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snooper007
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Is the square of Heaviside function equal to Heaviside?
[tex]H(t-t')\times H(t-t')=H(t-t')? [/tex]
Please help me on the above equation.
[tex]H(t-t')\times H(t-t')=H(t-t')? [/tex]
Please help me on the above equation.
mathman said:Quibble continued. We can define H(0) to be 0 or 1 (or anything else in between), since it doesn't affect its properties. If you want H(x)=H(x)2 for all values of x, let H(0)=0 or 1.
No. [tex]H(t-t')\times H(t-t')=R(t-t') [/tex]snooper007 said:Is the square of Heaviside function equal to Heaviside?
[tex]H(t-t')\times H(t-t')=H(t-t')? [/tex]
Please help me on the above equation.
The Heaviside function, also known as the unit step function, is a mathematical function that is defined as 0 for negative input values and 1 for positive input values.
The square of the Heaviside function is defined as the product of the Heaviside function with itself. It is often denoted as H(x)^2.
Yes, the square of the Heaviside function is equal to the Heaviside function for all values of x. This can be seen by substituting the definition of the Heaviside function into the square expression.
The square of the Heaviside function is often used in mathematical equations and models to represent the behavior of a system that has a discontinuity or sudden change. It is also useful in simplifying complex equations.
Yes, the square of the Heaviside function has several important properties, including being an even function, having a derivative of 0 everywhere except at x = 0, and being equal to the absolute value function. These properties make it a useful tool in mathematical analysis and applications.