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September 20

September 21

September 24

September 25

Linear equation perfect square solutions

I was reading about finding a solution to a quadratic equations which is a perfect square (the solution is easy to understand) and I started wondering about linear equations. Given a simple linear equation in the form y=ax+b where a and b are constants, what value of x will ensure that y is a perfect square? I've tried turning ax+b into (cx+d)(cx+d), but that hasn't worked. — Preceding unsigned comment added by 209.149.115.99 (talk) 18:19, 25 September 2014 (UTC)[reply]

If you want integer solutions {x,y} then these only exist if conditions are met for the constants, see Diophantine_equation#One_equation. Then you would just have to search among the (countable) solution set for perfect squares. You could also write y=z^2 and think of z^2=ax+b as a quadratic Diophantine equation, and solve using a Pell equation as described here [1]. Both of these methods would miss solutions where x is not an integer and y is a perfect square. SemanticMantis (talk) 19:12, 25 September 2014 (UTC)[reply]

I worked on this last night in earnest, but I don't see the relationship between y²=ax+b and the Diophantine ax²+by²=k. I changed my original request of y to y² to reinforce that I am looking for a square. However, the x is not a square. If I rearrange it, I can make, at best, ax+by²=k. Without the squared x, I cannot make any headway on the solutions provided. 209.149.115.99 (talk) 11:26, 26 September 2014 (UTC)[reply]

Sorry about that, I misread my mathworld link. I don't know much about this area, but you could try looking up the Ito (1987) paper cited at mathworld, and also see what other papers have cited it. SemanticMantis (talk) 14:51, 26 September 2014 (UTC)[reply]

If you're happy with just rational solutions then draw a circle around the origin intersecting where y=ax+b is zero. This will be a rational point since y is zero and a and b are rational so the other point on the circle where y=ax+b cuts it is also rational. Any choice of a rational number for x and 0 for y for the origin will do the same thing. Dmcq (talk) 12:03, 26 September 2014 (UTC) Oops silly me I misread the question, I'll have a think about it. Dmcq (talk) 17:08, 26 September 2014 (UTC)[reply]

If you are looking for integer solutions with a and b being integers then your question is equivalent to asking if

y^{2}=b\mod a

. See the article Quadratic reciprocity for how the Legendre symbol can be used to quickly check if there is a solution is a or b are large - though it won't give an actual solution unfortunately. The article quadratic residue gives a ways of calculating the residues without checking each possibility except when a prime factor is of the form

a=8n+1

. Dmcq (talk) 17:29, 26 September 2014 (UTC)[reply]

Thanks. I didn't look at it as a mod problem. Once you throw that in, I can see why there is no simple solution for x. It looks like you'd have to try every x in the range you like to see if any work. 209.149.115.99 (talk) 17:49, 26 September 2014 (UTC)[reply]

September 26

Approximation of the cosine function

In the Cosine article on Mathworld, an interesting approximation is mentioned:

A close approximation to cos(pix/2) for x in [0,1] is

${\begin{aligned}H(x)&=1-{\frac {x^{2}}{x+(1-x){\sqrt {\frac {2-x}{3}}}}}&\quad (8)\\&\approx \cos \left({\frac {\pi }{2}}x\right)&\quad (9)\end{aligned}}$

(Hardy 1959), where the difference between cos(pix/2) and Hardy's approximation is plotted above.

However, all it mentions is Hardy's work, and the work cited does not contain any other information about this approximation! The derivation and/or potential applications for this would be interesting, but no other information about this can be found on the internet. Do you know any other sources/information about this? Llightex (talk) 21:05, 26 September 2014 (UTC)[reply]

Here is a way you might reverse engineer the formula, though I have no idea how Hardy derived it. Let C(x) = cos(πx/2). We know from the Taylor series that C(x) = 1 - x²/constant + other terms. Rewrite this as C(x) = 1 - x²/K(x) where K is to be determined. We also know C(1) = 0 from which K(1)=1. Expand K at x=1 to get K(x)=1+constant⋅(x-1)+higher terms. Again, collecting the the constant and higher terms into a single function, write K(x)=1+(x-1)L(x). At this point you can get a fairly good approximation for C by plugging in a linear approximation for L. But we also know C(1/2)=√2/2 which would imply (after some computation) L(1/2) = 1 - √(1/2). So perhaps a better approximation of L would be L(x)≈ 1 - √(1/2+m(x-1/2)) for some constant m. If you plug in C(2/3)=1/2 you get m=-1/3 which produces the approximation given, but other values of m might work just as well or better. I found m=-.337 gives the lowest mean square error on the interval. Note that there are points in the derivation where different choices could be made, for example you could write C(x) = 1 - x²⋅K(x) or K(x)=1+(x-1)/L(x). It might be fun to explore these variations to see how they compare with the one given. --RDBury (talk) 00:01, 27 September 2014 (UTC)[reply]

September 27

Decimal exponents

[I really really ought to remember this!]

How do decimal exponents work? How do I calculate the result of x × 10^y.z (with a numeral in place of each variable), for example? To my surprise, decimal exponent redirects to scientific notation, which didn't even address the issue as far as I could tell, and the fractional "Rational exponents" section of exponentiation didn't either, unless it went over my head. Not trying to understand the theoretical basis for it — I just want to know how to solve x × 10^y.z, and I don't even know what to call it, since "decimal exponent" apparently means something completely different. Nyttend (talk) 11:40, 27 September 2014 (UTC)[reply]

If you have

a^{x}

and x is not a whole number, the decimal form for x isn't helpful in the computation. Write the exponent as a fraction x = n/d, and then

a^{x}=a^{n/d}=(a^{n})^{1/d}={\sqrt[{d}]{a^{n}}}=({\sqrt[{d}]{a}})^{n}

,

So for example

4^{1.5}

is

4^{3/2}=({\sqrt {4}})^{3}=2^{3}=8

. Staecker (talk) 12:22, 27 September 2014 (UTC)[reply]

For one thing, I'd forgotten that x^½ is equal to the square root of x. But what to do about a fraction that's not conveniently .5? I don't know how to interpret the "d" outside the √ following your third equals sign — I know it's the denominator, but what's the significance of a number in that position? It's been several years since I took a maths course. Nyttend (talk) 12:39, 27 September 2014 (UTC)[reply]

Take 100^0.1. I just discovered that the result is the tenth root of 100. But the result of 100^0.4 is not 4x the tenth root of 100, and it's not exactly what you get when you raise the tenth root of 100 to the fourth power. I can't yet follow your instructions to use the fractional method, since I don't understand what you're doing in the <math> section. Nyttend (talk) 12:46, 27 September 2014 (UTC)[reply]

Any difference between your calculation of 100^0.4 and the fourth power of 100^0.1 will just be rounding errors, because they are the same by definition. In the fraction n/d above, n is the nth power and d is the dth root (for example if d is 3 then you want the cube root). You can do the calculations in either order. Try working out ~~1028~~1024 to the power of 0.7 (=7/10) It is easiest to find the tenth root first (a whole number) then raise this to the seventh power. You should get 128. Dbfirs 13:01, 27 September 2014 (UTC)[reply]

Okay, now I think I remember better; thank you. What I needed was the simple examples such as 1028^0.1 and 1028^0.7. I was using Windows Calculator (Windows 8), typing 100, hitting the x^y button, typing 0.1, hitting enter, squaring the result twice, and adding this to memory. Next, I typed 100, hit the x^y button, typed 0.4, hit enter, and subtracted it from memory; there was a difference, with the latter method getting a result smaller by approximately 4.958 × 10^-37. (I do understand negative exponents). But then, I know quite well that 2 is the tenth root of 1028, but typing 1028, hitting the x^y button, typing 0.1, and hitting enter produces a result of 2.0007798800968566238308934329587. Is it a calculator error? If my experiment with 0.1⁴ and 0.4 had produced equal results, I would have understood much more easily. Nyttend (talk) 13:16, 27 September 2014 (UTC)[reply]

You're confusing 1024 with 1028. Try 2 [x^y] 10 to get 1024. Then 1024 [x^y] .1 to get 2 again. Generally arithmetic errors in computers, even in Microsoft products, are so rare as to be practically nonexistent. There was the infamous Pentium FDIV bug, but that's the exception rather than the rule. Not to say you don't have to watch out for rounding errors which are a different type of thing, --RDBury (talk) 15:32, 27 September 2014 (UTC)[reply]

Oops! that was my fault. Memory problems of the non-electronic variety! Dbfirs 16:37, 27 September 2014 (UTC)[reply]

My Mac calculator program gives the exact value '2' when you raise 1024 to the 0.1 power. 'Decimal exponents' isn't the best search term to use. Fractional exponents would be better. More background on fractional exponents is given in our logarithm article. When x = 10^y, x is the logarithm of y, y is the antilogarithm of x. EdJohnston (talk) 16:00, 27 September 2014 (UTC)[reply]

One has to be a bit careful when applying this to complex numbers since the complex logarithm is multi-valued. For example the complex square root can have multiple meanings depending on context.--Jasper Deng (talk) 16:42, 27 September 2014 (UTC)[reply]

September 28

September 29

September 30

Quadratic expression evaluates to perfect square

For example, 2x^2 - 238x + 14161. Is there any way, apart from just substituting a succession of x values, to determine which integers will give the square of an integer? If my calculation is correct, the first eight positive ones are 20, 35, 39, 51, 68, 80, 84 and 99, which give the squares of 101, 91, 89, 85, 85, 89, 91 and 101. I see no obvious pattern beyond the symmetry about x=59.5 for a minimum.←86.146.61.61 (talk) 16:42, 30 September 2014 (UTC)[reply]