Assume that k is positive (else the root of k will be a complex number), and the root is the positive square root.
Suppose this thing converges and the value it converges to is x. Then,
x^2 - k = x
So, x^2 - x - k = 0.
Solving, this has two solutions and one is negative. It can't be the negative one because everything is always positive by our assumption, so if x exists it must equal the positive root, i.e. x = (1+√(4k+1))/2.
It remains to show that the finite surds :√k,√(k+√k),... actually _do_ converge. Denote the n'th term by x_n. Observe x_n is increasing, so if we can show x_n is bounded above, we deduce convergence and the limit must be the x above.
Why is x_n bounded? Observe
(x_n)^2 - (x_n) < k (write out the left hand side)
Therefore, x_n < x because at x we have equality, x^2 - x = k.
So we're done!
If you drop the assumption and allow negative square roots and complex numbers, it's all going to depend on which of the two square roots you choose so it's not going to be well defined.
Infinite surd is a term used in mathematics. The definition of an infinite surd is a never ending irrational number with an exact value that would be left in square root form.
A surd in the form a√b cannot, in general, be simplified.
No.
If an infinite surd is like (√(6+√(6+√(6+√(6...),to solve it, follow these steps: Set x = √(6+√(6+√(6+√(6... therefore x2 = 6+√(6+√(6+√(6+√(6... therefore x2 = 6+x therefore 6+x-x2 = 0. Factorising the expression gives you (-x+3)(x+2)=0 Only the positive answer need be concerned with: -x+3=0 therefore -x = -3 therefore x = 3.
Normally they are.
Infinite surd is a term used in mathematics. The definition of an infinite surd is a never ending irrational number with an exact value that would be left in square root form.
2
the same way you find chuck Norris
A surd in the form a√b cannot, in general, be simplified.
Yes it is a surd
If the value of the surd is positive, then it will be another surd. Otherwise it will be a complex number.
A surd is the square root of an integer (whole number), e.g. square root of 2 is a surd. Pi is not the square root of an integer so not a surd, just an irrational number
I am pretty sure that the cube root of negative seven is a surd. I checked on the calculator.......and it showed a negative number.....??I think when it is not a surd...it's supposed to say error, so the number probably means that it is a surd..
No. It is not a criminal or civil offence!
No.
i did
In general, you cannot. An infinite number divided by any non-zero number is still infinite. An infinite number divided by another infinite number may or may not be infinite.