Community Profile # bym

### NCSU

2011 年からアクティブ

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linear system of equations
Your equations are not independent. The second is 2 times the first

9年弱 前 | 1

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Troubles with interaction problem
here is a partial answer, without using symbolic tool box...see my comment above and you should be able to adapt the code to wha...

9年以上 前 | 0

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Why is my Simpson's 3/8 code not producing the correct values?
your first loop should start at 2, second at 3, third at 4. Set end points for all loops at n-1. Make sure n is divisible by 3 ...

9年以上 前 | 1

how to create customized curve ( only know the shape )
maybe this or similar would help <http://www.mathworks.com/matlabcentral/fileexchange/7173-grabit>

10年弱 前 | 0

problem about ODE (Rayleigh-Plesset Equation) error in ode15s
check your call to ode15s [T,R] = ode15s('RP',[0 3e-8],[1e-6 70]); %your code what does the following give you? [T,...

Problem in reading data from Excel sheet and copying it in 2d array
reverse i & j ? %copying data from the Numeric array to P for i = 1: 1:1110 for j = 1: 1 : 6...

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Issue with lsqcurvefit and langmuir type equation
see comments in revised code below myfun = @(x,xdata) (x(1)*x(2).*xdata)./(1+x(2).*xdata); %added elementwise operators ./ ...

how do i save looped output into 1 variable matrix
c=19; % D=[]; p = zeros(c,1); % preallocate for k=1:c; Z=[X(:,1),Y(:,1)]; p(k)=anova1(Z); %D=save(p) unecessary %X(...

Plese help me. Matlab 'while' loop. A set of data is given..?
x =[ 8, 10, 12, 14, 16] find(x>=11,1) ans = 3 *edit*----------------------------- x = 8 10 ...

Finding coefficients and constant for multivariable linear equation.
You don't have enough points. 4 points < 5 unknowns

Vandermonde Matrix and an Error Vector
X = V\B

gaussian histogram so that area is equal to one.
it does not make sense to me to have the number of bins equal to your length of data. The purpose of a histogram is to describe...

error message using polyfit (nonlinear regression)
I think it is obvious from the message; you don't have a enough data points to fit the polynomial uniquely. Like fitting a line ...

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Don't define wn as symbolic vo=input('vo='); xo=input('xo='); wn=input('wn='); syms t; %syms wn; comment this...

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Plot inside a "for" loop
Please see comments in code. I checked a couple of values and they agree with your excel table, but it is up to you to verify t...

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finding the breakeven point
Are you sure your equation is right? Anyway here is an example with a modified equation: f = @(c) log((c-1)./log(1+c)) ...

sum(w) and ones(1,size(w,2))*w' results totally different numbers
try ones(size(w))*w.'

Ttest problem with alpha
you do not need to pass 'Alpha' to the function, you are getting the error because you are trying to pass a string when it is ex...

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%{ were you thinking of multiple lines of comments? (curly braces,not tildes) }%

Curve fitting to get gamma curve parameter
If you have the statistics toolbox, you can use: gamfit()

to find area of parabola?
x=[7.8 8.25 8.55]; y=[0.96 0.99 0.94]; p=polyfit(x,y,2); %f=polyval(x,p); x & p should be reversed ...

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Plot empties after using datetick
clc;clear; close all startDate = datenum('01-01-2009'); endDate = datenum('12-31-2009'); xData = linspace(startDa...

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How to plot in 4_D?
look at the documentation for slice by typing in the command window doc slice also, look at the documentation f...

Simpson's Rule Illustration - How to create those parabolas?
here is what I have done <<http://snag.gy/cjsli.jpg>> clc;clear; close all x = linspace(0,4*pi); % create data ...

How to solve max load of P in Matlab?
So, when I see a problem statement that says "solve the above system of equations" I think of formulating the equations in matri...

Want to make linear fitting and plot both graph on the same curve
you are going to have to transform your data: x=[0;0.100000000000000;0.200000000000000;0.300000000000000;0.40000000000000...

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Solving for a data point
pt2pt1 = @(M)(((y+1)*M^2)/((y-1)*M^2+2))^(y/(y-1))*((y+1)/(2*y*M^2-(y-1)))^(1/(y-1))-.6925 fsolve(pt2pt1,2)

Creating a for loop that adds a number to a letter
for insight into what is going on try char(1:99) does it look familiar to what you are getting?

How to obtain values symbolically
alternative: c =sym(1:10) f = @(x) sum(c.*x.^(1:length(c)));

10年以上 前 | 0