- #1
sparkle123
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Do you just replace the x's with (x-3)'s? Since e^(-x^2) is defined as the taylor series though, it seems like the answer should be the same as the series about x=0.
Thanks!
P.S. does anyone know how to resize images? :$
Where exactly are you getting stuck?sparkle123 said:Thanks, but how would you do that?
sparkle123 said:Do you just replace the x's with (x-3)'s? Since e^(-x^2) is defined as the taylor series though, it seems like the answer should be the same as the series about x=0.
Thanks!
P.S. does anyone know how to resize images? :$
The Taylor series for e^(-x^2) is an infinite series representation of the function e^(-x^2) centered at x=0. It can be written as:
e^(-x^2) = 1 - x^2 + (x^4/2!) - (x^6/3!) + (x^8/4!) - ...
The Taylor series for e^(-x^2) can be solved by using the general formula for a Taylor series and plugging in the derivatives of e^(-x^2) at x=0. This results in the infinite series shown in question 1.
No, the Taylor series for e^(-x^2) can be centered at any value of x. However, using x=0 is often the most convenient choice because it simplifies the calculations.
The Taylor series for e^(-x^2) is important because it allows us to approximate the value of e^(-x^2) at any point, not just at whole numbers. This is especially useful in mathematical and scientific calculations that involve the function e^(-x^2).
Yes, the value of e^(-x^2) at x=0 is equal to the sum of the Taylor series. However, since the series is infinite, the sum is an approximation of the actual value, which means it will only be equal if all terms are included.