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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)
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What is x to the power of the number that it is divided by?
This is represented as the algebraic expression xn/n or xn ÷ n.
What number should come next 1 7 14 22?
31. The pattern is likely xn = xn-1 + n + 6.
Fixed point iteration methods plus algorithm?
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)
All polynomials have at least one minimum?
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.
Newtons iteration formula for finding square root of N?
xn+1 = 1/2 ( xn + N/xn )
What is the only perfect number of the form Xn Yn?
The answer is 28.
The only perfect number of the form x thothe n power y to the n power?
What is the only perfect number of the form Xn + Yn
What monomial term has the form xn where the coefficient is a and the degree is n?
If by "xn" you mean ax^n then the answer is "a"
What is the process for estimating the square root of a number that is not a perfect square?
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]
What is x to the power of the number that it is divided by?
This is represented as the algebraic expression xn/n or xn ÷ n.
What is the closed form for a linear recurrence sequence xn 11.2xn-14?
25.2
Why when you have a nonzero number and its exponent is 0 why is the number equal to 1?
x0 = x(n -n), which is equal to xn/xn by the law of powers. This obvoiusly = 1
What number should come next 1 7 14 22?
31. The pattern is likely xn = xn-1 + n + 6.
Fixed point iteration methods plus algorithm?
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)
Sum of series 1 plus x2 plus x4.....plus xn?
(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)
All polynomials have at least one minimum?
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.
What is the formula for this sequence 1 1 2 3 4 6 9 13 19 28 41 60 88 129 The pattern is you add the first and third number to get the next and then u add the 2nd a 4th to get the 5th?
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.
