Suppose f is a function. How can we solve the equation f = x? I think I've found an interesting new way, I'm not sure it's correct however. Say we want to solve the equation cos(x)=x, then a suitable way to solve this equation is to use Newton-Raphson to get a solution. But I think I've found a method that also converges, not as quickly however. Make a guess x0 for cos(x)=x. Calculate cos(x0) = x1, calculate cos(x1) = x2, calculate cos(x2) = x3 and so on. So we use x_(n + 1) = f(n). To demonstrate this method I solve the following equation:(adsbygoogle = window.adsbygoogle || []).push({});

1. cos(x)=x, start with a seed, I chose x0=-100 just to show that it does in deed seem to converge. This gives with the first 10 calculations:

0.84598898198358890274

0.78815472769682836497

0.76151115452367370362

0.74929854146798181945

0.74372766089317781727

0.74119341653274063166

0.74004213182177068154

0.73951944851492685807

0.73928222045897252958

0.73917456532965709331

cos(x)-x=0.00011648598787101138

2. ln(x)=x, start with a seed, I chose x0 = 1000. This gives:

0.31725931762133114611+1.33153507953496342211i

0.31718629567450052325+1.34014616816598811224i

0.31927011778936558593+1.33608733612343809223i

0.31731905625121883267+1.3375069028202560968i

0.3185799875460203639+1.33730100833347268442i

0.3179351563786906755+1.3370979004383313528i

0.31819146448512503102+1.33734761595800791012i

0.31812984714983422095+1.33716852678175367383i

0.31811627833445380074+1.33726784220316782635i

0.31814636098265470986+1.33722414212673179521i

This is quite good x-ln(x) = 0.00095307601525015884-0.01187533532322322971i

3. cos(x)^2=x, seed x=1000, gives:

~0.90326410456638764967

~0.86019033654796716573

~0.82962596896246371156

~0.80621728441276723759

~0.78742570290033394307

~0.77185503812233617185

~0.75865872241297805902

~0.74728626630218568701

~0.73736019802076276074

~0.72861058786690100970

This isn't good, since cos(x)^2-x = -0.1719 ..., however calculating more terms, like 100 terms gives cos(x)^2 - x = 0.00004448970614357726

I want to know if this is new and if it does in fact converge to a root of the equation?

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# On f(x)=x

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