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No.

Because 7 is not even a perfect number; it is conjectured (not proven) that there are no odd perfect numbers only even ones and 7 is an odd number - it has been proven that if there is an odd perfect number it has to be extremely large, much-much-much larger than 7)

Q: Is 7 the only perfect number on the form xn yn?

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This is represented as the algebraic expression xn/n or xn ÷ n.

31. The pattern is likely xn = xn-1 + n + 6.

There are 2 main methods for fixed point iteration, the Newton-Raphson method, and the Secant method.This method uses the formula, xn+1 = xn - f(xn) / f'(xn),where xn is the initial point, f(xn) is the value of the function at that point, and f'(xn) is the value of the differentiated function at that point. Plug all these values into the above equation to get xn+1, which then becomes the next initial point. Repeat until you get a point within an acceptable degree of error.the formula for this method is, xn+1= xn - (f(xn)(xn - xn-1)) / (f(xn) - f(xn-1)).With this formula you do not need the differentiated form of the function, making this a better method than the N-R for functions difficult to differentiate. However you do need two initial points for this method to work (xn and xn-1). Again just plug the appropriate values into the formula to generate the next point approximation.NB: with both of these methods be careful which initial point you choose, especially with the N-R method, as depending on the function the approximation iterations can go out of control and zoom away from the point you're trying to find.Also, a note on errors, you can sometimes get a better approximation by fiddling with your function a bit and reducing the amount of calculations needed. For example if you have two equivalent functions, but one takes 3 calculations to get a value and the other takes 6, the latter takes more work and will generally give a bigger error than the previous. Usually this error increase is only marginal, but depending on the function and the values used, the potential error can be huge. This only needs to be taken into account though if you want extremely accurate results. (Things to look out for if trying to reduce error: subtracting near equal numbers, dividing by a small number or multiplying by a large number, cancellation of significant figures)

No. For example all polynomials of the form y=xn (or sums of such positive terms) where n is a positive odd number do not have a minimum.

xn+1 = 1/2 ( xn + N/xn )

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The answer is 28.

What is the only perfect number of the form Xn + Yn

If by "xn" you mean ax^n then the answer is "a"

The Newton-Raphson method is a pretty efficient process. See link for details. If you want the square root of 34, say, define f(x) = x2 - 34 so that when x is the square root of 34, f(x) is zero. That would imply f'(x) = 2x So start with x0 and use the iteration xn+1 = xn - f(xn)/f'(xn) = xn - [xn2 - 34]/[2*xn]

This is represented as the algebraic expression xn/n or xn ÷ n.

25.2

x0 = x(n -n), which is equal to xn/xn by the law of powers. This obvoiusly = 1

31. The pattern is likely xn = xn-1 + n + 6.

There are 2 main methods for fixed point iteration, the Newton-Raphson method, and the Secant method.This method uses the formula, xn+1 = xn - f(xn) / f'(xn),where xn is the initial point, f(xn) is the value of the function at that point, and f'(xn) is the value of the differentiated function at that point. Plug all these values into the above equation to get xn+1, which then becomes the next initial point. Repeat until you get a point within an acceptable degree of error.the formula for this method is, xn+1= xn - (f(xn)(xn - xn-1)) / (f(xn) - f(xn-1)).With this formula you do not need the differentiated form of the function, making this a better method than the N-R for functions difficult to differentiate. However you do need two initial points for this method to work (xn and xn-1). Again just plug the appropriate values into the formula to generate the next point approximation.NB: with both of these methods be careful which initial point you choose, especially with the N-R method, as depending on the function the approximation iterations can go out of control and zoom away from the point you're trying to find.Also, a note on errors, you can sometimes get a better approximation by fiddling with your function a bit and reducing the amount of calculations needed. For example if you have two equivalent functions, but one takes 3 calculations to get a value and the other takes 6, the latter takes more work and will generally give a bigger error than the previous. Usually this error increase is only marginal, but depending on the function and the values used, the potential error can be huge. This only needs to be taken into account though if you want extremely accurate results. (Things to look out for if trying to reduce error: subtracting near equal numbers, dividing by a small number or multiplying by a large number, cancellation of significant figures)

(xn+2-1)/(x2-1)ExplanationLet Y=1+x2+x4+...+xn. Now notice that:Y=1+x2+x4+...+xn=x2(1+x2+x4+...+xn-2)+1Y+xn+2=x2(1+x2+x4+...+xn-2+xn)+1Y+xn+2=x2*Y+1Y+xn+2-x2*Y=1Y-x2*Y=1-xn+2Y(1-x2)=1-xn+2Y=(1-xn+2)/(1-x2)=(xn+2-1)/(x2-1)

No. For example all polynomials of the form y=xn (or sums of such positive terms) where n is a positive odd number do not have a minimum.

Each number in the series is the sum of the preceding number and the number two numbers back from the preceding number. Xn = Xn-1+Xn-3 where the number that started should be zero.