{"id":21,"date":"2004-04-09T14:35:59","date_gmt":"2004-04-09T14:35:59","guid":{"rendered":"http:\/\/www.sixthform.info\/maths\/?p=21"},"modified":"2004-04-09T14:35:59","modified_gmt":"2004-04-09T14:35:59","slug":"the-lambertw-function","status":"publish","type":"post","link":"https:\/\/www.sixthform.info\/maths\/?p=21","title":{"rendered":"The LambertW Function"},"content":{"rendered":"

Many equations cannot be solved exactly without using special functions. For example, to solve 2^x=3 requires the use of the \\ln function (or similar). This function is sometimes defined in terms of an integral from which their properties can be deduced. Thus \\ln is defined by \\displaystyle \\ln x=\\int ^x _1 \\dfrac{dt}{t}\\text{ for } x>0 and it is then clear that, for example, \\ln 1=0<\/p>\n

There are many equations that can only be solved in terms of newly-defined functions. One such function that isn’t all that well known is the LambertW function<\/b> where w\\left(x\\right) is defined as a solution (for w) of we^w=x. This allows you to solve equations like 2^x=x^8 which was asked about on the S.O.S. Mathematics CyberBoard<\/a><\/p>\n

To solve 2^x=x^8 let w=-\\ln x so that x=e^{-w}. Then<\/p>\n

2^x=x^8 \\Longrightarrow 2^{e^{-w}}=e^{-8w} \\Longrightarrow e^{-w}\\ln 2=-8w \\Longrightarrow -\\frac{1}{8}\\ln 2=we^w<\/p>\n

Thus w=w(\\frac{1}{8}\\ln 2) and so x=e^{-w}=-8\\dfrac{w(\\frac{1}{8}\\ln 2)}{\\ln 2} which is our answer.<\/p>\n

Using tables or software this gives 1.100.<\/p>\n

But hang on, is that the only solution? No, because 2^x<x ^8 for small values of x and 2^x grows much faster than x^8 so 2^x>x^8 for large values of x. Since both x \\mapsto 2^x and x \\mapsto x^8 are continuous on \\mathbb{R} there is another value of x for which 2^x=x^8. A quick fiddle with a calculator gives x\\approx43.5.<\/p>\n

Research into the LambertW function to find out how this other solution can be given in terms of this function.<\/x><\/p>\n","protected":false},"excerpt":{"rendered":"

Many equations cannot be solved exactly without using special functions. For example, to solve requires the use of the function (or similar). This function is sometimes defined in terms of an integral from which their properties can be deduced. Thus is defined by and it is then clear that, for example, There are many equations […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-21","post","type-post","status-publish","format-standard","hentry","category-articles"],"_links":{"self":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts\/21","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=21"}],"version-history":[{"count":0,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts\/21\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=21"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=21"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=21"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}