how to solve equation

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Ls 2021-8-17

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评论： Ls 2021-8-18

when

x=1
y=x.^2+3 /(x-1)

and when

x=2:100, y=x.^3+55

how do you solve this in matlab using vectorization method

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Answer 1

Image Analyst 2021-8-18

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Try this:

% When x=1,  y=x.^2+3 /(x-1) and when x=2:100, y=x.^3+55,
x = 1 : 100;
% In general
y = x .^ 3 + 55;
% Special case for x = 1
y(1) = 1 .^ 2 + 3 ./ (1 - 1); % y will be infinity of course.
% Plot non-infinite results.
plot(x(2:end), y(2:end), 'b-', 'LineWidth', 2);
grid on;
fontSize = 15;
title('y = x .^ 3 + 55', 'FontSize', fontSize, 'Interpreter', 'none');
xlabel('x', 'FontSize', fontSize);
ylabel('y', 'FontSize', fontSize);

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Ls 2021-8-18

Sorry In the case of y(1), the equation is x.^2+3/(x+1)

Image Analyst 2021-8-18

在 MATLAB Online 中打开

@Ls, then the line of code would of course be

y(1) = 1 .^ 2 + 3 ./ (1 + 1);

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Answer 2

David Hill 2021-8-17

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y=zeros(size(x));
y(x==1)=3;
y(x~=1)=x(x~=1).^3+55;

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Image Analyst 2021-8-18

Yes. Originally it said

"when x=1, y=x.^2+3 /(x-1) and when x=2:100, y=x.^3+55,"

as I captured in the comment in my code. Then in his comment to me (after your and my Answers were posted) he changed it to x.^2+3/(x+1). And I added a comment to update it to the new formula.

Ls 2021-8-18