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Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_ode_ivp_rk_reset_tend (d02pw)

## Purpose

nag_ode_ivp_rk_reset_tend (d02pw) resets the end point in an integration performed by nag_ode_ivp_rk_onestep (d02pd).
Note: this function is scheduled to be withdrawn, please see d02pw in Advice on Replacement Calls for Withdrawn/Superseded Routines..

## Syntax

[ifail] = d02pw(tendnu)
[ifail] = nag_ode_ivp_rk_reset_tend(tendnu)

## Description

nag_ode_ivp_rk_reset_tend (d02pw) and its associated functions (nag_ode_ivp_rk_onestep (d02pd), nag_ode_ivp_rk_setup (d02pv), nag_ode_ivp_rk_interp (d02px), nag_ode_ivp_rk_diag (d02py) and nag_ode_ivp_rk_errass (d02pz)) solve the initial value problem for a first-order system of ordinary differential equations. The functions, based on Runge–Kutta methods and derived from RKSUITE (see Brankin et al. (1991)), integrate
 y′ = f(t,y)  given  y(t0) = y0 $y′=f(t,y) given y(t0)=y0$
where y$y$ is the vector of n$n$ solution components and t$t$ is the independent variable.
nag_ode_ivp_rk_reset_tend (d02pw) is used to reset the final value of the independent variable, tf${t}_{f}$, when the integration is already underway. It can be used to extend or reduce the range of integration. The new value must be beyond the current value of the independent variable (as returned in tnow by nag_ode_ivp_rk_onestep (d02pd)) in the current direction of integration. It is much more efficient to use nag_ode_ivp_rk_reset_tend (d02pw) for this purpose than to use nag_ode_ivp_rk_setup (d02pv) which involves the overhead of a complete restart of the integration.
If you want to change the direction of integration then you must restart by a call to nag_ode_ivp_rk_setup (d02pv).

## References

Brankin R W, Gladwell I and Shampine L F (1991) RKSUITE: A suite of Runge–Kutta codes for the initial value problems for ODEs SoftReport 91-S1 Southern Methodist University

## Parameters

### Compulsory Input Parameters

1:     tendnu – double scalar
The new value for tf${t}_{f}$.
Constraint: sign(tendnutnow) = sign(tendtstart)$\mathrm{sign}\left({\mathbf{tendnu}}-{\mathbf{tnow}}\right)=\mathrm{sign}\left({\mathbf{tend}}-{\mathbf{tstart}}\right)$, where tstart and tend are as supplied in the previous call to nag_ode_ivp_rk_setup (d02pv) and tnow is returned by the preceding call to nag_ode_ivp_rk_onestep (d02pd). tendnu must be distinguishable from tnow for the method and the machine precision being used.

None.

None.

### Output Parameters

1:     ifail – int64int32nag_int scalar
${\mathrm{ifail}}={\mathbf{0}}$ unless the function detects an error (see [Error Indicators and Warnings]).

## Error Indicators and Warnings

Errors or warnings detected by the function:
ifail = 1${\mathbf{ifail}}=1$
On entry, an invalid input value for tendnu was detected or an invalid call to nag_ode_ivp_rk_reset_tend (d02pw) was made, for example without a previous call to the integration function nag_ode_ivp_rk_onestep (d02pd). You cannot continue integrating the problem.

Not applicable.

None.

## Example

This example integrates a two body problem. The equations for the coordinates (x(t),y(t))$\left(x\left(t\right),y\left(t\right)\right)$ of one body as functions of time t$t$ in a suitable frame of reference are
 x ′ ′ = − x/(r3) $x′′=-xr3$
 y ′ ′ = − y/(r3),   r = sqrt(x2 + y2). $y′′=-yr3, r=x2+y2.$
The initial conditions
 x(0) = 1 − ε, x′(0) = 0 y(0) = 0, y′(0) = sqrt((1 + ε)/(1 − ε))
$x(0)=1-ε, x′(0)=0 y(0)=0, y′(0)= 1+ε 1-ε$
lead to elliptic motion with 0 < ε < 1$0<\epsilon <1$. ε = 0.7$\epsilon =0.7$ is selected and reposed as
 y1 ′ = y3 y2 ′ = y4 y3 ′ = − (y1)/(r3) y4 ′ = − (y2)/(r3)
$y1′=y3 y2′=y4 y3′=- y1r3 y4′=- y2r3$
over the range [0,6π]$\left[0,6\pi \right]$. Relative error control is used with threshold values of 1.0e−10$\text{1.0e−10}$ for each solution component and compute the solution at intervals of length π$\pi$ across the range using nag_ode_ivp_rk_reset_tend (d02pw) to reset the end of the integration range. A high-order Runge–Kutta method (method = 3${\mathbf{method}}=3$) is also used with tolerances tol = 1.0e−4${\mathbf{tol}}=\text{1.0e−4}$ and tol = 1.0e−5${\mathbf{tol}}=\text{1.0e−5}$ in turn so that the solutions may be compared. The value of π$\pi$ is obtained by using nag_math_pi (x01aa).
Note that the length of tol = 1.0e−4${\mathbf{tol}}=\text{1.0e−4}$ and work is large enough for any valid combination of input arguments to nag_ode_ivp_rk_setup (d02pv).
```function nag_ode_ivp_rk_reset_tend_example
% Initialize variables and arrays.
neq = 4;
lenwrk = 32*neq;
method = 3;
ecc = 0.7;

tstart = 0.0;
ystart = [1.0-ecc; 0.0; 0.0; sqrt((1.0+ecc)/(1.0-ecc))];
tend = 6.0*pi;
thres = [1e-10; 1e-10; 1e-10; 1e-10];
errass = false;
hstart = 0.0;

fprintf('nag_ode_ivp_rk_reset_tend example program results \n\n');

% We run through the calculation twice: once to output the results at
% a collection of points, and again to accumulate a series of results for
% plotting.
nstep = [6; 96];

% Prepare to accumulate results.  The first and last points should be the
% same - i.e. so the plot shows a closed trajectory.
xarray = zeros(nstep(2)+1, 1);
yarray = zeros(nstep(2)+1, 4);

for icalc = 1:2
tinc  = tend/nstep(icalc);

% For each calculation, we use two tolerances (and we'll plot the
% results corresponding to the second one).
tol = [1.0e-4; 1.0e-5];
for itol = 1:2;

istep = 1;
twant = tinc;

% nag_ode_ivp_rk_setup is a setup routine to be called prior to nag_ode_ivp_rk_onestep.
[work, ifail] = nag_ode_ivp_rk_setup(tstart, ystart, twant, tol(itol), thres, ...
int64(method), task, errass, int64(lenwrk), 'neq', ...
int64(neq), 'hstart', hstart);
if ifail ~= 0
% Unsuccessful call.  Print message and exit.
error('Warning: nag_ode_ivp_rk_setup returned with ifail = %1d ',ifail);
end

if icalc == 1
% Output initial results.
fprintf('Calculation with tol = %1.1e\n\n',tol(itol));
fprintf('   t         y1        y2        y3        y4\n');
fprintf(' %6.3f', tstart);
for ieq = 1:neq
fprintf('  %8.4f', ystart(ieq));
end
fprintf('\n');
else
% Store current results.
xarray(1) = tstart;
for ieq = 1:neq
yarray(1, ieq) = ystart(ieq);
end
end

tnow = 0.0;
while tnow < tend
if icalc == 1
% For the first calculation, keep integrating till the
% (current value for the) endpoint is reached.
while tnow < twant
[tnow, ynow, ypnow, work, ifail] = nag_ode_ivp_rk_onestep(@f, ...
int64(neq), work);
if ifail ~= 0
% Unsuccessful call.  Print message and exit.
error(...
'Warning: nag_ode_ivp_rk_onestep returned with ifail = %1d ',...
ifail);
end
end
% Output current result.
fprintf(' %6.3f', tnow);
for ieq = 1:neq
fprintf('  %8.4f', ynow(ieq));
end
fprintf('\n');
else
% For the second calculation, just take a single step.
[tnow, ynow, ypnow, work, ifail] = nag_ode_ivp_rk_onestep(@f, ...
int64(neq), work);
if ifail ~= 0
% Unsuccessful call.  Print message and exit.
error(...
'Warning: nag_ode_ivp_rk_onestep returned with ifail = %1d ',...
ifail);
end
% Store current result.
xarray(istep+1) = tnow;
for ieq = 1:neq
yarray(istep+1, ieq) = ynow(ieq);
end
end

% Update the endpoint, and call nag_ode_ivp_rk_reset_tend to reset it.
istep = istep + 1;
twant = istep*tinc;
[ifail] = nag_ode_ivp_rk_reset_tend(twant);
end

if icalc == 1
% nag_ode_ivp_rk_diag is a diagnostic routine.
[totfcn, stpcst, waste, stpsok, hnext, ifail] = nag_ode_ivp_rk_diag;
fprintf(['\nCost of the integration in evaluations of ', ...
'F is %1.0f\n\n\n'], totfcn);
end
end
end

% Use the first two components of the solution, and calculate the deviation
% from a true ellipse.
x = yarray(:,1);
y = yarray(:,2);
xdev = zeros(nstep(2)+1, 1);
ydev = zeros(nstep(2)+1, 1);
for i = 1:nstep(2)+1
fac = abs((x(i) + ecc)*(x(i) + ecc) + y(i)*y(i)/(ecc*ecc) - 1.0);
xdev(i) = fac*cos(xarray(i));
ydev(i) = fac*sin(xarray(i));
end
% Plot results.
fig = figure('Number', 'off');
display_plot(x, y, xdev, ydev);

function [yp] = f(t, y, yp)
% Evaluate derivative vector.

yp = zeros(4, 1);
r = sqrt((y(1)^2 + y(2)^2));
yp(1) =  y(3);
yp(2) =  y(4);
yp(3) = -y(1)/r^3;
yp(4) = -y(2)/r^3;
function display_plot(x1, y1, x2, y2)
% Formatting for title and axis labels.
titleFmt = {'FontName', 'Helvetica', 'FontWeight', 'Bold', 'FontSize', 14};
labFmt = {'FontName', 'Helvetica', 'FontWeight', 'Bold', 'FontSize', 13};
set(gca, 'FontSize', 13); % for legend, axis tick labels, etc.
% Plot the results.  First, get the list of default colours (we have to
% specify the plot colours by hand).
cols = get(gca, 'ColorOrder');
% Plot the first curve.
hline1 = line(x1, y1, 'Color', cols(1,:));
ax1 = gca;
set(ax1, 'XColor', cols(1,:), 'YColor', cols(1,:));
% Label these axes.
xlabel('Orbit - x', labFmt{:});
ylabel('Orbit - y', labFmt{:});
% NB We don't give a title to this plot, because it would collide with the
% the xlabel for the second set of axes (see below).
% Set up the second set of axes and plot the second curve.
ax2 = axes('Position', get(ax1,'Position'), ...
'XAxisLocation', 'top', 'YAxisLocation', 'right', ...
'Color','none', 'XColor',cols(2,:), 'YColor', cols(2,:), ...
'XLim', [-0.25 0.06], 'YLim', [-0.1 0.1], 'FontSize', 13);
hline2 = line(x2, y2, 'Color', cols(2,:), 'Parent', ax2);
% Label these axes.
xlabel('x Deviation from True Ellipse', labFmt{:});
ylabel('y Deviation from True Ellipse', labFmt{:});
% Add a legend, specifying the lines explicitly.
legend([hline1, hline2],'Orbit','Deviation','Location','Best');
% Set some features of the two lines.
set(hline1, 'Linewidth', 0.25, 'Marker', '+', 'Line', '-');
set(hline2, 'Linewidth', 0.25, 'Marker', 'x', 'Line', '--');
```
```
nag_ode_ivp_rk_reset_tend example program results

Calculation with tol = 1.0e-04

t         y1        y2        y3        y4
0.000    0.3000    0.0000    0.0000    2.3805
3.142   -1.7000    0.0000   -0.0000   -0.4201
6.283    0.3000   -0.0000    0.0001    2.3805
9.425   -1.7000    0.0000   -0.0000   -0.4201
12.566    0.3000   -0.0003    0.0016    2.3805
15.708   -1.7001    0.0001   -0.0001   -0.4201
18.850    0.3000   -0.0010    0.0045    2.3805

Cost of the integration in evaluations of F is 571

Calculation with tol = 1.0e-05

t         y1        y2        y3        y4
0.000    0.3000    0.0000    0.0000    2.3805
3.142   -1.7000   -0.0000    0.0000   -0.4201
6.283    0.3000    0.0000   -0.0000    2.3805
9.425   -1.7000    0.0000   -0.0000   -0.4201
12.566    0.3000   -0.0001    0.0004    2.3805
15.708   -1.7000    0.0000   -0.0000   -0.4201
18.850    0.3000   -0.0003    0.0012    2.3805

Cost of the integration in evaluations of F is 748

```
```function d02pw_example
% Initialize variables and arrays.
neq = 4;
lenwrk = 32*neq;
method = 3;
ecc = 0.7;

tstart = 0.0;
ystart = [1.0-ecc; 0.0; 0.0; sqrt((1.0+ecc)/(1.0-ecc))];
tend = 6.0*pi;
thres = [1e-10; 1e-10; 1e-10; 1e-10];
errass = false;
hstart = 0.0;

fprintf('d02pw example program results \n\n');

% We run through the calculation twice: once to output the results at
% a collection of points, and again to accumulate a series of results for
% plotting.
nstep = [6; 96];

% Prepare to accumulate results.  The first and last points should be the
% same - i.e. so the plot shows a closed trajectory.
xarray = zeros(nstep(2)+1, 1);
yarray = zeros(nstep(2)+1, 4);

for icalc = 1:2
tinc  = tend/nstep(icalc);

% For each calculation, we use two tolerances (and we'll plot the
% results corresponding to the second one).
tol = [1.0e-4; 1.0e-5];
for itol = 1:2;

istep = 1;
twant = tinc;

% d02pv is a setup routine to be called prior to d02pd.
[work, ifail] = d02pv(tstart, ystart, twant, tol(itol), thres, ...
int64(method), task, errass, int64(lenwrk), 'neq', ...
int64(neq), 'hstart', hstart);
if ifail ~= 0
% Unsuccessful call.  Print message and exit.
error('Warning: d02pv returned with ifail = %1d ',ifail);
end

if icalc == 1
% Output initial results.
fprintf('Calculation with tol = %1.1e\n\n',tol(itol));
fprintf('   t         y1        y2        y3        y4\n');
fprintf(' %6.3f', tstart);
for ieq = 1:neq
fprintf('  %8.4f', ystart(ieq));
end
fprintf('\n');
else
% Store current results.
xarray(1) = tstart;
for ieq = 1:neq
yarray(1, ieq) = ystart(ieq);
end
end

tnow = 0.0;
while tnow < tend
if icalc == 1
% For the first calculation, keep integrating till the
% (current value for the) endpoint is reached.
while tnow < twant
[tnow, ynow, ypnow, work, ifail] = d02pd(@f, ...
int64(neq), work);
if ifail ~= 0
% Unsuccessful call.  Print message and exit.
error(...
'Warning: d02pd returned with ifail = %1d ',...
ifail);
end
end
% Output current result.
fprintf(' %6.3f', tnow);
for ieq = 1:neq
fprintf('  %8.4f', ynow(ieq));
end
fprintf('\n');
else
% For the second calculation, just take a single step.
[tnow, ynow, ypnow, work, ifail] = d02pd(@f, ...
int64(neq), work);
if ifail ~= 0
% Unsuccessful call.  Print message and exit.
error(...
'Warning: d02pd returned with ifail = %1d ',...
ifail);
end
% Store current result.
xarray(istep+1) = tnow;
for ieq = 1:neq
yarray(istep+1, ieq) = ynow(ieq);
end
end

% Update the endpoint, and call d02pw to reset it.
istep = istep + 1;
twant = istep*tinc;
[ifail] = d02pw(twant);
end

if icalc == 1
% d02py is a diagnostic routine.
[totfcn, stpcst, waste, stpsok, hnext, ifail] = d02py;
fprintf(['\nCost of the integration in evaluations of ', ...
'F is %1.0f\n\n\n'], totfcn);
end
end
end

% Use the first two components of the solution, and calculate the deviation
% from a true ellipse.
x = yarray(:,1);
y = yarray(:,2);
xdev = zeros(nstep(2)+1, 1);
ydev = zeros(nstep(2)+1, 1);
for i = 1:nstep(2)+1
fac = abs((x(i) + ecc)*(x(i) + ecc) + y(i)*y(i)/(ecc*ecc) - 1.0);
xdev(i) = fac*cos(xarray(i));
ydev(i) = fac*sin(xarray(i));
end
% Plot results.
fig = figure('Number', 'off');
display_plot(x, y, xdev, ydev);

function [yp] = f(t, y, yp)
% Evaluate derivative vector.

yp = zeros(4, 1);
r = sqrt((y(1)^2 + y(2)^2));
yp(1) =  y(3);
yp(2) =  y(4);
yp(3) = -y(1)/r^3;
yp(4) = -y(2)/r^3;
function display_plot(x1, y1, x2, y2)
% Formatting for title and axis labels.
titleFmt = {'FontName', 'Helvetica', 'FontWeight', 'Bold', 'FontSize', 14};
labFmt = {'FontName', 'Helvetica', 'FontWeight', 'Bold', 'FontSize', 13};
set(gca, 'FontSize', 13); % for legend, axis tick labels, etc.
% Plot the results.  First, get the list of default colours (we have to
% specify the plot colours by hand).
cols = get(gca, 'ColorOrder');
% Plot the first curve.
hline1 = line(x1, y1, 'Color', cols(1,:));
ax1 = gca;
set(ax1, 'XColor', cols(1,:), 'YColor', cols(1,:));
% Label these axes.
xlabel('Orbit - x', labFmt{:});
ylabel('Orbit - y', labFmt{:});
% NB We don't give a title to this plot, because it would collide with the
% the xlabel for the second set of axes (see below).
% Set up the second set of axes and plot the second curve.
ax2 = axes('Position', get(ax1,'Position'), ...
'XAxisLocation', 'top', 'YAxisLocation', 'right', ...
'Color','none', 'XColor',cols(2,:), 'YColor', cols(2,:), ...
'XLim', [-0.25 0.06], 'YLim', [-0.1 0.1], 'FontSize', 13);
hline2 = line(x2, y2, 'Color', cols(2,:), 'Parent', ax2);
% Label these axes.
xlabel('x Deviation from True Ellipse', labFmt{:});
ylabel('y Deviation from True Ellipse', labFmt{:});
% Add a legend, specifying the lines explicitly.
legend([hline1, hline2],'Orbit','Deviation','Location','Best');
% Set some features of the two lines.
set(hline1, 'Linewidth', 0.25, 'Marker', '+', 'Line', '-');
set(hline2, 'Linewidth', 0.25, 'Marker', 'x', 'Line', '--');
```
```
d02pw example program results

Calculation with tol = 1.0e-04

t         y1        y2        y3        y4
0.000    0.3000    0.0000    0.0000    2.3805
3.142   -1.7000    0.0000   -0.0000   -0.4201
6.283    0.3000   -0.0000    0.0001    2.3805
9.425   -1.7000    0.0000   -0.0000   -0.4201
12.566    0.3000   -0.0003    0.0016    2.3805
15.708   -1.7001    0.0001   -0.0001   -0.4201
18.850    0.3000   -0.0010    0.0045    2.3805

Cost of the integration in evaluations of F is 571

Calculation with tol = 1.0e-05

t         y1        y2        y3        y4
0.000    0.3000    0.0000    0.0000    2.3805
3.142   -1.7000   -0.0000    0.0000   -0.4201
6.283    0.3000    0.0000   -0.0000    2.3805
9.425   -1.7000    0.0000   -0.0000   -0.4201
12.566    0.3000   -0.0001    0.0004    2.3805
15.708   -1.7000    0.0000   -0.0000   -0.4201
18.850    0.3000   -0.0003    0.0012    2.3805

Cost of the integration in evaluations of F is 748

```


text:
# Multivariable piecewise functions: Julia-basics help

Hi, I was wondering how to create a piecewise function of two variables

r(x,y) = \begin{cases} f(x,y) & d(x,y)\geq 0 \\ g(x,y) & d(x,y)<0\end{cases} and use it to plot the coordinate pair (x',y')= (m(x,y), n(x,y)) on xy- plane where both m and n are scalar [but also functions of r(x,y)].

So far I thought of creating x= -10:10 and y : -10:10 and to store their values in arrays and then just output array of m and n values. Is it the correct thought process? End goal is to plot (m,n) based on (x,y) based on different values of a. Sorry for the basic doubts. Any help is appreciated. Thanks.

a= 1/2
g(x) = a*x^2

function p(y)
return -(2a*y-1)^2
end

function q(x,y)
return ((2a*y-1)^3-(27a^2)*(x^2))
end

function del(x,y)
return (a^2*x^2-2(a*y-1)^3)*x^2
end

function r(x,y)
if all(del(x,y) .>= 0)
return -(a*y+1)./(3a)
else
return (a*y+1)./(3a)
end
end

x = range(0, 50, length=100);
y = range(0, 50, length=100);
#plot(m,n)??

I wouldn’t do it this way. Just define a function to work on scalar inputs, and then broadcast it. (And it’s good style to avoid non-constant globals, i.e. pass a as a parameter.)

e.g.

del(x,y,a=1/2) = ((a^2*x^2-2(a*y-1)^3)*x^2
r(x,y,a=1/2) = del(x,y,a) ≥ 0 ? -(a*y+1)/(3a) : (a*y+1)/(3a)

x = y = range(-10,10, length=100)
z = del.(x, y') # 100×100 grid of values for plotting

Get out of the habit (from Matlab or Python or R) of trying to define “vectorized” functions for things that are logically scalar. Scalar functions are fast in Julia, and are easier to work with.

1 Like

Thanks a lot for your answer, it helped. Thank you for the advice as well [beginning to learn Julia].

format: plain


text:
## anonymous one year ago Lim x goes to 1 x^2-1/|x-1| anyone can find this limit?

1. anonymous

$\lim_{x\to 1}\frac{x^2-1}{|x-1|}$

2. anonymous

seems unlikely since this is is a piecewise function it is one thing if $$x>1$$ and quite another if $$x<1$$

3. anonymous

if $$x>1$$ then $$|x-1|=x-1$$ when you factor and cancel you get $x+1$

4. anonymous

you get something else if $$x<1$$ try it and see

5. anonymous

I wonder if I should use left and right limit to do this

6. anonymous

you have no choice but to do that, since $$|x-1|$$ is a piecewise function which changes definition at $$x=1$$

7. anonymous

lets take it step by step

8. anonymous

if $$x>1$$ then $$|x-1|=x-1$$ right?

9. anonymous

did i lose you there?

10. anonymous

im here

11. anonymous

if $$x>1$$ then $$|x-1|=x-1$$so your function is $\frac{x^2-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1$ making the limit as $$x\to 1^+=1+1=2$$

12. anonymous

I see, now the left limit I guess

13. anonymous

left limit is different because if $$x<1$$ then $$|x-1|=1-x$$

14. anonymous

so the limit doesnt exist right

15. anonymous

no the two sided limit does not exist

16. anonymous

i see

17. anonymous

Thank you


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A number when divided by 36 leaves a reminder of 23. Which o  [#permalink]

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Updated on: 05 Oct 2013, 04:29
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A number when divided by 36 leaves a reminder of 23. Which of the following could be a reminder when it is divided by72.

A. 13
B. 59
C. 37
D. 21

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Originally posted by cleetus on 24 Nov 2011, 10:37.
Last edited by Bunuel on 05 Oct 2013, 04:29, edited 1 time in total.
Renamed the topic and edited the question.
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25 Nov 2011, 11:11
5
liftoff wrote:
I got B... Used the long approach
List multiples of 36
liftoff wrote:
I got B... Used the long approach
List multiples of 36
thisiszico2006 wrote:
Answer is B , but is there a generic formula or approach for this kind of problem instead of trail and error approach ?

U need to have an approach to solve this. Lets solve this logiccally

Here is the Approach
Given that N=36k+23 Here k could be either odd or even.
we got to find the reminder when 36 multiplied by a number, is divided by 72. So in the above equation K could be of odd or even numbers. give k values starting from 0 to 4

Case 1. When k=0 (36*0)+23 /72 gives reminder 23
Case 2. when k=1 (36*1)+23 /72 gives reminder 59.How we got59 is that-36/72gives reminder36.This 36+ 23=59. )
Case 3.when k= 2 (36*2)+23 /72 gives reminder 23
Case 4.When k=3 (36*3)+23 /72 gives reminder 59
So from this it can be understood that reminder can be either 23 or 59 based on the value of k if its even or odd.
In the choice only 59 is given.
So the answer is B 59 BINGO!!
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24 Nov 2011, 12:41
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N = 36*a+23. Try a = 1, then N = 59. 59/72 = 0 with remainder 59. Answer is B.
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24 Nov 2011, 23:11
Answer is B , but is there a generic formula or approach for this kind of problem instead of trail and error approach ?
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h(n) defined as product of even integers from 2 to n
Number N divided by D leaves remainder R
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25 Nov 2011, 04:22
I got B... Used the long approach
List multiples of 36
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25 Nov 2011, 12:57
2
So the learning from this problem can be summarized below :

If a number "N" is divided by "D" and leaves a remainder "R" , then if that same number "N" is divided by "k*D" where "k" is a positive integer , then the number of remainders that are possible are R , D+R , 2D+R , .... , (k-1)D + R.

So if the question was that what is the remainder when the number was divided by 288 , we can approach it in a mechanical way.

The possible remainders are 23 , 36+23 , 72+23 , .... , 252+23 . We can straightaway reject answer choices that are not in this list.

Sample question : A number N divided by D gives a remainder of 7 , the same number divided by 5D gives a remainder of 24. Find D [N and D are both positive integers].

Ans: The possible remainder for the 2nd case are 7 , D+7 , 2D+7 , 3D+7 , 4D+7.
So 7<> 24, case closed for 1st possibility ,
D+7 = 24 or D=17 (ok)
2D+7= 24, not ok as D cannot be fraction.
3D + 7=24, not ok as D cannot be fraction
4D+ 7 = 24, not possible as D cannot be fraction.

So possible values of D is only 17. So 17 must be present among the answer choices.
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h(n) defined as product of even integers from 2 to n
Number N divided by D leaves remainder R
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Re: A number when divided by 36 leaves a reminder of 23. Which o  [#permalink]

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15 Nov 2013, 03:14
Hi guys,

N = 36 A + 23
N = 72 B + X ( A, B and X are integers)

This implies that : 36 A + 23 = 72 B + X or 36 (A - 2B) = X - 23

X - 23 must be a multiple of 36. The only option is 59.

Not sure whether this is a good approach. Anyway it took me almost 2 min.
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Re: A number when divided by 36 leaves a reminder of 23. Which o  [#permalink]

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14 Dec 2013, 05:37
2
It is given that
$$n = 36p+23$$

let r be remainder when n is divided by 72 so
$$n = 72q+r$$ where p and q will be integers.

from the two we have

$$36p+23 = 72q+r$$
=> $$r = 23+36p-72q$$
=> $$r = 23+36(p-2q)$$

since p and q are integers $$p-2q$$ is also an integer so it can be ...-3,-2-1,0,1,2,3...

If p-2q = 0 then r =23
If p- 2q = 1 then r = 23+36 = 59 we can stop here and pick B.
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Re: A number when divided by 36 leaves a reminder of 23. Which o  [#permalink]

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02 Feb 2014, 16:53
cleetus wrote:
A number when divided by 36 leaves a reminder of 23. Which of the following could be a reminder when it is divided by72.

A. 13
B. 59
C. 37
D. 21

You only need to know that n = 36p + 23

Now when 36p+23 / 72 what is the remainder?

Well if p = 1 then remainder is 36+23 =59, which is answer choice B

So B

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Re: A number when divided by 36 leaves a reminder of 23. Which o  [#permalink]

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05 Jun 2016, 14:03
cleetus wrote:
A number when divided by 36 leaves a reminder of 23. Which of the following could be a reminder when it is divided by72.

A. 13
B. 59
C. 37
D. 21

n= 36x + 23

If we substitute number x= 1

we get n= 59, and when n is divided by 72 we get 59 as the remainder.

We have 59 as one option, and hence we don't need to do any further calculations

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Re: A number when divided by 36 leaves a reminder of 23. Which o  [#permalink]

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07 Jul 2018, 18:25
cleetus wrote:
A number when divided by 36 leaves a reminder of 23. Which of the following could be a reminder when it is divided by72.

A. 13
B. 59
C. 37
D. 21

We can let the number be 36 + 23 = 59. When 59 is divided by 72 the remainder is 59.

Alternate Solution:

We observe that 72 is exactly twice 36. If the remainder when a number is divided by 36 is 23, then the remainder when the same number is divided by 72 can be either 23 or 23 + 36 = 59.

If this is hard to see, let’s illustrate the same phenomenon in a simple example. Let’s divide some numbers by 4 and 8 and compare the remainders:

3/4 = 0 R 3, 3/8 = 0 R 3

5/4 = 1 R 1, 5/8 = 0 R 5 (which is 1 + 4)

9/4 = 2 R 1, 9/8 = 1 R 1

14/4 = 3 R 2, 14/8 = 1 R 6 (which is 2 + 4)

The only possible remainder in the answer choices is 59.

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Re: A number when divided by 36 leaves a reminder of 23. Which o   [#permalink] 07 Jul 2018, 18:25
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text:
# Thread: Find the Midpoint Between Two Points

1. ## Find the Midpoint Between Two Points

The question is worded as follows:

find the midpoint (a,b) between the two points (-5,5),(-4,2).

Then it has two answer fields. one for A= and another for B=. How am I supposed to answer this?

2. ## re: Find the Midpoint Between Two Points

Hi Champloo11,
The question is for you to calculate the values of x-coordinate 'a' and the y -coordinate 'b' of the mid-point.
Is the question clear to you now.

Kalyan

3. ## re: Find the Midpoint Between Two Points

The midpoint betweens two others is defined as the one with equal distance to both. That would mean that if $X=(-5,5)$, $Y=(-4,2)$ and $M$ is the midpoint between those, we have:

$d(X,M) = d(Y,M)$

Since this is $\mathbb{R}^2$, the midpoint would be middleway the line connecting $X$ and $Y$.

4. ## re: Find the Midpoint Between Two Points

Thank you everyone, I kept misreading the question. I thought it was asking for 2 individual points, and not the assignment of the points x and y as a and b.


text:
# Velocity of Simple Pendulum Formula?

#### Jumb0

5+ Year Member
Ok, so the velocity of pendulum bob at some angle θ made with respect to the vertical is supposed to be given by :

v = sqrt [2gL(1−cosθ)]

Which makes sense until you think of the fact that θ ought to be ZERO when the pendulum is at it's lowest point. This does not agree with the formula, however, because cos(0)=1, therefore 1-cos(0) = 0, therefore the velocity is ZERO at the bottom of the swing...but everyone knows that velocity is actually at a MAXIMUM at the bottom of the swing....

Maybe they meant that θ is the angle made by the pendulum with respect tot he HORIZONTAL...That way it would be cos(90) at the bottom, therefore velocity would be at a maximum...

Is there something simple I am overlooking here?

#### Cawolf

##### Medical Student
5+ Year Member
You seemed to have it figured out. Theta can be set however you like, but it the formula that you wrote out, assume that theta = 90 at the point where all energy is kinetic, straight down.

If you set the sum of PE and KE = 0 and then derive the velocity formula you can see how theta is described.

OP

#### Jumb0

5+ Year Member

1. This formula tells you the MAXIMUM velocity ONLY. It is only good for the velocity at the very bottom.

2. The angle θ in this case refers to the angle made with the vertical IN THE INITIAL CONDITION. It is the angle that you displace the bob from equilibrium to begin the periodic motion...

It is not a general formula for the velocity of the bob at any point in its trajectory like I had erroneously assumed at first.

#### Cawolf

##### Medical Student
5+ Year Member
The formula can describe the velocity at any angle theta.

It only depends on the height of the bob, which is described by the angle.


text:
# Simple Explanation of Baum Welch/Viterbi

I'm looking for a very simple explanation as possible for Baum Welch and Viterbi for HMMs with a straightforward well annotated example. Almost all of the explanations I find on the net invariably jump from the beginning into an almost completely jargon/symbol based explanation. The few times they have an example it is usually shoved to the side or poorly annotated (ie its unclear how the symbols and the example are related to each other). I've heard stuff like some sources claiming Baum Welch is based on the EM algorithm and others saying its based off of the forward backward algorithm and some say both so I'd like to get details like that cleared up.

The closest I found so far is a 1hr video posted here that people said was great but is unplayable because flash is banned and an explanation on wikipedia that has a note next to it saying its wrong.

I'm looking for a very simple explanation as possible for Baum Welch and Viterbi for HMMs...

• Given the observed data, the Baum-Welch algorithm finds the likelihood-maximizing parameters.

• Given the observed data, and the parameters, the Viterbi algorithm finds the most likely sequence of hidden states.

I've heard stuff like some sources claiming Baum Welch is based on the EM algorithm...

The Baum-Welch algorithm is a special case of the EM algorithm.

...and others saying its based off of the forward backward algorithm

The forward-backward decomposition is how you do the "E-" step (using the EM terminology). It's specific to hidden Markov models.

The few times they have an example it is usually shoved to the side or poorly annotated (ie its unclear how the symbols and the example are related to each other)

I don't know what examples you're looking at, but I can tell you that there are many good ones out there. Remember that these methods were invented a very long time ago, and they're still widely used. The Rabiner one linked to above is probably the most famous, but you have to find one you like that you'll keep working through. I would recommend you find one with an example that has data you're interested in, and notation that you feel is learn-able. Whatever you find will have a lot of notation. There is no way around that, so you've just got to find something you can settle into.

I have the perfect paper for you: http://www.stat.columbia.edu/~liam/teaching/neurostat-fall17/papers/hmm/rabiner.pdf

Note afaik there are newer BW methods.

Attempted simple explanation for Viterbi: The goal of viterbi is to find an optimal sequence of states during some discrete period of time.
Start off with the initial probabilities of being in a state $s$ at time $t=0$. Then increase $t$ and for each state at that time ($t=1$), find the best state $s$ at $t-1$ (best as in the sequence which gives the highest probability). Save this information. That means for each state $s$ at $t=1$ save what the best state $s$ at $t=0$ is. Also save the probability for that (optimal) sequence.
Then increase $t$ to $t=2$, and repeat the procedure (with the probabilities for each state $s$ at $t-1$ being the just calculated best probabilities for that state). At some point you'll reach $t=T$ (last point in time). Then you just pick the state with the highest probability, and remember for each state at each time you have saved the best state for $t-1$, so take the best state at $t=T$ and then work your way backwards always picking that saved optimal state.

To elaborate on what it does on a more general level: If you have a set of states $S$ which can all be picked over some duration of time $T$, a naive solution to find the best sequence would be to enumerate all possible sequences and then calculate the probability for each of them (that means calculating the product of the state and transition probabilities). The mistake one is doing here is that a lot of the calculations are duplicated, as an example you could have two sequences $p_1, p_2$ ($p$ is a sequence of states) where the only difference is the final state, yet with this approach you are calculating the probabilities of the shared states for both $P(p_1)$ and $P(p_2)$.
By taking the iterative approach used in viterbi, there are no duplicated calculations, so obviously it will be a lot more efficient.


text:
# kwant.physics.magnetic_gauge¶

class kwant.physics.magnetic_gauge(syst)[source]

Fix the magnetic gauge for a finalized system.

This can be used to later calculate the Peierls phases that should be applied to each hopping, given a magnetic field.

This API is currently provisional. Refer to the documentation for details.

Parameters
syst

Examples

The following illustrates basic usage for a scattering region with a single lead attached:

>>> import numpy as np
>>> import kwant
>>>
>>> def hopping(a, b, t, peierls):
>>>     return -t * peierls(a, b)
>>>
>>> syst = make_system(hopping)
>>> syst = syst.finalized()
>>>
>>> gauge = kwant.physics.magnetic_gauge(syst)
>>>
>>> def B_syst(pos):
>>>    return np.exp(-np.sum(pos * pos))
>>>
>>> peierls_syst, peierls_lead = gauge(B_syst, 0)
>>>
>>> kwant.hamiltonian_submatrix(syst, params=params)

Methods

Return the Peierls phase for a magnetic field configuration.

Parameters
syst_fieldscalar, vector or callable

The magnetic field to apply to the scattering region. If callable, takes a position and returns the magnetic field at that position. Can be a scalar if the system is 1D or 2D, otherwise must be a vector. Magnetic field is expressed in units $$φ₀ / l²$$, where $$φ₀$$ is the magnetic flux quantum and $$l$$ is the unit of length.

The magnetic fields to apply to each of the leads, in the same format as ‘syst_field’. In addition, if a callable is provided, then the magnetic field must have the symmetry of the associated lead.

tolfloat, default: 1E-8

The tolerance to which to calculate the flux through each hopping loop in the system.

averagebool, default: False

If True, estimate the magnetic flux through each hopping loop in the system by evaluating the magnetic field at a single position inside the loop and multiplying it by the area of the loop. If False, then scipy.integrate.quad is used to integrate the magnetic field. This parameter is only used when ‘syst_field’ or ‘lead_fields’ are callable.

Returns
phasescallable, or sequence of callables

The first callable computes the Peierls phase in the scattering region and the remaining callables compute the Peierls phases in each of the leads. Each callable takes a pair of Site (a hopping) and returns a unit complex number (Peierls phase) that multiplies that hopping.

#### Previous topic

kwant.physics.DiscreteSymmetry

#### Next topic

4.1. kwant.digest – Random-access random numbers


text:
# expected length of broken stick

You break a stick of unit length in two. You then subsequently break the biggest of the resulting two sides in two, thus obtaining three pieces.

What is the expected length of the smallest of the three?

(Each breaking of a stick is assumed to be at a random point in that stick, uniformly distributed.)

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Is it possible to answer this? Could it be any value less than $1/3$? – Sigur Dec 8 '12 at 14:40
It certainly is possible - and the result will have to be consistent with a monte carlo simulation. For each particular experiment, it can indeed be any value less than 1/3, but I am asking for the average across a large set of experiments. – alexandreC Dec 8 '12 at 14:43
It depends on the assumptions on where the stick breaks. You said nothing about this so the question cannot be answered. – tst Dec 8 '12 at 14:44
What do you mean by "breaking a stick" in two? I mean what will be the distribution? If you fix it at both ends and bend, it will more likely break in the middle, maybe following a gaussian. Or do you assume it might brake anywhere with the same probability? – Simon Markett Dec 8 '12 at 14:46
Also in the first scenario, the parameters will be different for a smaller stick I guess, so I the first outcome will influence the second one. In short, you need to describe better what you want. – Simon Markett Dec 8 '12 at 14:47

If the shorter stick after the first break has length $S$ then $S$ uniformly distributed on $[0,\frac{L}{2}]$.

If $S=s$ then the shorter part of the longer stick will have length $T$ uniformly distributed on $[0,\frac{L-s}{2}]$. You will certainly have $T \lt s$ if $\frac{L}{3} \lt s \le \frac{L}{2}$.

So the expectation of the minimum of $S$ and $T$ is

$$\int_{s=\frac{L}{3}}^\frac{L}{2} \int_{t=0}^\frac{L-s}{2} t \frac{2}{L-s} \frac{2}{L} \,dt\,ds + \int_{s=0}^\frac{L}{3} \int_{t=0}^s t \frac{2}{L-s} \frac{2}{L} \,dt\,ds + \int_{s=0}^\frac{L}{3} \int_{t=s}^\frac{L-s}{2} s \frac{2}{L-s} \frac{2}{L} \,dt\,ds$$ $$= \frac{7 L}{144} +2L\log\left( \frac{3}{2}\right)-\frac{7L}{9} +\frac{5L}{3} -4L\log\left( \frac{3}{2}\right)$$ $$= L\left(\frac{15}{16} -2\log\left( \frac{3}{2}\right) \right)$$ the same as did's answer

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Hi. Aren't you missing a contribution from the possibility that is the shorter part the one that breaks again? – drake Jun 16 at 17:00
@drake - the question says "You then subsequently break the biggest of the resulting two sides in two" – Henry Jun 16 at 17:07
Oh, so sorry, I was trying to solve a different problem and though this one was the same. I didn't read the question carefully. I agree your answer is right. – drake Jun 16 at 18:37

The lengths of the two pieces of a stick of length $L$ are $\frac12(1\pm U)L$, where $U$ is uniform on $(0,1)$. Hence the lengths of the three pieces are $$\frac12(1-U),\quad \frac14(1+U)(1+V),\quad\frac14(1+U)(1-V),$$ where $U$ and $V$ are i.i.d. uniform on $(0,1)$. The smallest $S$ of these three lengths is $S=\frac12(1-U)$ on $A=[V\lt a(U)]$ and $S=\frac14(1+U)(1-V)$ on $B=[V\gt a(U)]$, where $a$ is defined by $$a(u)=\frac{(3u-1)^+}{u+1}.$$ Since $\mathbb E(A\mid U)=a(U)$, $$\mathbb E(S\mathbf 1_A\mid U)=\frac12(1-U)a(U).$$ Likewise, $$\mathbb E(S\mathbf 1_B\mid U)=\frac14(1+U)\mathbb E((1-V)\mathbb 1_{V\gt a(U)}\mid U)=\frac14(1+U)\frac12(1-a(U))^2.$$ Summing these yields $$\mathbb E(S\mid U)=\frac12(1-U)a(U)+\frac18(1+U)(1-a(U))^2,$$ which can be simplified into $$\mathbb E(S\mid U)=\frac18(1+U)-\frac18\frac{((3U-1)^+)^2}{1+U}.$$ Thus, $\mathbb E(S)=\frac3{16}-\tfrac18t$, with $$t=\int_{1/3}^1\frac{(3u-1)^2}{1+u}\mathrm du=\int_0^{1}\frac{4x^2}{2+x}\mathrm dx,$$ that is, $$t=\left[2x^2-8x+16\log(2+x)\right]_{0}^1=-6+16\log\left(\frac32\right).$$ Finally, $$\mathbb E(S)=\frac{15}{16}-2\log\left(\frac32\right)=0.12657\ldots$$

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good job! anyone for an alternative answer? – alexandreC Dec 8 '12 at 16:01
??     – Did Dec 8 '12 at 16:14
Your reasoning was excellent!! result agrees with MC simulation (unsurprisingly) - was just wondering if someone can come up with a different reasoning, that would lead to your correct answer! :) – alexandreC Dec 8 '12 at 16:26
Oh oops, somehow the website didn't show me you had already posted your answer... I need not have spent my time calculating. :-) – ShreevatsaR Dec 8 '12 at 17:36
alexandreC: How is the reasoning in the two other correct answers, different? – Did Dec 8 '12 at 18:27

When you make the first cut, the bigger piece has length $p$ that is uniformly distributed in $[\frac12, 1]$. (And so the smaller piece has length $1-p$ uniformly distributed in $[0, \frac12]$, but of course not independent of the length $p$ of the bigger piece.)

Similarly, when you cut this piece of length $p$ at a uniformly chosen point, the smaller piece of the two has length $q$ uniformly distributed in $[0, \frac{p}{2}]$.

The smallest of the three pieces is therefore of length $\min(q, 1-p)$. We can calculate its expected value as $$\int_{1/2}^{1} \int_{0}^{p/2} \min(q, 1-p) \;(\frac{1}{p/2}dq) \;(\frac{1}{1/2}dp) = \int_{1/2}^{1} \frac{4}{p} \int_{0}^{p/2} \min(q, 1-p) \;dq \;dp$$

If $p/2 \le 1-p$ (which happens when $p \le 2/3$), then as $q \le p/2 \le 1-p$, it is always the case that $\min(q, 1-p) = q$, and so the inner integral becomes $$\int_{0}^{p/2} \min(q, 1-p) \;dq = \int_{0}^{p/2} q \;dq = \frac{p^2}{8}.$$

Else, for $p/2 > 1-p$, the inner integral can be split as \begin{align} \int_{0}^{p/2} \min(q, 1-p) \;dq &= \int_{0}^{1-p} q \;dq + \int_{1-p}^{p/2} (1-p) \;dq \\ &= (1-p)^2/2 \;\;+\;\; p(1-p)/2 - (1-p)^2 \\ &= p(1-p)/2 - (1-p)^2/2 \end{align}

So the outer integral is \begin{align} &\int_{1/2}^{1} \frac{4}{p} \int_{0}^{p/2} \min(q, 1-p) \;dq \;dp \\ &=\int_{1/2}^{2/3} \frac{4}{p}\frac{p^2}{8} \;dp + \int_{2/3}^{1} \frac{4}{p}(\frac{p(1-p)}{2} - \frac{(1-p)^2}{2})\;dp \\ &=\int_{1/2}^{2/3} p/2 \;dp + \int_{2/3}^{1} \left(2(1-p) - \frac{2(1-p)^2}{p}\right) \; dp \\ &= \frac{7}{144} + \frac{1}{9} - (\log(9/4) - 7/9)\\ &= \frac{15}{16} - \log\left(\frac94\right) \approx 0.12657 \end{align}

This looks like a really strange answer, but it is roughly confirmed by the following simulation:

#!/usr/bin/env python
import random
num_samples = 0
sum_samples = 0
for num_samples in xrange(1, 100000000):
r1 = random.uniform(0, 1)
p = max(r1, 1 - r1)
r2 = random.uniform(0, p)
q = min(r2, p - r2)
cur_sample = min(q, p - q, 1 - p)
sum_samples += cur_sample
average = sum_samples / num_samples
if num_samples % 100000 == 0:
print 'Average is %.5f after %d samples' % (average,num_samples)

-


text:
### Anything Goes

While the math in my previous post was correct, some of the assumptions I made were not. Definitely not. For instance, one of the initial equations I derived from the WAI document was wrong.

$299|r_1-r_2|+587|g_1-g_2|+114|b_1-b_2|\ge125000$

should have been

$299(r_1-r_2)+587(g_1-g_2)+114(b_1-b_2)\ge125000$

And in this case it makes an enormous difference. So it turns out that neither set of equations I presented before actually works. If you use those equations, the contrast difference will be correct (the equation I did have right) but the brightness? Not so much. So, back to the scratch paper…

My gnomic warlock (is that the correct…er…ethnonym?) is finally a member of the guild, so I won’t be bugged by other guild requests anymore. All I’ll say about this tonight is that I hate escort quests. Failed one twice in a row, gained a couple of levels, then finally managed to complete it, but then only just. I’m beginning to think that power-leveling is best done with a partner. So I’m going to spend a bit of time working on professions before going back to leveling. 31 is sufficient for now, seeing as my skills in cooking, fishing, herbalism, and inscription are 1/75.

Got the symphony chorus schedule in the mail today! I’m going to use an old stamp to send this postcard. 32¢…”I remember when…”, I won’t go there, but I have some old 37¢ first class stamps that should do the trick. The interesting thing this year is that we’re doing three concerts instead of the typical two. Friday evening we’re doing just Carmina Burana, with the Singing Quakers and the Wichita Community Children’s Choir. Saturday we’ve got Haydn’s Symphony no. 90 (yes, ninety) as well as the Orff, with the same chorus. Sunday again is the full concert, this time with Bethel College Concert Choir in place of the Singing Quakers. This means a really crazy schedule there at the end. Rehearsals on the 13th, 14th, and 15th, concerts on the 16th, 17th, and 18th. And somehow those rehearsals we’ll have to fit all four chorus groups on stage at once. Fun. But anyway, I’m looking forward to it!

I have a new idea for a sort of social application. One creates tasks which they may or may not be otherwise inclined to complete, validates them with a friend (so they’re less inclined to cheat and put down silly stuff like “walk to the car” or “pet the dog”), and they compete with friends on who can have either the longest streak of completing these chores or who can complete the biggest percentage in a day, a week, a month. Say I put down “trim ear hairs” as a daily chore. I complete it, I get a gold star (for instance)! I don’t, it’s reflected in the score. That’s not something I struggle with, but someone might. More apt for me might be “do the dishes” or “top off fishtank water”. The advantage of making it social is the motivation of friends or family. At the same time, events or tasks might be private, and posting all over Facebook that you won your game by not [insert inappropriate behavior here] all week may be a bad thing. So, no autoposting of accomplishments. On the other hand, if you’re trying to quit smoking, you might be eager to say, “Smoke free for 7 days!”. Or for your kids, it’s a 21st century replacement to the gold star charts I had in school classrooms growing up. Set rewards for completing milestones, for picking up a good habit or breaking a bad one. Kids brush their teeth and eat their vegetables: maybe by competing with their siblings (or cousins, or friends) in a little friendly rivalry, they’ll both do better. Just an idea for the pile. Someday I’m going to have to start working on that, and I could make it one of my items on the list!


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# If k is a positive integer and m is the product of the first 40 positi

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10 Jul 2018, 21:37
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If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

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10 Jul 2018, 21:48
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Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Basically $$m=40!$$

Statement 1: $$10^k$$ is a factor of $$40!$$. but $$40!$$ has $$\frac{40}{5}+\frac{40}{25}=8+1=9$$ trailing $$0$$s. so $$k$$ could be $$1$$ or $$2$$ or any other number. insufficient

Statement 2: $$n=9!=1*2*3*4*5*6*7*8*9$$. this number has only one trailing $$0$$ which will be a multiple of $$2*5=10$$. hence $$k=1$$. Sufficient.

Option B
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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10 Jul 2018, 21:58
From statement 1: k could be 0,1,2.. Insufficient

From statement 2: k could be 0 or 1. Question stem says k is positive. Hence sufficient

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If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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10 Jul 2018, 22:43
Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Given m=40!
Question stem, k=?

We need only 1 value of k to prove sufficiency & a minimum of 2 values of k to prove insufficiency.

St1:- $$10^k$$is a factor of 40!
40!=1*2*3*4*5*6*7*8*9*10*...*40
So $$10^k$$ is a factor of 10 & 10^2 where k is 1 and 2 respectively.

Insufficient.
st2:-$$10^k$$ is a factor of 9!
9!=1*2*3*4*5*6*7*8*9
So $$10^k$$ is a factor of 10 when k is 1 only.
Hence, sufficient.

Ans. (B)
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 00:31
Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Ans is B...(1) says k<= 8
(2) k=1
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 06:55
Can we look at it as

1) since there is 10 and 100 in 40!, we unable to determine k
2) there is only 2*5=10 in 9!, hence we can determine k=1
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If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 07:06
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ETLim wrote:
Can we look at it as

1) since there is 10 and 100 in 40!, we unable to determine k
2) there is only 2*5=10 in 9!, hence we can determine k=1

Hi ETLim,
In your quotation (1), we are able to determine more than one value of 'k'. Hence given statement-1 is insufficient to answer the question stem. Please note: In DS questions those ask the value of variable or things, if we get one value from the statement then that statement is sufficient and if we get more than one value from the statement, then that statement is insufficient. For both st1 & st2 ,individually to be sufficient, we must obtain an unique value from both the statements.

In your quotation (2), since we obtained one value of k, hence st2 is sufficient.

Hope it helps.
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 08:02
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
m=1*2*3*.....*40=40!
Number of 10s will depend on the bigger prime factor that is 5.
# of 5s =$$\frac{40}{5}+\frac{40}{25}= 8+1=9$$
So k can take any values till 9 as 10^1, 10^2,....10^9 all are factors of 40!
Insuff

(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.
Now n=9! And number of 5s are 9/5..=1
So k can take Only one value 1
Sufficient

B
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 18:54
chetan2u niks18 pushpitkc pikolo2510

How about below analysis of option A?

Quote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

$$k\geq{0}$$
m = 40!

Quote:
(1) 10^k is a factor of m.

So m also contains 2*5 ie 10 divisible by 10^1
and m also contains 25*4 which is divisible by 10^2
So k can be 1 or 2.

I could not understand above approach used for examinating statement 1
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If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 22:22
chetan2u niks18 pushpitkc pikolo2510

How about below analysis of option A?

Quote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

$$k\geq{0}$$
m = 40!

Quote:
(1) 10^k is a factor of m.

So m also contains 2*5 ie 10 divisible by 10^1
and m also contains 25*4 which is divisible by 10^2
So k can be 1 or 2.

I could not understand above approach used for examinating statement 1

The analysis of statement 1 as already explained by chetan2u is as follows

There is a formula for finding the highest power of a prime number in a factorial
If p is a prime number, then the highest power of p in a factorial n is {$$\frac{n}{p}$$} + {$$\frac{n}{p^2}$$} + {$$\frac{n}{p^3}$$}.....
where {$$\frac{a}{b}$$} is the quotient when integer b divides another integer a

Now coming to the problem at hand. Since p is 10, which can be prime-factorized as 2*5, it is
enough if we are able to find the highest power of 5(since it is the biggest prime number). n = 40

Therefore, the highest power of p(10) is nothing but the highest power of 5.
Substituting values in the formula $${\frac{40}{5^1}} + {\frac{40}{5^2}} = {\frac{40}{5}} + {\frac{40}{25}} = 8 + 1 = 9$$
This means that the 10^9 will divide 40!. So even the smaller powers of 10 will divide 40!

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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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15 Jul 2018, 03:55
PKN wrote:
Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Given m=40!
Question stem, k=?

We need only 1 value of k to prove sufficiency & a minimum of 2 values of k to prove insufficiency.

St1:- $$10^k$$is a factor of 40!
40!=1*2*3*4*5*6*7*8*9*10*...*40
So $$10^k$$ is a factor of 10 & 10^2 where k is 1 and 2 respectively.

Insufficient.
st2:-$$10^k$$ is a factor of 9!
9!=1*2*3*4*5*6*7*8*9
So $$10^k$$ is a factor of 10 when k is 1 only.
Hence, sufficient.

Ans. (B)

If the question would have asked for highest value of k, then A would be the answer. Right?
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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15 Jul 2018, 05:10
AkshdeepS wrote:
PKN wrote:
Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Given m=40!
Question stem, k=?

We need only 1 value of k to prove sufficiency & a minimum of 2 values of k to prove insufficiency.

St1:- $$10^k$$is a factor of 40!
40!=1*2*3*4*5*6*7*8*9*10*...*40
So $$10^k$$ is a factor of 10 & 10^2 where k is 1 and 2 respectively.

Insufficient.
st2:-$$10^k$$ is a factor of 9!
9!=1*2*3*4*5*6*7*8*9
So $$10^k$$ is a factor of 10 when k is 1 only.
Hence, sufficient.

Ans. (B)

If the question would have asked for highest value of k, then A would be the answer. Right?

I don't think so. Each statement has its own greatest value of k. $$(st1(k_{max}=9, st2(k_{max}=1)$$

In DS value questions, each statement must produce same value. Even answer can't be D.

So, test maker would set question accordingly.
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Re: If k is a positive integer and m is the product of the first 40 positi &nbs [#permalink] 15 Jul 2018, 05:10
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9

Since you're working on $S_6$ for class, I won't answer that one for you. But I have another easy example of an outer automorphism. The Quaternion group $Q_8$ can be represented as $\{1, -1, i, -i, j, -j, k, -k\}$, where $ij=k$, $jk=i$, $ki=j$, and $ji=-k$, $kj=-i$, $ik=-j$. The $-1$ element acts pretty obviously on the rest, e.g. $(-1)j=-j$. I think ...

8

Well, the category of schemes is a completely different category than the category of locally ringed spaces. It doesn't have colimits (including quotients by group actions), limits, whereas the category of locally ringed spaces has these. Don't forget the forgetful functor here. If you have a group acting on a scheme, then the quotient of the underlying ...

8

Here is a context of group action where ergodicity arises naturally. As pointed out by Martin, the definition given in the grey box applies to more general situations where the transformations are not required to be measure-preserving, and the group is not countable discrete. Take a probability measure space $(X,\mathcal{A},\mu)$ and let a countable ...

8

We have $\lvert G\rvert = 28 = 2^2\cdot 7$. Let $a_7$ be the number of $7$-Sylow groups in $G$. By Sylow $$a_7\equiv 1\mod 7\qquad\text{and}\qquad a_7 \mid 4\text{.}$$ This implies $a_7 = 1$, so there is a unique subgroup $H$ of $G$ of order $7$. For all $g\in G$, $gHg^{-1}$ is again a subgroup of order $7$, which forces $gHg^{-1} = H$ for all $g\in G$. So ...

7

For a finite group acting on a scheme, if each orbit is contained in an open affine, then one can form the quotient scheme, by reducing to the affine case (where one just takes invariants). For example, since any finite set of points in a projective space are contained in the complement of a(n appropriately chosen) hyperplane, we see that for finite groups ...

7

Two examples of purely group-theoretic theorems proved using representation theory are: Burnside's theorem, which states that a finite group having no more than 2 distinct prime divisors must be solvable. (Much later, a proof without representation theory was found in the 1970's.) Theorem about Frobenius groups: Assume a finite group $G$ contains a ...

7

This is tricky - the two cases look the same, but they're not. The first one is a right action $v \cdot (p_1 \cdot p_2) = (v \cdot p_1) \cdot p_2$, while the second one is a left action $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. To see why, consider these 2 permutations: $$p_1(1) = 1, p_1(2) = 3, p_1(3) = 2$$ and $$p_2(1) = 3, p_2(2) = 2, ... 6 That's a feature, not a bug. Strong continuity is actually supposed to be a weaker condition than continuity (see also strong operator topology, which is weaker than the norm topology). For example, the action of \mathbb{R} on L^2(\mathbb{R}) by translation is strongly continuous if L^2(\mathbb{R}) is given the norm topology but not continuous. 6 Another different proof: since as you said you know that G has an element of order 7, by Cauchy, you also know that there's at least one subgroup of order 7, let's call it H. Suppose that K \leq G is another subgroup of order 7, then we can consider the subset HK that has order |HK|=|H||K|/|H \cap K|. If H and K were distinct then H ... 5 If you didn't invert and you tried to turn x \mapsto xg into a left-action then you'd get into trouble, because (if say g \star x = xg) then you'd have$$g \star (h \star x) = (xh)g = x(hg) = (hg) \star x \ne (gh) \star x$$however if you had g \star x = xg^{-1} then you'd have$$g \star (h \star x) = (xh^{-1})g^{-1} = x(h^{-1}g^{-1}) = x(gh)^{-1} = ...

5

Let $S_1$ and $S_2$ be two of the Sylow $3$-subgroups. If $a \in \ker \phi$, then in particular $$aS_1a^{-1} = S_1 \Rightarrow a \in N_G(S_1).$$ The same holds for $S_2$, so $$\ker \phi \subset N_G(S_1) \cap N_G(S_2).$$ Now, since the Sylow $3$-subgroups aren't normal, what is $N_G(S_i)$?

5

I'd say originally the action of a group on a set is what motivates many groups. Thus the symmetric group $S_n$ is in fact defined by its action on the set $\{1,\ldots ,n\}$. Or the symmetry group of an object (a regular dodecahedron, say) is inherently given by the way how this (abstract) group acts on the (more concrete) object. Often one gets insights out ...

5

If you have a subgroup $\overline{M}$ of a quotient group $G/N$, the lift of $\overline{M}$ is a subgroup $M$ of $G$ such that the $\overline{M}$ is the image of $M$ under the projection homomorphism $G\rightarrow G/N$. (This is guaranteed to exist by the correspondence theorem.) So speaking of the lifted action of some group action implies that you are ...

5

To address the stated subtext of your post: many people, myself included, take the position that groups are important because they act on things. A representation is just a group action on a vector space (by linear operators). And whenever you have a group action, even if it isn't on a vector space, there is often a closely related representation lurking ...

5

Just a few unsorted comments to add to what's been said above... The classification of finite simple groups relied heavily on modular representation theory, so there is that. Many of the sporadics are only described reasonably through their representations. Symmetric groups have nicely describable representations that correspond to a combinatorics ...

5

The group action is faithful if the action of an element $x$ is trivial iff $x=1_G$. That is, if $$(xgH=gH\text{ for each }g\in G)\iff x=e$$ Of course, the condition $xgH=gH$ can be rewritten as $$g^{-1}xgH=H\iff g^{-1}xg\in H\iff x\in gHg^{-1}.$$ Is the path forward clear now?

5

$\def\R{\mathbb{R}} \def\SL{\text{SL}} \def\SO{\text{SO}}$Often you use the group action to study $G$ and not just to study $X$. Here is an example: what does $\SL_2(\R)$ look like as a manifold? You can solve this by thinking of the group directly, but an easier way is to note that it acts transitively on the upper half plane by Mobius transformations. ...

5

I believe it's more the other way around: a group action of $G$ on a space $X$ allows to construct a new space, the quotient $G\backslash X$ (of course the quotient space is "nice" only under some technical conditions). Recognizing that a space $Y$ is actually realized as the orbit space $G\backslash X$ of some simpler space $X$ can help understand better ...

5

Gromov in his original 1987 book (Section 3.1) wrote a classification for arbitrary isometric group actions on hyperbolic spaces (with no further assumption) into 5 main classes. It goes at follows (the terminology is borrowed from here) 1: bounded: orbits are bounded 2: horocyclic: orbits are unbounded, $G$ acts with no hyperbolic isometry (hence there's ...

5

You can indeed work by analogy with the real case. To do so, you must think of $\mathbb{CP}^1$ as the quotient of the complex sphere by the aciton of the unit length complex numbers: $$\mathbb S^3\times \mathbb S^1\to\mathbb S^3, ((z,z'),\lambda)\mapsto (z\lambda,z'\lambda)$$ There is an obvious identification $$\mathbb S^3/\mathbb S^1\simeq\mathbb{CP}^1$$ ...

5

There is no direct relation between the uses of "transitive" in "transitive relation" and in "transitive action". The relation of being in the same orbit is transitive, but that is true for any group action. However both senses are derived from the latin verb "transire", meaning something like "going across" or "going through". A transitive action allows ...

5

What normality of the stabiliser says is exactly that every group element $g\in G$ that fixes $s$ also fixes the entire orbit $Gs$ pointwise. Conversely any $g\in G$ that fixes any element of the orbit $Gs$ will also fix $s$. These two parts are equivalent, although the first sentence says that every conjugate of $G_s$ contains $G_s$, while the second ...

4

In the semidirect product $N \rtimes \operatorname{Inn} N$, the normal subgroup $N \times 1$ is centralized by the subgroup $\{ (h^{-1},h) \mid h \in N \}$, which is isomorphic to $N$. So this subgroup is equal to the second factor $1 \times N$ in the direct product $1 \times N$. The isomorphism $N \times N \to N \rtimes \operatorname{Inn} N$ is given by ...

4

The fundamental lemma for a group action is the following. Let $x$ be in $X$, a set on which $G$ acts. There is a map $f: G \rightarrow X$ defined by $g \mapsto g.x$. By definition, the image of this map is the orbit of $x$, denoted by $G.x$. Morever, let $H$ be the subset of $G$ of elements $h$ satisfying $h.x = x$; $H$ is called the stabilizer of $x$ and ...

4

The general theorem that you need here, and which Steve aludes, is: Theorem: Let $\,P\,$ be a finite $\,p-\,$ group , $\,X\,$ a finite set, and suppose $\,P\,$ acts on $\,X\,$ . Define $\,X_P:=\{x\in X\;:\;ax=x\,\,\forall\,a\in P\}\,$ . Then, $\,|X_P|=|X|\pmod p\,$ Proof: We know that $\,|X|=\sum|\mathcal Orb(x)|\,$ , where the sum runs over different (and ...

4

I think, it is quite important to carefully write down what one is supposed to do in a situation like this one, as it is really easy to get confused. You are given a subbasis element $S(C,U)$ with $C\subset X$ compact, and $U\subset X$ open. To prove continuity of $\gamma_2$ it suffices to show that for any $g\in G$ satisfying $g(C)\subset U$ there exists a ...

4

Yes if you assume a group has a normal subgroup $Z$ of finite index which is infinite cyclic then it's much easier. If $Z$ is central, then a classical theorem shows that the derived subgroup is finite. So modding out by the derived subgroup, you get an abelian group, and eventually get that the group has a homomorphism onto $\mathbf{Z}$ (with finite kernel, ...

4

You aren't computing $\sigma(\tau(v_1 \otimes ... \otimes v_4))$ correctly. We have $$\tau(v_1 \otimes ... \otimes v_4) = v_3 \otimes v_2 \otimes v_4 \otimes v_1.$$ Now rename $w_1 = v_3, w_2 = v_2, w_3 = v_4, w_4 = v_1$. Then $$\sigma(w_1 \otimes ... \otimes w_4) = w_2 \otimes w_3 \otimes w_1 \otimes w_4$$ which is equal to v_2 \otimes v_4 \otimes ...

4

One important thing that comes out of this argument is that $SO(4)$ almost factors as a direct product, namely its universal (double) cover Spin$(4)$ factors as the direct product of Spin$(3)$ (which is $S^3$, or equivalently $SU(2)$) with itself. So regarding question (3), in some sense your are asking whether there is a way to see that $SO(4)$ should ...

4

The proof is correct, but one can generalize and shorten it as follows: Let $G$ be a finite group and assume that the intersection of all non-trivial subgroups of $G$ is non-trivial, i.e. contains some $g \neq 1$. If $G$ acts on a set $A$ with $|A|<|G|$, then for every $a \in A$ we have $|G:G_a|=|Ga| \leq |A| < |G|$. Therefore, $G_a$ is non-trivial, ...

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1 Faculty of Medicine, University of Coimbra, 3000-370 Coimbra, Portugal
2 Department of Mechanical Engineering, ADAI, University of Coimbra, 3030-788 Coimbra, Portugal
3 Research Unit for Sport and Physical Activity, Faculty of Sport Sciences and Physical Education, University of Coimbra, 3040-256 Coimbra, Portugal
4 Faculty of Sport Sciences and Physical Education, ADAI, University of Coimbra, 3040-256 Coimbra, Portugal
5 Centro Hospitalar do Tâmega e Sousa, 4560-136 Penafiel, Portugal
6 Research Unit for Sport and Physical Activity, Faculty of Medicine, University of Coimbra, 3000-370 Coimbra, Portugal
*Correspondence: kikalaia@hotmail.com (Francisca D Laia)
J. Mens. Health 2022, 18(8), 169; https://doi.org/10.31083/j.jomh1808169
Submitted: 3 November 2021 | Accepted: 21 December 2021 | Published: 8 August 2022
(This article belongs to the Special Issue Sports training, recovery and nutrition in male athletes)
This is an open access article under the CC BY 4.0 license.
Abstract

Background: Cooling vests are popular and widely used in many sports. There are several studies about their efficacy but there is not much evidence about their impact on the recovery of kayak sprint male athletes. Methods: Ten elite male sprint kayakers completed an exercise protocol on a kayak ergometer on two separate days. They were randomized to wear a cold vest during a 15 min active rest on one of these days. Following the pre designed protocol, all sessions were performed in a climate chamber with controlled temperature and humidity conditions. At both moments, the blood lactate concentration was determined, as well as urine temperature. Skin temperature in the latissimus dorsi muscle area was assessed several times along the protocol, using infrared thermography. Borg’s rating of perceived exertion (RPE Scale) was applied to the protocol after each exercise bout and the perceived recovery scale was applied before the last moment of the protocol (500 m test). Results: The subjective perception of recovery was significantly higher during the cooling vest condition (p $<$ 0.01). There were no significant differences on performance in the 500 m maximal test when comparing both days. On the other hand, after the active recovery with cold vest, the skin temperature (p $<$ 0.01), urine temperature (p $<$ 0.05) and blood lactate concentration were all significantly lower (p $<$ 0.05). Discussion: Results revealed that the cooling vest is a suitable instrument to induce a decrease in skin and core temperature while paddling and also to reduce more quickly the blood lactate concentration. The vest should be tested under on-water conditions in order to consider other factors such as wind and solar radiation. Conclusions: Wearing a cooling vest during active rest after a training session or a competition seems to improve the recovery of male kayakers.

Keywords
Cooling vest
Post-exercise recovery
Kayak sprint
Male athlete
Thermography
Cryotherapy
1. Introduction

During kayak sprint competitions, athletes often compete in 2 to 4 events a day, and, most of the time, there is only a short period for them to recover between races. Therefore, a complete recovery is unlikely to occur. On this account, finding a way to quickly cool down the athletes’ bodies is a major request, so that they can recover better in the brief interval between races. This will prevent heat stress from happening, which can impair aerobic performance when hyperthermia occurs [1]. Since the major international kayak sprint championships take place in the summer season this has an even greater impact. This type of competition schedule promotes incomplete athletes’ recoveries between races, which are associated with the levels of blood lactate concentration (BLC) above the rest values [2]. Therefore, these values are positively correlated with muscular fatigue, and a greater probability of injury occurrences [3, 4]. It is common to use active recovery (AR) after submaximal and maximal intensity efforts in order to decrease BLC levels, thus reducing the muscular fatigue [5] and, consequently, the rate of injuries.

Cryotherapy is a recovery modality that is usually applied to speed up the regeneration of tissues and to prevent swelling of acute musculoskeletal injuries [6, 7, 8], especially for the athletes who have a training and competition schedule that requires several bouts of exercises within one day or in environments of extreme heat and humidity [6]. Cryotherapy can be used with two different purposes that are related between each other: to recover between exercise bouts or to recover during heat stress [9]. In kayak sprint events both situations have to be considered, so cryotherapy is a valuable tool in this sport. Cold stimulation shows positive effects on muscle enzymes, creatine kinase and lactate dehydrogenase [1, 10]. Furthermore, there is an increase in $\beta{}$-endorphin secretion by the pituitary gland and a decrease in nociceptive response, mainly in C fibers. On that account, cryotherapy can reduce the fibers damage, the delayed onset muscle soreness (DOMS), the pain’s perception and it may also have a placebo effect on the sense of recovery [11, 12]. Most scientists and practitioners in health sciences agree that it is important to understand subjective symptoms and how they relate to objective findings. In sport the same principle can be applied, considering that perceived exertion is closely related to the heart rate, among other strain variables such as blood lactate levels [13]. The Borg’s rate scale (6 to 20) is considered to be the most appropriate to be applied in studies of perceived exertion, both for exercise testing and prediction [14].

According to Racinais et al. [12] the decrease in the core temperature is smaller with a cooling vest than with cold water immersion or mixed-cooling methods. However, without directly reducing muscle temperature, cooling garments present the advantages of lowering the skin temperature, thus reducing cardiovascular strain and heat storage. Since they provide a fast reduction of the skin temperature, athletes can wear them during the recovery breaks. A delayed effect on temperature reduction in deeper tissues can be expected due to the lowering of the skin temperature.

In recent years some elite teams began to wear cooling vests combined with the active rest after competition, in order to accelerate recovery and prevent injuries in hot environments. However, there are few studies approaching the evaluation of their efficacy in kayak sprint. Finding an effective strategy is important to avoid the potential maladaptive physiological and psychological effects of overtraining [15]. Therefore, the main goal of this study was to evaluate the cooling vest’s impact on the recovery process of kayak sprint male athletes, by means of the analysis of the BLC, Borg’s rate of perceived exertion (RPE), performance and differences in infrared thermographic imaging distributions.

2. Methods
2.1 Participants

Ten kayak sprint elite male athletes took part in this study. The subjects were characterized by the following biometric characteristics (mean $\pm$ SD): age, 21.63 $\pm$ 2.85; height, 177.9 cm $\pm$ 7.33; body mass, 75.18 $\pm$ 7.33 kg; arm span, 180.92 $\pm$ 9.5 cm; fat mass %, 10.19 $\pm$ 4.65. The inclusion criteria were: (1) participation in at least one international event in the last two years, in order to be considered an elite athlete; (2) be included in a training plan which aimed the participation in international competitions. The exclusion criterion was: having a fat mass percentage higher than 20%, considering that fat can interfere in the cold impact on deeper tissues and, consequently, in the skin temperature measurements performed [16]. Moreover, each athlete was his own control.

2.2 Experimental protocol

The study protocol was completed in a climate chamber with controlled temperature and humidity, using a wind-braked kayak ergometer (Dansprint® PRO kayak, Hvidovre, Denmark) that has already been used on previous studies [17, 18]. The participants reported to the laboratory on three occasions with a week in between. On the first occasion each subject performed a 4 minutes test to determine his maximal average power (MAP). This test defined the power output of the training session (TS) and active recovery (AR) performed during the protocol. On the second and third occasions, the participants reported to the laboratory in order to follow the same protocol, with a single difference: the randomized wearing of a cold vest during the AR on one of the occasions.

The protocol (Fig. 1) consisted of a 15 min acclimatization, a warm-up (5 min at 40% MAP), an aerobic TS (5 times 3 min work with 3 min of passive rest) at 75% of the MAP, similar to what the athletes were used to do in their daily training routine. After it, they had a 5 min passive rest (PR) and then paddled for 15 min (40% MAP) as an AR. Depending on the randomization, the AR was performed wearing or not a colling vest (Inuteq Siku PAC${}^{\textregistered{}}$ & H2O${}^{\textregistered{}}$, Deventer, The Netherlands). This cooling vest had pockets to put 4 cold accumulators inside, 2 pockets in the front side and 2 in the back. The recommendations of the cooling vest brand, on how to get the most out of the garment’s wearing, were followed. After resting passively for 15 min, athletes finally performed a 500 m race simulation in order to evaluate their performance. The distance of 500 m was chosen because it is the distance used on national team selection races. The wearing of the cooling vest was decided randomly by means of a calculator in an Excel 2013 spreadsheet.

Fig. 1.

Study’s protocol: expressing the procedures and the assessments made in each phase. IRT, infrared thermography; BLC, blood lactate concentration; TS, training session.

Prior to the test, the body characteristics such as height, body mass, arm span and skinfolds were measured to estimate the body density (BD) based on Jackson and Pollock’s equation. This is a generalized equation for the calculation of the BD of men aged between 18 and 61 years old, and it uses the sum of seven skinfolds (subscapular, triceps, abdominal, suprailiac, thigh, chest and medium axillar). The BD’s result enabled the estimation of the fat mass percentage with Siri’s formula [19].

During the protocol the skin temperature was assessed five times, the BLC was measured four times, while the urine temperature was measured twice. In each phase of the study the perceived exertion was questioned by using Borg’s RPE. Prior to the 500 m race simulation, all kayakers were asked to classify their perceived recovery, in a scale ranging from 0 to 10 (in which 10 represents completely recovered).

The air temperature of the chamber throughout the three phases of the test varied between 20.9 °C and 21.2 °C. The optimal temperature for the use of the infrared camera in the study of the human body is 21 °C [20]. The relative humidity during all the tests ranged from 66.0% to 68.9%. The usual values for the use of the infrared camera ranged between 40% and 70% of relative humidity, but there are not established norms yet [20].

2.3 Skin temperature and core temperature

Since the latissimus dorsi along with the upper and lower abdominals, pectoralis magnus, anterior serratus and other torso muscles are some of the primary and supporting muscles of the paddling technique [21]. The radiant temperature of the human body can be measured by means of an infrared thermographic imaging (IRT), which is a contact-free technology used to monitor the skin temperature [22]. The temperature distribution of large areas of skin can be quantified quickly and accurately [23], thus enabling the transformation of body’s radiation into temperature values [22]. The use of IRT needs to follow some recommendations [20, 24]. Therefore, an infrared camera (FLIR® SC660; FLIR Systems, Notting Hill, VIC, Australia) on a tripod was used to take infrared pictures of the posterior part of the upper body of each subject. The region of interest (ROI) in the pictures was the skin area above the latissimus dorsi muscle (one of the most important muscles in the paddling technique). Special attention was given to five different moments: the initial resting conditions after acclimatization, TS, AR, and before and after the 500 m race simulation. During heavy physical exercise, there is an increased metabolic rate and, consequently, a rise in the internal heat production. With the continuity of the exercise, an increase of the blood flow rate between the human body core and the skin occurs, in order to exchange more heat with the surrounding environment [25]. All participants were aware of the recommendations before the IRT screening. During the assessment, the subjects were on the upright position and placed 1.5 m away from the camera. The calculated value was an average of the thermograms of the ROI.

A non-invasive evaluation of the core temperature based on skin surface measurements may be biased by multiple factors such as the measurement location, local evaporation and clothing conditions [26]. As a result, the urine temperature (Checktemp® digital thermometer 1HI98509, Hanna Instruments, Woonsocket, RI, USA) was determined in order to reduce error in the evaluation of the core temperature. Using the urinary outpour to measure the urine temperature is a less invasive technique and is better accepted by the athletes than other techniques, such as the rectal or esophageal temperature. According to Lefrant et al. [27], the urinary bladder technique is more accurate than others to measure the core temperature. Kawanami et al. [28] state that measuring the urine’s temperature is a good method to assess the human body core temperature and that there are small differences when comparing it with other techniques. In our study the urine temperature was assessed twice, firstly prior to the TS and secondly before the 500 m simulation race.

2.4 Blood lactate concentration

The BLC was measured after the TS, before the AR and before and after the 500 m race simulation. Capillary finger blood samples were collected and the BLC was determined by means of a spectrophotometric method that has a high degree of specificity for lactic acid [29, 30]. The instrument used was the Miniphotometer plus LP20, Dr. Lange, Bavaria, Germany.

2.5 Statistical analysis

The body mass percentage was calculated by using Excel 2013 for Windows 7. The statistical analysis was done using the IBM SPSS Statistics version 24 (IBM Corp., Chicago, IL, USA) for Windows 7. The descriptive statistics were performed for all the variables. The comparison between the two moments of the study was performed using Wilcoxon’s test and the correlation was tested with Spearman’s test. Both with and without the cooling vest, five time points of analysis of the independent variables were considered (IRT, core temperature, BLC, Borg RPE scale, average power and distance), has presented on Fig. 1. Values of p $\leq$ 0.05 were considered statistically significant.

3. Results

Both on the protocol with and without the cooling vest, the average skin temperature above the latissimus dorsi muscle dropped after the AR. Comparing the thermogram after the TS with the one after the AR (Fig. 2a and b), the temperature decrease (Table 1) was higher with the wearing of the cooling vest than without it. The ROI temperature after the AR with the wearing of the cooling vest was significantly lower on both sides (z = –2.803; p $<$ 0.01 (0.005)) (Fig. 3 – asterisk). There were no significant differences in the skin temperature after the 500 m race simulation (Fig. 2).

Fig. 2.

Infrared thermograms along the protocol – example of two sequences of thermographic images collected in different moments of the protocol, with and without the cooling vest. (a) Without cooling vest. (b) With cooling vest. TS, training session; AR, active recovery.

Table 1.Skin temperature above the right (RL) and left (LL) latissimus dorsi muscle (ROI) and the core temperature registered during the protocol (before and after the training session, after the active recovery and before and after the 500 m test) for the two active AR conditions.
 With cooling vest (temperature °C) Without cooling vest (temperature °C) RL LL Core RL LL Core Before TS 35.28 $\pm$ 0.79 35.29 $\pm$ 0.93 36.3 $\pm$ 0.22 34.61 $\pm$ 0.89 34.74 $\pm$ 0.77 36.7 $\pm$ 0.26 After TS 34.93 $\pm$ 0.89 34.95 $\pm$ 0.90 - 34.84 $\pm$ 1.26 34.65 $\pm$ 1.39 - After AR 30.09 $\pm$ 2.12 29.15 $\pm$ 2.39 - 34.22 $\pm$ 0.94 34.19 $\pm$ 1.05 - Before 500 m 33.9 $\pm$ 0.97 33.64 $\pm$ 1.01 37.1 ± 0.46 34.67 ± 0.86 34.55 $\pm$ 0.81 37.4 ± 0.32 After 500 m 33.62 $\pm$ 0.88 33.64 $\pm$ 1.02 - 33.51 $\pm$ 0.82 33.62 $\pm$ 0.88 - TS, training session; AR, active recovery; RL, right latissimus dorsi; LL, left latissimus dorsi; underlined for differences in the skin temperature (p $<$ 0.01); the statistical differences in the core temperature after AR are indicated in bold (p $<$ 0.05).
Fig. 3.

Skin temperature along the protocol – variation during the whole protocol in both situation of the study (AR with and without the cooling vest). Grey lines represent the protocol where the kayaker performed the AR with the cooling vest and black lines the AR without wearing it. Continues and dashed lines represent right or left lattisimus dorsi, respectively. The asterisk is indicating that after the AR there is a significant difference (p $<$ 0.01) in the temperature values between AR conditions. RL, right latissimus dorsi; LL, left latissimus dorsi.

Regarding the core temperature, there were no significant differences between the measures before the TS. On the other hand, after the AR the core temperature was significantly lower, 15 minutes after the wearing of the cooling vest (z = –2.014; p $<$ 0.05 (0.044)), when compared with the moment at which the athletes did not wear the cooling vest. All temperature values are registered in Table 1.

The values, which are statistically different when comparing the wearing of the cooling vest with the ones at which it was not being worn, concerning the skin temperature (p $<$ 0.01) are underlined and, in bold regarding the core temperature (p $<$ 0.05). Mean $\pm$ SD.

There were no significant differences in the values of the BLC after the TS when comparing the moments wearing the cooling vest with the ones without wearing it. The changes in the BLC at all the moments of the protocol are presented in Fig. 4.

Fig. 4.

Blood lactate concentration along the protocol for both AR conditions, with and without wearing the cooling vest. The asterisk indicates that there was a significant difference (p $<$ 0.05), for the respective measurement time point. BLC, blood lactate concentration; TS, training session; AR, active recovery.

After the AR with the cooling vest, the BLC was significantly lower (z = –2.191; p $<$ 0.05 (0.028)) comparing with the values obtained at the same moment without the cooling vest (Table 2) (Fig. 4asterisk).

Table 2.BLC values during the protocols with and without the cooling vest.
 BLC BLC With cooling vest (mmol/L) Without cooling vest (mmol/L) After TS 6.78 $\pm$ 2.61 6.86 $\pm$ 2.86 After AR 2.29 ± 1.10 2.77 ± 1.11 Before 500 m 1.89 $\pm$ 1.01 2.13 $\pm$ 1.13 After 500 m 12.13 $\pm$ 2.70 14.31 $\pm$ 3.87 TS, training session; AR, active recovery; BLC, blood lactate concentration. The values which are statistically different (p $<$ 0.05) are underlined. Mean $\pm$ SD.

Although there were no statistical differences between the BLC before the 500 m race simulation (z = –1.481; p = 0.139), after the AR wearing the cooling vest the average BLC was lower, compared with the protocol at which the cooling vest was not worn. The same happened with the BLC after the 500 m test (z = –1.836; p = 0.066), where, although the mean value was lower when the cooling vest was worn during the AR comparing with no cooling vest, it was not verified a significant statistical difference (Table 2). The variation of the BLC values before and after the 500 m for each athlete are specified in Fig. 5.

Fig. 5.

Variation of BLC values before (black) and after (grey) the 500 m race simulation for each athlete (1 to 10) at which moment of study considering the two AR conditions. Variation = (BLC with vest) – (BLC without vest).

The analysis of the perceived exertion (Borg RPE Scale) and the power registered in the ergometer showed no appreciable correlation in any period of the study when comparing the two cooling vest conditions. However, the athletes’ subjective perception of recovery before the 500 m simulation race was significantly higher after performing the AR wearing the cooling vest (z = –2.831; p $<$ 0.01 (0.005)), compared with the protocol of AR with no vest (Fig. 6).

Fig. 6.

Athlete’s subjective perception of recovery before the 500 m test. Mean $\pm$ SD with vest (6.85 $\pm$ 0.94) and without vest (5.25 $\pm$ 1.09) during the active recovery.

No statistical difference was observed in the average power of the maximal 500 m test between the two conditions. The same happened regarding the performance time in the 500 m race simulation test, which was not statistically different when comparing the protocols of AR wearing cooling vest and without wearing it. At both conditions, no correlation was not found between the BLC before the 500 m test and the performance time in the race simulation test.

4. Discussion

The results of the carried out study revealed that the cooling vest is a suitable instrument to induce a lowering in the skin temperature while paddling (Fig. 3), being at the same time light and not limiting the movement of low-intensity paddling. As described in other studies [11, 23] the IRT can be used to monitor the skin temperature during physical activity and cold treatments. In Table 1 it is patent that the skin temperature above the latissimus dorsi, for the athletes wearing the cooling vest during the AR, was almost 4 degrees lower when comparing with the AR without wearing the cooling vest. After the application of the cooling vest the body needs to increase the blood flow in the covered area in order to return to the homeostasis point. The increased blood flow enables to eliminate more quickly the waste products such as lactate, thus promoting a better and faster recovery [11, 27]. If it is worn regularly in the hot days, the cooling vest will reduce fatigue, by decreasing lactate levels and, consequently, the probability of developing an injury or reach an overtraining level is lower [6].

When analyzing Fig. 5 it can be verified that, before and after the 500 m race simulation, eight athletes presented lower BLC values when they wore the cooling vest during the AR, in comparison with the AR without wearing the cooling vest. One athlete (number 5 – Fig. 5) presented higher BLC values after performing the AR with the wearing of the cooling vest, and another athlete (number 6 – Fig. 5), after wearing of the cooling vest presented higher BLC values before the race simulation, but lower values after it. These changes were more evident in the values after the race simulation as indicated by the size of the grey columns, and it clearly suggests a difference in the kayakers’ physiologic response to this recovery method. This fact shows a singularity of high-performance sports, the need of individualization. In addition, a different response to the same recovery method means that what is an advantage for a given athlete may not be necessarily an advantage for another, although, on overall, wearing a cooling vest presented itself as an advantage.

Both the RPE and power had similar values in the two occasions of the study, which means that the kayakers performed the TS and the AR with similar intensity and the same amount of physical and motivational resources [31]. As a result, it can be assumed that the differences found in the BLC values were independent of these variables. Since the RPE has a correlation with lactate levels [14], it was expected that the RPE of the AR with the cooling vest would be lower, but this fact did not occur.

As it is described in literature [11, 12], cold increases the athletes’ subjective perception of recovery with an effect similar to a placebo. This effect is linked to dopamine release from the nucleus accumbens, a central component of the brain reward system, and it is known that the expectation of an analgesic effect can trigger endogenous opioid systems [15]. In the current study, similar results were obtained and they confirmed that, when kayakers performed the active recovery wearing the cooling vest, they felt more recovered than when they did not wear it. The individual perception of recovery is related to the subsequent performance, and the psychometric measures can be more sensitive than the physiological markers in determining the recovery state of the athletes [15].

At the two protocol conditions the performance has not been different. This means that it was not affected by the use of the cooling vest, despite the difference in the physiological response to cold. The reason for this result may have been the small number of participants on the study. Therefore, performing the same protocol with a larger number of athletes should be considered in future studies.

5. Limitations of this study

The core temperature measurement was not assessed with a sensor inside the body because this is an invasive method and it is not practical for the performance during a protocol with intense physical activity. It was assumed that the temperature of the urinary outpour is the same as the one inside the urinary bladder. In further studies, either parameter needs to be fixed or a correlation between the urine temperature inside and outside the bladder should be done prior to the study.

This study was conducted without taking into account the wind, solar radiation and other variables that exist in real weather conditions. For that reason, there is a need to confirm these results, of the recovery method in the real environment of racing, in different weather conditions, it would be interesting to carry out this study analyzing more muscles and with other cooling vests, in order to see if there is consistency in the results. Testing the muscles and comparing their impact, by taking into account the types of muscle fibers and their less or more intense participation in the paddling technique, it would be also of interest.

6. Conclusions

A cooling vest may lead to a faster physiological recovery by reducing the core and skin temperature and promoting a faster lowering of the BLC levels on male kayakers. This vest can be a useful tool to apply during the AR training sessions or races with a short period to recover in between them. Nevertheless, an evident correlation between cold effects and performance improvements was not found in the present study. Regarding the used settings, the IRT revealed to be a reliable method to monitor the skin temperature before and after the exercise.

Author contributions

FL, BG, AS and JP designed the research study. AS and MS prepared the set-up of the experiment. FL and NS performed the data collection. FL analyzed the data. FL, BG, AS and JP wrote the manuscript. All authors contributed to editorial changes in the manuscript. All authors read and approved the final manuscript.

Ethics approval and consent to participate

The study was approved by the Ethics Committee of the Faculty of Medicine of the University of Coimbra, with the official reference 131-CE-2017. Each subject signed a written informed consent.

Acknowledgment

Not applicable.

Funding

The research work here presented was partially supported by the Associated Laboratory of Energy, Transports and Aeronautics Projects with reference Nos.FCT/UIDB/50022/2020 and FCT/UIDP/50022/2020.

Conflict of interest

The authors declare no conflict of interest.

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Publisher’s Note: IMR Press stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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# What is G and what is H? What is the difference in the way they are defined?

Both ${\displaystyle G(s)\,}$ and ${\displaystyle H(s)\,}$ are notations for transfer functions. Sometimes we use ${\displaystyle G(i\omega )\,}$ (or ${\displaystyle H(i\omega )\,}$) to refer to the "frequency response" if the system is asymptotically stable; it can be obtained by evaluating ${\displaystyle G(s)\,}$ at ${\displaystyle s=i\omega \,}$.