Writing Coefficients of a Summation?

Question

amateurintraining 2017-9-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/357971-writing-coefficients-of-a-summation

回答： Image Analyst 2017-9-22

So, I have a quadratic function that is a summation. And the coefficients are the same, save for the first one but the powers are the powers from n_t to n_t-1.

To clarify: P*x^n_t - R*x^0 - R*x^1 ... - R*x^(n_t-1)

with coefficients: [P, -R, -R, ... , -R]

and powers: [n_t:n_t-1]

How do I write the coefficients in a way Matlab recognizes it?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

James Tursa 2017-9-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/357971-writing-coefficients-of-a-summation#answer_282748

编辑：James Tursa 2017-9-22

在 MATLAB Online 中打开

For use with MATLAB functions like polyval and friends, the highest power coefficient is the first element on down to the constant term which is the last element. So you should build your coefficient vector as

n_t = whatever;
coef = [P,-R*ones(1,n_t)];

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

Star Strider 2017-9-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/357971-writing-coefficients-of-a-summation#answer_282749

在 MATLAB Online 中打开

See if this does what you want:

n = 6;                                      % Arbitrary Constant
P = 3;                                      % Arbitrary Constant
R = 5;                                      % Arbitrary Constant
coefv = [P -R*ones(1, n)];                  % Coefficient Vector
xpntv = [n  0:n-1];                         % Exponent Vector
x = 4.2;                                    % Scalar Value For ‘x’
S = (x.^xpntv) .* coefv;                    % Series

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 3

Image Analyst 2017-9-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/357971-writing-coefficients-of-a-summation#answer_282754

在 MATLAB Online 中打开

Let's say, n_t is 5. Then (n_t - 1) is 4. Since you always go from n_t down to n_t-1, you always have only two terms, in this case an x^5 term and an x^4 term. But this does not make it a quadratic just because it has two terms. To be a quadratic, the highest term should be 2. So you can just make your coefficient array like

coefficient = [P, R];

And your output, for my example would be

y = P * x^5 + R * x^4;

I don't see any real need to make an array for the coefficients when you have only two of them.