regression
reg.1 <- lm(sales ~ temps, icecream)
summary(reg.1)

##
## Call:
## lm(formula = sales ~ temps, data = icecream)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -1088.72  -287.65    16.32   242.20  1049.45
##
## Coefficients:
##             Estimate Std. Error t value    Pr(>|t|)
## (Intercept)   496.46     342.71   1.449       0.153
## temps          81.97      13.79   5.944 0.000000169 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 422.6 on 58 degrees of freedom
## Multiple R-squared:  0.3785, Adjusted R-squared:  0.3678
## F-statistic: 35.33 on 1 and 58 DF,  p-value: 0.0000001688

Note the R-Squared on this output, but don’t detain yourself here. Let’s make these calculations by hand, so that we can see how this works.

Let’s add the model’s predicted values to the “icecream” data frame:

icecream$predicted <- fitted(reg.1)  #### Calculating Redisduals Residuals (or “error”) refer to the difference between our model’s predicted outcome values and the data’s observed outcome values. $resid = y_{observed} - y_{predicted}$ Let’s create a new variable in our “icecream” data frame that contains our model’s residual: icecream$resid = icecream$sales - icecream$predicted

##    sales temps predicted      resid
## 1   2356    24  2463.684 -107.68390
## 2   2997    25  2545.652  451.34841
## 3   3269    25  2545.652  723.34841
## 4   2729    31  3037.458 -308.45775
## 5   3180    32  3119.425   60.57455
## 6   3313    32  3119.425  193.57455
## 7   2092    29  2873.522 -781.52237
## 8   1965    25  2545.652 -580.65159
## 9   2179    26  2627.619 -448.61928
## 10  2500    24  2463.684   36.31610

#### Sum of Squared Residuals

These residual scores are variance estimates, like the ones with which we worked in the earlier section total sum of squares. They are just capturing the difference between predicted and observed values, instead of observed individual and overall mean values above.

To calculate the residual sum of squares ($SS_{resid}$):

$SS_{residual} = \sum(y_{observed} - y_{predicted})^2$

We calculate below:

#Calculate the residual sum of squares
SS.resid <- sum((icecream$sales - icecream$predicted)^2)
round(SS.resid, 3)

## [1] 10357420

#And the total sum of squares, while we're at it:
SS.tot <- sum((icecream$sales - mean(icecream$sales))^2)
round(SS.tot, 3)

## [1] 16665834

#### Residuals and Overall Variance

If we take the ratio of $SS_{resid}$ and $SS_{total}$:

round(SS.resid / SS.tot, 3)

## [1] 0.621

The resulting percentage gives us the percentage of the total variance that we can attribute to model error, the degree to which observed values are different from predicted values.

Note that, if you subtract this figure from one, the result is the R-Squared of our regression model above:

round(1 - SS.resid / SS.tot, 3)

## [1] 0.379

The explanation follows.

## Regression Sum of Squares

The regression sum of squares ($$SS_{reg}$$) is a metric that tries to capture how much of the variance estimates by $$RSS_{tot}$$ is predicted by our model. This is the conceptual framework by which we estimate this metric.

1. The total sum of squares ($$RSS_{total}$$) measures variation in the dependent variable.
2. The residual sum of squares ($$RSS_{residual}$$) is the amount of dependent variable variance attributable to the model’s failure to predict values.
3. The remainder is presumed to be the amount of variance that is predicted by the model:
$RSS_{model} = 1 - \frac{RSS_{residual}}{RSS_{total}}$

## Deconstructed Example

#### Generate Data

Imagine a hypothetical sample in which we asked 100 people to rate Funzo’s Burger House on a scale of zero to 100. Assume that Funzo’s is for kids, and people like it less aas they age. We assume the true model underlying the data is:

$RATING = 100 - 2 * AGE + \epsilon$

Where $RATING$ is the person’s rating of Funzo’s, $AGE$ is their age in years, and $\epsilon$ is the error term.

#Create the data

age <- runif(100, 5, 70)   #Generate 100 values from a uniform distribution between 5 and 70
error <- rnorm(100,0, 15)  #Set expected model error at 15 points
rating <- 100 - (2 * age) + error
funzo <- data.frame(cbind(rating, age))

Here is a scatterplot, to depict the bivariate regression. Play with the numbers above and see the changes…

ggplot(funzo, aes(x = age, y = rating)) + geom_point() + geom_smooth(method = "lm", se=F)

#### Regression

Next, run the regression:

reg.2 <- lm(rating~age, funzo)
summary(reg.2)

##
## Call:
## lm(formula = rating ~ age, data = funzo)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -37.902  -9.897  -0.289   7.944  39.270
##
## Coefficients:
##             Estimate Std. Error t value            Pr(>|t|)
## (Intercept) 95.99133    3.18496   30.14 <0.0000000000000002 ***
## age         -1.89573    0.07581  -25.00 <0.0000000000000002 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 14.44 on 98 degrees of freedom
## Multiple R-squared:  0.8645, Adjusted R-squared:  0.8631
## F-statistic: 625.3 on 1 and 98 DF,  p-value: < 0.00000000000000022

Note the R-Squared. Calculate the fitted values and error terms, and then add them to the data set:

funzo$predicted <- fitted(reg.2) funzo$error <- funzo$rating - funzo$predicted

##       rating      age predicted       error
## 1  -11.58271 46.50208   7.83606 -19.4187687
## 2  -22.28820 66.67941 -30.41465   8.1264496
## 3   60.53316 14.76919  67.99298  -7.4598183
## 4   19.73782 32.07206  35.19145 -15.4536250
## 5  -34.78134 68.10289 -33.11319  -1.6681523
## 6   25.30219 37.20698  25.45703  -0.1548443
## 7   37.16856 41.00807  18.25121  18.9173463
## 8   35.28165 30.14312  38.84820  -3.5665496
## 9  -47.37257 67.86045 -32.65358 -14.7189878
## 10 -17.70322 45.29438  10.12554 -27.8287609

#### Calculation

Now we can calculate our variance and fit statistics:

#Calculate our sums of squares:
ss.total <- sum((funzo$rating-mean(funzo$rating))^2)
ss.total

## [1] 150803.8

ss.residual <- sum((funzo$rating - funzo$predicted)^2)
ss.residual

## [1] 20433.77

#And we calculate the R-Squared:

r.squared <- 1 - (ss.residual/ss.total)
r.squared

## [1] 0.8645009

Which should match the R-Squared given in the example above. The results will be differrent depending on your session’s random seed (see the command set.seed())

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text: Persuasive writing is the main style of writing you will use in academic papers. Abhimanyu on August 18, 2019: It is a easy way to remember. The following paragraphs have been devised in an attempt to emphasize the characteristics of each mode of writing. The Read/Write Logic performs several operations on the written/read data, and a set of internal registers called the latches. What are the Modes of Writing Mode generally means type or purpose of writing. Mode: Personal Writing Narrative Writing Explanatory Writing Persuasive Writing Response to Literature Research Writing Creative Writing There's no reason why art can't also make money, but things often don't work out that way. When an author writes in a descriptive style, they are painting a picture in words of a person, place, or thing for their audience. All types of fiction (e.g., novels, short stories, novellas) There are many different models of reflection – you will find some examples in this guide. The author might describe the scene in terms of all five senses. Create. Each of these writing styles is used for a specific purpose. Modes Of Writing Showing top 8 worksheets in the category - Modes Of Writing . A number of statements in this paragraph are opinion rather than fact: that the bike trail is a glittering gem, that the facilities are clean and convenient, and that life in Happyville is idyllic. Answer. The purpose of narrative writing is to tell a story, whether that story is real or imaginary. In many ways it is the easiest kind of writing because it comes so naturally to most people. Expository – Writing in which author’s purpose is to inform or explain the subject to the reader. STUDY. Textbooks Models of communication are conceptual models used to explain the human communication process. Four Major Types of Essays Distinguishing between types of essays is simply a matter of determining the writer’s goal. Ionian. 5 modes of writing. Descriptive writing allows the writer a great deal more artistic freedom than expository writing does. The written form's most important characteristic is that, unlike the immediacy of speech, there is the opportunity to revise and correct. How many vibrational modes are there in the linear $$CO_2$$ molecule ? ... there are some cases that fall in between. Some of them are major modes, some are minor, and some are ambiguous. I really wanted to try to list them here briefly but there are far too many groupings and sub-groupings. Writing, grammar, and communication tips for your inbox. When an author writes in a persuasive style, they are trying to convince the audience of a position or belief. An example journal entry is “What I Miss Most About My Home Country”: Although I miss many things about my home country, I would have to say I miss my family most of all. Created by. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Through description, this passage paints a vivid picture of a scene on the new bike trail. 7 MODES OF WRITING: KEY POINTS TO REMEMBER. Mood: How to Use Tone and Mood in Your Writing, 5 Writing "Rules" That Are Really Guidelines, Beware of These Common Consistency Issues in Writing. Thanks very much. We use communication every day in nearly every environment, including in the workplace. Additionally, there are questions to other types such as Mythos. Answer the question with both modes.e.g. There are four modes or genres of writing: description, narration, exposition, and argumentation. Journalism (except for opinion and editorial articles) However, rest assured, the number is actually more manageable. We’ll also compare writing style examples. They do not require materials such as paper and ink. Advertising In this mode, the writer paints a picture of something with words—an object, person, place, experience, emotion, situation, etc. Omotayo on August 13, 2019: Thank U Sir. Three little girls sat in the grass by a big oak tree. Opinion and editorial pieces There is alignment in concept across the board between the CCSS and the traits, although some of the language changed. In this type of essay, you are required to discuss two opinions given in the question statement and also you have to give your personal opinion. There are 7 modes in music: Ionian; Dorian; Phrygian; Lydian; Mixolydian; Aeolian; Locrian; If you’ve been writing a lot using the vanilla major and minor scales, your next song can … Does the written piece inform? The following descriptions are from PUSD Writing Style and Manual. Languages are systems of symbols, and writing is a system for symbolizing these symbols. But there are only four different types of writing styles. The story could be fact or fiction. To remember the seven modes of music, many musicians use the following … Practically everyone enjoys telling and hearing stories. Creative Commons Attribution 4.0 International License, Business, technical, or scientific writing. For that matter, what is style in writing? Audience – Know to whom your words will be directed. . The greater the distance between “be” and the entity following it, the more likely speakers are to ignore the plural rule. Introduction. Narrative writing can also include dialogue. Writing is long lasting while speech is ephemeral. Pieces in a narrative style will have characters, and through the narrative, the reader learns what happens to them. The author might employ metaphor or other literary devices in order to describe the author’s impressions via their five senses (what they hear, see, smell, taste, or touch). This helps you understand the meanings of the things you read and why they were written. If you’re not quite ready for the writing process, consider keeping a diary and/or journal so you have plenty of notes when you go back to write yo… Among the introductory activities that might be used by teachers is one that focuses on the purposes of writing… Sweet-scented wildflowers brought an array of color to the gently undulating landscape, tempting the children to dismount now and then so they could lay down in the springy, soft grass. One of the most common is the comparison/contrast essay, in which you focus on the ways in which certain things or ideas—usually two of them—are similar to (this is the comparison) and/or different from (this is the contrast) one another. You can have multiple modes in a data set. Inside that subfolder, players should see a folder called Mods.If there is not one already installed, players can create a new folder with that name. However, rest assured, the number is actually more manageable. There are only 9 rhetorical modes. the six modes of language arts are reading, writing, listening, speaking, viewing and representing How many rhetorical modes are there? Poetry Be careful not to oversimplify your cause and effect statements, be cautious about saying that a cause always … So, the list goes: Ionian, Dorian, Phrygian, Lydian, Mixolydian, Aeolian and Locrian. Reviews For example, there is a genre called "historical fiction" that may more or less accurately portray an historical event while making up a story about real or fictional characters involved in that event. If you are not used to being reflective it can be hard to know where to start the process. This is one of the common IELTS writing task 2 essay questions. A single text may include more than one writing style. Types of novels - overview The publishing world tends to classify fiction as either commercial, meaning it's built to make money, or as literary, meaning that it's a work of art. Date: 10/08/2002 at 14:16:35 From: Doctor Achilles Subject: Re: Statistical mode Hi Robin, Thanks for writing back to Dr. Math. Technical writing Read more on creative writing character sketches with this article. Test. Grants the capability to modify, or remove the content of the file. In a future article we will look at various forms of writing. Modes of Writing Chapter Exam Take this practice test to check your existing knowledge of the course material. 5. If you’re interested learning how to write one, consider this article on writing an autobiography. There are many kinds of definitions and many forms that they can take ... You can organize a definition essay according to one or more of the patterns of development (i.e. Santosh on August 14, 2019: it is bad letter. Directory Access Modes. About Writing: A Guide by Robin Jeffrey is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. In an expository piece, a topic will be introduced and laid out in a logical order without reference to the author’s personal opinions. However, there are actually just four main types of writing styles, and together they cover practically all the writing you see, from textbooks to novels, to billboards and more. : The modes are 16 as well as 19rather than:The mode is 16If you are writing it without the words, just put a comma between them.e.g. Example, writing a journal of how you feel about your home country. Descriptive writing is often found in fiction, though it can make an appearance in nonfiction as well (for example, memoirs, first-hand accounts of events, or travel guides). There are equally different reasons for choosing a particular mode of writing. By knowing the different types of writing you will start to recognize them in everything you read. a sonnet is a poem of an expressive thought idea made up of 14 lines each being 10 syllables long. Clearly, the author’s aim here is to use these depictions to persuade readers to use the bike trail. Gravity . Even if they eventually specialize in a particular type of writing, there is great benefit in learning to write broadly. One way to examine writing is to divide writing into specific types or genres. Essentially there are four major types of essays, with the variations making up the remainder. CAUSE AND EFFECT ARGUMENT AND PERSUASION Points to Remember Points to Remember 1. Models of reflection; Reflective writing style; Models of reflection Models and stages of reflection. Essays Persuasive – Writing that states the opinion of the writer and attempts to influence the reader. And impressions by narration whether fiction or nonfiction: 10/08/2002 at 14:16:35 from: Doctor Achilles:! You can almost never prove direct, simple, cause and EFFECT ARGUMENT and PERSUASION to. Or modes, some are minor, and descriptive more entertaining but no less style... 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Here is an explicit proof that a standard Chernoff bound is tight up to constant factors in the exponent for a particular range of the parameters. (In particular, whenever the variables are 0 or 1, and 1 with probability 1/2 or less, and $\epsilon\in(0,1/2)$, and the Chernoff upper bound is less than a constant.) If you find a mistake, please let me know. ...

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The Berry-Esseen theorem can give tail probability lower bounds, as long as they are higher than $n^{-1/2}$. Another tool you can use is the Paley-Zygmund inequality. It implies that for any even integer $k$, and any real-valued random variable $X$, $$\Pr[|X| >= \frac{1}{2}(\mathbb{E}[X^k])^{1/k}] \geq \frac{\mathbb{E}[X^k]^2}{4\mathbb{E}[X^{2k}]}$$ ...

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If you are indeed okay with bounding sums of Bernoulli trials (and not, say, bounded random variables), the following is pretty tight. Slud's Inequality*. Let $\{X_i\}_{i=1}^n$ be i.i.d. draws from a Bernoulli r.v. with $\mathbb{E}(X_1) = p$, and let integer $k\leq n$ be given. If either (a) $p\leq 1/4$ and $np \leq k$, or (b) $np \leq k \leq n(1-p)$, ...

7

The de Moivre-Laplace Theorem shows that variables like $|T\cap S_1|$, after being suitably normalised and under certain conditions, will converge in distribution to a normal distribution. That's enough if you want constant lower bounds. For lower bounds like $n^{-c}$, you need a slightly finer tool. Here's one reference I know of (but only by accident - I'...

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For concreteness, say that the pdf of the r.v. $X_i$ is $$p(X_i = x) = \frac{1}{2} \lambda_i e^{-\lambda_i|x|}.$$ This is the Laplace distribution, or the double exponential distribution. Its variance is $\frac{2}{\lambda_i^2}$. The cdf is $$\Pr[X_i \leq x] = 1 - \frac{1}{2}e^{-\lambda_i x}$$ for $x \geq 0$. The moment generating function of $X_i$ ...

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If $X_1, \cdots, X_n$ are $k$-wise independent random variables on $[0,1]$, then $$\text{Pr} \left[ \left| \sum_i X_i - \text{Ex}\left[ \sum_i X_i \right] \right| \geq n \cdot \varepsilon \right] \leq \left( \frac{k^2}{4 n \varepsilon^2} \right)^{\lfloor k/2 \rfloor}.$$ Note that this bound does not necessarily improve with greater $k$. So you may want to ...

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This is the first link on Google for the search "chernoff limited independence": "Chernoff-Hoeffding Bounds for Applications with Limited Independence" by Schmidt et al.

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I'll take as given the existence of Chaitin's constant $\Omega\in[0,1]$, and that knowing its first $k$ bits is equivalent to be able to decide the halting problem for all Turning machines of size up to $k$: https://en.wikipedia.org/wiki/Chaitin%27s_constant Given access to an $\Omega$-biased coin, one can use Chernoff bounds to compute $\hat\Omega_k$, ...

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The Generalized Littlewood-Offord Theorem isn't exactly what you want, but it gives what I think of as a "reverse Chernoff" bound by showing that the sum of random variables is unlikely to fall within a small range around any particular value (including the expectation). Perhaps it will be useful. Formally, the theorem is as follows. Generalized ...

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For the Laplace distribution, if you use the Bernoulli bound you can write $$Ee^{u\sum_i X_i} = \prod_i \frac1{1-u^2/\lambda_i^2} \le \frac1{1-u^2\sigma^2/2},$$ where $\sigma^2=2\sum_i\lambda_i^{-2}$. Then the classical Chernoff method to gives $$\Pr[\sum_i X_i \ge t\sigma]\le \tfrac{1+\sqrt{1+2t^2}}{2} e^{1-\sqrt{1+2t^2}} \le\cases{ (et/\sqrt2+1) e^{-\... 3 An alternative reference could be Lemma 1.19 in B. Doerr, Analyzing randomized search heuristics: Tools from probability theory, Theory of Randomized Search Heuristics (A. Auger and B. Doerr, eds.), World Scientific Publishing, 2011, pp. 1-20. In simple words, it shows that when X_i=1 with probability p_i no matter what you condition X_1, \dots, X_{i-... 3 You have \mathbb{E}[y_i]=\epsilon q(x_i) + (1-\epsilon)/2 and 0 \leq y_i \leq 1, with all they y_is being independent. Thus the Chernoff-Hoeffding bound gives$$\mathbb{P}\left[\left|\frac{1}{n} \sum_{i=1}^n y_i -\epsilon q(x_i) - \frac{1-\epsilon}2\right| \geq \lambda\right] \leq 2 \cdot e^{-2\lambda^2 n}$$for all \lambda>0. Multiply through by ... 3 The exponent in the standard Chernoff bound as it is stated on Wikipedia is tight for 0/1-valued random variables. Let 0<p<1 and let X_1,X_2,\ldots  be a sequence of independent random variables such that for each i, \Pr[X_i=1]=p and \Pr[X_i=0]=1-p. Then for every \varepsilon>0, \frac{2^{-D(p+\varepsilon\| p)\cdot n}}{n+... 2 You need McDiarmid's inequality, or in particular the Hoeffding inequality, which is a simplification: Given n independent X_i with the property that X_i \in [a_i, b_i], and X = (1/n)\sum X_i, then$$ \text{Pr}(|X - EX| \ge t) \le 2 \exp(\frac{2t^2n^2}{\sum (b_i - a_i)^2}) $$2 The Kearns-Saul inequality states that if X\sim Ber(p) then$$ E[\exp(t(X-p))] = (1-p)e^{-tp}+pe^{t(1-p)} \le \exp\left(\frac{1-2p}{4\log((1-p)/p)}t^2\right). The subgaussian constant $\frac{1-2p}{4\log((1-p)/p)}$ is optimal. See http://ecp.ejpecp.org/article/view/2359 and especially the appendix in http://www.jmlr.org/papers/v16/berend15a.html for ...

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One way to approach this problem is via the CDF transformation. Consider $Z=F(X)$. We know $Z$ is uniformly distributed in $[0,1]$. Let $Z_{(1)},...,Z_{(m)}$ be the order statistics of these $m$ samples (after transforming to Z). It can be shown that $Z_{(t)}\sim \text{Beta}(t,m+1-t)$. Using this, $Z_{(m\cdot k/n)}=F(X_{(m\cdot k/n)})$ is a consistent ...

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I'm posting a suitable answer that I found for this problem. The approach linked here does not exploit the fact that no $t$-tuples of elements of $\Omega$ are present in too many subsets $Y$ (for arbitrary $t$), but it does end up giving me a good enough bound anyway.

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I think that in principle it would take exponential time to compute this probability exactly. Hence sampling would be the only option here, although you would have to settle for additive and multiplicative approximation here. Can I ask you about the problem in which you need this value to be computed?

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# Rational Equations Using Proportions

## Cross-multiply to solve

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Rational Equations Using Proportions

Suppose you were traveling on a paddle boat at a constant speed. In 6 minutes, you traveled \begin{align*}x\end{align*} meters, and in 10 minutes, you traveled \begin{align*}x+4\end{align*} meters. Could you find the value of \begin{align*}x\end{align*} in this scenario? If so, how would you do it? After completing this Concept, you'll be able to solve rational equations using proportions so that you can handle this type of problem.

### Solution of Rational Equations

You are now ready to solve rational equations! There are two main methods you will learn to solve rational equations:

• Cross products
• Lowest common denominators

In this Concept you will learn how to solve using cross products.

Solving a Rational Proportion

When two rational expressions are equal, a proportion is created and can be solved using its cross products.

#### Example A

For example, to solve \begin{align*}\frac{x}{5}=\frac{(x+1)}{2}\end{align*}, cross multiply and the products are equal.

Solve for \begin{align*}x\end{align*}:

#### Example B

Solve \begin{align*}\frac{2x}{x+4}=\frac{5}{x}\end{align*}.

Solution:

Notice that this equation has a degree of two; that is, it is a quadratic equation. We can solve it using the quadratic formula.

#### Example C

Solve \begin{align*}\frac{3x}{5x+2}=\frac{1}{x}\end{align*}.

Solution:

Start by cross multiplying:

\begin{align*}\frac{3x}{5x+2}=\frac{1}{x} \Rightarrow 3x^2=5x+2 \Rightarrow 3x^2-5x-2=0\end{align*}

Since this equation has a squared term as its highest power, it is a quadratic equation. We can solve this by using the quadratic formula, or by factoring.

1. Since there are no common factors, start by finding the product of the coefficient in front of the squared term and the constant:

\begin{align*}3\cdot -2=-6\end{align*}

2. What factors of -6 add up to 5? That would be -6 and 1, since -6+1=-5.

3. Factor, beginning by breaking up the middle term, \begin{align*}-5x\end{align*}, as above:

4. Use the Zero Product Principle:

\begin{align*}(3x+1)(x-2)=0 \Rightarrow 3x+1=0 \text{ or } x-2=0 \Rightarrow x=-\frac{1}{3} \text{ or } x=2\end{align*}

-->

### Guided Practice

Solve \begin{align*} -\frac{x}{2}=\frac{3x-8}{x}\end{align*}.

Solution:

### Explore More

Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Solving Rational Equations (12:57)

Solve the following equations.

1. \begin{align*}\frac{2x+1}{4}=\frac{x-3}{10}\end{align*}
2. \begin{align*}\frac{4x}{x+2}=\frac{5}{9}\end{align*}
3. \begin{align*}\frac{5}{3x-4}=\frac{2}{x+1}\end{align*}
4. \begin{align*}\frac{7x}{x-5}=\frac{x+3}{x}\end{align*}
5. \begin{align*}\frac{2}{x+3}-\frac{1}{x+4}=0\end{align*}
6. \begin{align*}\frac{3x^2+2x-1}{x^2-1}=-2\end{align*}

Mixed Review

1. Divide: \begin{align*}-2 \frac{9}{10} \div - \frac{15}{8}\end{align*}.
2. Solve for \begin{align*}g: -1.5 \left(-3 \frac{4}{5}+g \right)=\frac{201}{20}\end{align*}.
3. Find the discriminant of \begin{align*}6x^2+3x+4=0\end{align*} and determine the nature of the roots.
4. Simplify \begin{align*}\frac{6b}{2b+2}+3\end{align*}.
5. Simplify \begin{align*}\frac{8}{2x-4}- \frac{5x}{x-5}\end{align*}.
6. Divide: \begin{align*}(7x^2+16x-10) \div (x+3)\end{align*}.
7. Simplify \begin{align*}(n-1)*(3n+2)(n-4)\end{align*}.

### Vocabulary Language: English Spanish

proportion

proportion

A statement in which two fractions are equal: $\frac{a}{b} = \frac{c}{d}$.
Zero Product Property

Zero Product Property

The only way a product is zero is if one or more of the terms are equal to zero: $a\cdot b=0 \Rightarrow a=0 \text{ or } b=0.$
Rational Expression

Rational Expression

A rational expression is a fraction with polynomials in the numerator and the denominator.


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# Improving dV/dt turn-on immunity of a mosfet without increasing turn-off time

I'm designing a solid state battery disconnect switch and I want to improve the mosfets' turn-on immunity (high dV/dt on Vds) without impacting the turn-off time. I don't really care about the turn-on time, it can be 100x slower than turn-off time, no problem.

I came up with this solution:

C1 obviously improves the immunity and D1 should leave the turn-off time the same. Did I miss something? is there a better way to do this?

NB: mosfet's datasheet is available here and the driver's datasheet is here. I'll use a +/- 12 V symmetric supply and the driver will drive 10 mosfets in parallel.

• How does D1 prevent C1 from slowing turn off? – evildemonic Mar 17 at 22:27
• What exactly do you mean by mosfet's dynamic turn-on immunity? Do you mean unwanted switchg on of the MOSFET via noise? – vtolentino Mar 17 at 22:34
• @evildemonic It avoids C1 discharging into the driver's output when it goes low. – Biduleohm Mar 17 at 22:35
• @vtolentino I mean unwanted turn-on if Vds rises quickly. – Biduleohm Mar 17 at 22:37
• infineon.com/dgdl/… page 11 and 12 – Biduleohm Mar 17 at 22:45

I simulated your circuit using a similar FET with a transient suppressor added, switching a 'short circuit' drawing 1000 amps from a 50 V battery with 10 nH of wiring inductance. It 'worked' in that it had no effect on the switch-off time or power dissipated during the transition. However in this scenario the transition is slowed down by Miller effect, so (assuming adequate transient suppression) you should not have to worry about Dv/Dt causing the MOSFET to turn on again.

So when might Dv/Dt be a problem? One possibility could be if the battery is plugged in while the FET driver is not powered. In this case there could be a fast rising Drain voltage with only weak resistance pulling the Gate down. But would this cause the FET to turn on? Looking at the MOSFET's capacitance graph we see that Gate-Drain capacitance is ~800 pF at 50 V, while Gate-Source capacitance is ~12 nF. That corresponds to a division ratio of 16:1, so 50 V on the Drain could produce ~3.1V on the Gate. As the threshold voltage is between 2 and 4 V the FET might partially turn on for a few microseconds.

By adding extra capacitance to the Gate your circuit increases the capacitive division ratio which makes the power-on transient Gate voltage lower, (hopefully) ensuring that the FET stays turned off when an external voltage is applied.

• I was somewhat concerned about the inductive spike at turn-off but after a few calculations I deemed it not a problem, not enough inductance. Connecting the battery to something (while the mosfet is or isn't powered, even if when it is it should not be a problem as the gate will be at a negative potential) is effectively my main concern. I'll have a Vds of 64 V maximum and I use the worst case which is Vgsth = 1.2 V @ 120 °C and Crss = 1.7 nF @ a low Vds. Also, I wonder what might be adequate transient suppression because I already thought about it and I'm not sure a TVS diode will be enough. – Biduleohm Mar 18 at 13:00
• If the Gate is pulled negative then it shouldn't be a problem because capacitance will be discharged quickly. If switching off very high current you need very high current transient suppression - I used 6 5W Zeners in parallel in my simulation and it was barely enough. A good TVS should do it. – Bruce Abbott Mar 18 at 20:21
• My worry is more if the control circuitry isn't powered. Ok, I guess this one isn't gonna be good enough (I already use it somewhere elsein the project), maybe this one then? – Biduleohm Mar 18 at 20:36

Your MOSFET driver 1EDI has excellent description !

"The 1EDI EiceDRIVER™ Compact is a general purpose MOSFET gate driver. .... The separated rail-to-rail driver outputs simplify gate resistor selection, save an external high current bypass diode and enhance dV/dt control."

So You can control On and Off time simply by selecting two resistors: Ron from Out+(ON) to Q1gate and Roff from Out-(OFF) to Q1gate;

There is no need of any other components!

time on = Ron.Ciss; time of = Roff*Ciss;

{FDBL86361-F085 Ciss Input Capacitance VDS = 40 V, VGS = 0 V, f = 1 MHz − 12800 − pF}

Aprox. calculations for single MOSFET: Ron=100R; Ciss=13nF=12e-9F; Ton=1.3e-6s = 1.3uS
Roff=10R; Ciss=13nF=12e-9F; Toff=1.3e-7s = 130ns

If you will drive 10 mosfets in parallel then Ciss will increase 10 times.

About driver supply: It is better to use bipolar to "prevent a dynamic turn on of the MOSFET". P.S. If you want to decrease dV/dt simply add an inductance to the load in series. (with all the consequences)

• That's not what the question is about and I already have different resistors for the turn-on and turn-off outputs of the driver. The driver can even be left out of the equation here. Given a high enough dV/dt of Vds the mosfet will turn-on, that's what want to prevent (even when the driver isn't powered). – Biduleohm Mar 27 at 21:33
• What is the type of the load and why are these precautions? – Peter MP Mar 28 at 1:19
• Typical loads would be inverters, lights, motors. I'm not trying to protect against loads but against the act of connecting the battery to something or even if there's a short circuit. You can consider the following: battery negative is connected to the source, the driver isn't powered (so you can ignore it), the mosfet should not turn on if you connect the battery positive to the drain instantly. – Biduleohm Mar 28 at 2:08

Besides your suggestion, there is something else you can do in order to reduce the unwanted gate voltage rise due to the parasitic leakage current.

Consider the following circuit: The mosfet is permanently grounded through the resistor R1 and a square wave with a slew rate of 100n is applied to its drain.

How it works: the leakage current is mostly a function of the parasitic capacitance and the slew rate dV/dt of the drain voltage. Thus, once the drain voltage starts to increase, part of it is coupled through the capacitor C2 and used to turn on the BJts. Due to their high speed, a few bipolars are connected in a cascade fashion in order to increase even more their gain. For a short moment, the gate of the mosfet is shorted and the parasitic capacitance is discharged.

In the waveform you see the effect of not using the circuit (C2 ~ 1p) and with a coupling capacitor (10n). Using this approach the gate voltage dropped by half. I must say that I never saw something like that in any application. The slew rate used in the simulation is extremely high (fast), and probably does not represent a real load.

P.S.: I did not consider any side effect of this proposal.

• That's an interesting solution. But I can't see how Q2 and Q3 can amplify anything as they're not connected to anything else than the signal they should amplify; but I see the idea. I'm not a huge fan of connecting the power side to the gate, there's too many variables (like in my case Vds can be negative for example) to take into account and I'm afraid to forget something and have big problems. – Biduleohm Mar 18 at 0:56

I used the solution I gave in the question with:

C1 = 100 nF (min 90.7 nF calculated with Crss = 1.7 nF, Vds = 64 V and Vgsth = 1.2 V)
R1 = 47 k (value not critical, BoM reuse)
D1 = STPS1L30

Also be warned the inductive voltage spike at turn-off is actually a significant problem contrary to what I thought, you must do something about it if your application is similar to mine. NB: TVS/zener diodes aren't good enough if your working voltage is 64 V and the mosfets have a Vdss of only 80 V.

Why not simply increase the switch-on resistance by adding R3?

simulate this circuit – Schematic created using CircuitLab

• That's not what the question is about and I already have different resistors for the turn-on and turn-off outputs of the driver. The driver can even be left out of the equation here. Given a high enough dV/dt of Vds the mosfet will turn-on, that's what want to prevent (even when the driver isn't powered). – Biduleohm Mar 27 at 21:29
• @Biduleohm I misunderstood your question then. For protection against $dV_{DS}/dt$ you have to add a snubber between D and S. – skvery Mar 28 at 5:50
• I can't rely on a snubber because the power source is a battery capable of delivering kA so a snubber would be either too big and expensive or uneffective. – Biduleohm Mar 28 at 15:24


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Conservation of Angular Momentum Video Lessons

Concept

# Problem: A solid disk of mass m1 = 9.1 kg and radius R = 0.18 m is rotating with a constant angular velocity of ω = 35 rad/s. A thin rectangular rod with mass m2 = 3.6 kg and length L = 2R = 0.36 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.1).What is the initial angular momentum of the rod and disk system?2).What is the initial rotational energy of the rod and disk system?3). What is the final angular velocity of the disk?4).What is the final angular momentum of the rod and disk system?5).What is the final rotational energy of the rod and disk system?6)The rod took t = 5 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?

###### FREE Expert Solution

1.

Angular momentum:

$\overline{){\mathbf{L}}{\mathbf{=}}{\mathbf{I}}{\mathbf{\omega }}}$ where I is the moment of inertia and ω is the angular velocity.

Moment of inertia of solid disk:

$\overline{){\mathbf{I}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{M}}{{\mathbf{R}}}^{{\mathbf{2}}}}$

Moment of inertia of a rod:

$\overline{){\mathbf{I}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{12}}{\mathbf{M}}{{\mathbf{R}}}^{{\mathbf{2}}}}$

The initial angular momentum of the rod is zero because its velocity is zero.

The initial angular momentum of the rod and disk system is the sum of the angular momentum of the two objects.

82% (343 ratings)
###### Problem Details

A solid disk of mass m1 = 9.1 kg and radius R = 0.18 m is rotating with a constant angular velocity of ω = 35 rad/s. A thin rectangular rod with mass m2 = 3.6 kg and length L = 2R = 0.36 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.

1).What is the initial angular momentum of the rod and disk system?

2).What is the initial rotational energy of the rod and disk system?

3). What is the final angular velocity of the disk?

4).What is the final angular momentum of the rod and disk system?

5).What is the final rotational energy of the rod and disk system?

6)The rod took t = 5 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?


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# Capacitor Correcting Power Factor

#### jegues

Joined Sep 13, 2010
733
I feel fairly comfortable with this question, however I have some concerns about the signs for certain values throughout the problem.

First concern, the negative sign on the angle θ from the PF, does it matter that I put it to -36.87°?

The voltage across the load sits at 0°, and since the PF is lagging, the current should be behind the voltage, hence my reason for the negative sign.

Note that I did this for the angle for the new PF in part 2), which brings my next question.

Since I made θ and α negative angles, my reactive portions of the power come out as negative, which is okay but how do I get rid of the j in the impeadance for the capacitor in the expression,

$$Q_{cap} = V^{2}(j\omega C)$$

Are we only considering the magnitude?

If that's the case then the j disapeers, but how do I get rid of the negative on the LHS?

Is this the right way of thinking about it?

EDIT: I realized I've made mistakes in the first portion of the question.

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Last edited:

#### mik3

Joined Feb 4, 2008
4,843
Note that the angle of I is negative but not the angle used in cos() and sin() since S=VI*.

To get rid of j you can set it as a 90 degrees angle.


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1. [SOLVED] inverse of angles

Hey everyone,
I have a test tomorrow and having some difficulty with inverses. so if anyone can help, I'd really appreciate it. Also, I am a visual learner so if it is shown I understand a lot more and can able to apply the same principles I have seen to other problems.

So can someone show me how to do these:

1) Find the inverse function for

, with .

2) Evaluate each of the inverse trig functions in radians rounded to 4 decimal places

2. Hello lsnyder
Originally Posted by lsnyder
1) Find the inverse function for

, with .

$0\le x\le\frac{\pi}{14}\Rightarrow 0\le 7x\le \frac{\pi}{2}$ and $f(x)$ is therefore is a one-to-one function, with values in the range $0\le f(x) \le 1$. So we can find its inverse, using $\arccos$, taking the angle in the range $0$ to $\frac{\pi}{2}$.

So, let $y = f(x) = \cos(7x)$

Then $\arccos(y) = 7x$

$\Rightarrow x = \tfrac17\arccos(y)$

So the inverse function (replacing $y$ by $x$) is $f^{-1}(x) = \tfrac17\arccos(x)$

2) Evaluate each of the inverse trig functions in radians rounded to 4 decimal places

I assume that you have a calculator that will give you inverse tangents and inverse sines. Then:

$x = \text{arccot} (1.5)$

$\Rightarrow\cot(x) = 1.5$

$\Rightarrow \tan(x) = \frac{1}{1.5} = \frac23$

$\Rightarrow x = \arctan\Big(\frac23 \Big) = 0.5880$ to 4 d.p.

And $x = \csc^{-1}(8)$

$\Rightarrow \csc(x) = 8$

$\Rightarrow \sin(x) = \tfrac18$

$x = 0.1253$ to 4 d.p.