{"id":3713,"date":"2013-11-09T19:17:50","date_gmt":"2013-11-09T13:47:50","guid":{"rendered":"http:\/\/www.bhagwad.com\/blog\/?p=3713"},"modified":"2017-10-06T01:58:01","modified_gmt":"2017-10-05T20:28:01","slug":"missing-steps-in-theorem-1-21-of-rudins-principles-of-mathematical-analysis","status":"publish","type":"post","link":"https:\/\/www.bhagwad.com\/blog\/2013\/personal\/missing-steps-in-theorem-1-21-of-rudins-principles-of-mathematical-analysis.html\/","title":{"rendered":"Missing Steps in Theorem 1.21 of Rudin’s “Principles of Mathematical Analysis”"},"content":{"rendered":"

Since Rudin frequently skips steps either for compactness, or to make the student work out the answers themselves, I thought I’d post my workings online so that I don’t forget them later on. I like to have everything spelled out for me, so I’ll be as detailed as possible. Here are the workings for Theorem 1.21 where I feel Rudin achieves results without telling us how he got there.<\/p>\n

Statement:<\/strong><\/p>\n

$latex \\text{If}\\;t=\\frac{x}{(1+x)}\\;\\text{then}\\;0\\le{t}<1$<\/p>\n

Proof:<\/strong><\/p>\n

$latex \\text{Since}\\;0<1\\;,\\text{we have}\\;x<x+1\\\\\\\\<\/p>\n

\\text{and}\\;0<x<x+1,\\;\\text{since}\\;x>0\\;\\text{by definition}$<\/p>\n

By Proposition 1.18(e), we have<\/p>\n

$latex 0<\\frac{1}{x+1}<\\frac{1}{x}\\\\\\\\<\/p>\n

\\text{Since} \\;\\frac{1}{x+1}>0\\;\\text{and}\\;x>0,\\;\\text{By Definition 1.17(ii), we have}\\\\\\\\<\/p>\n

x.\\frac{1}{x+1}=\\frac{x}{1+x}=t>0$<\/p>\n

So the first part of the inequality is satisfied.<\/p>\n

$latex \\text{Again, since}\\;x+1>x,\\;\\text{and}\\;\\frac{1}{x+1}>0\\;\\text{we have}\\;\\frac{1}{x+1}.(x+1)>\\frac{1}{x+1}.x\\;\\text{by proposition 1.18(b)}\\\\\\\\<\/p>\n

\\text{Therefore}\\;\\frac{x}{x+1}<1\\;\\text{Which is the second half of the inequality. So we have}\\linebreak0<t<1$<\/p>\n

Statement:<\/strong><\/p>\n

<\/strong>$latex t^n\\le t<x$<\/p>\n

Proof:<\/strong><\/p>\n

From the above result:<\/p>\n

[latex] 0<t<1\\;\\therefore\\;t.t<t.1\\implies t^2<t.\\;\\text{Also}\\;t.0<t.t\\implies t^2>0[\/latex] (Again from Proposition 1.18(b))<\/p>\n

And so,<\/p>\n

[latex]0<t^2<t<1[\/latex]<\/p>\n

Similarly, we can obtain:<\/p>\n

[latex]0<t^n\\le t<1\\;\\text{Since n can be equal to 1}[\/latex] So we have the first half of the inequality.<\/p>\n

Now from Proposition 1.18(d), we have\u00a0[latex]x^2>0\\;\\text{Since}\\;x>0\\implies x\\neq0[\/latex]<\/p>\n

From Definition 1.17(i),<\/p>\n

$latex x+x^2>x+0\\\\\\\\\\implies x(1+x)>x<\/p>\n

\\\\\\\\\\text{Since}\\;\\frac{1}{x+1}>0,\\;\\frac{1}{x+1}.x(1+x)>\\frac{x}{1+x}&s=1\\\\\\\\<\/p>\n

\\implies x>\\frac{x}{1+x}\\\\\\\\<\/p>\n

\\implies x>t$<\/p>\n

Which is the second part of the inequality. So we obtain:<\/p>\n

$latex t^n\\le t<x$<\/p>\n

Later, Rudin introduces this identity:<\/p>\n

$latex b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+….+a^{n-1})\\\\\\\\<\/p>\n

\\text{when }0<a<b$<\/p>\n

This is not proved directly, but we can prove it from right to left by expanding the right hand term like so:<\/p>\n

[latex](b-a)(b^{n-1}+b^{n-2}a+….+a^{n-1}) = b(b^{n-1}+b^{n-2}a+….+a^{n-1})-a(b^{n-1}+b^{n-2}a+….+a^{n-1})=(b^{n}+b^{n-1}a+….+ba^{n-1}) – (b^{n-1}a+b^{n-2}a^2+….+a^{n})=b^n-a^n[\/latex]<\/p>\n

(Thank you to Axiomer<\/a> from the physicsforums.com for supplying me with this). Don’t ask me how someone first came up with this formula!<\/p>\n

Statement:<\/strong><\/p>\n

[latex]b^n-a^n<(b-a)nb^{n-1}[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

$latex \\text{Since }0<a<b,\\;0<a^n<b^n\\\\\\\\<\/p>\n

\\implies (b^{n-1}.a^n).b>(b^{n-1}.a^n).a\\\\\\\\<\/p>\n

\\implies b^n.a^n>b^{n-1}.a^{n+1}\\\\\\\\<\/p>\n

\\text{So every successive term of }(b^{n-1}+b^{n-2}a+….+a^{n-1})\\text{ is lesser than }b^{n-1}\\\\\\\\<\/p>\n

\\text{since each term reduces the power of “b”by 1 and increases the power of “a” by 1.}\\\\\\\\<\/p>\n

\\implies nb^{n-1}>(b^{n-1}+b^{n-2}a+….+a^{n-1})\\\\\\\\<\/p>\n

\\text{Since }b>a,\\;(b-a)nb^{n-1}>(b-a)(b^{n-1}+b^{n-2}a+….+a^{n-1})\\\\\\\\<\/p>\n

\\implies b^n-a^n<(b-a)nb^{n-1}$<\/p>\n

Statement:<\/strong><\/p>\n

$latex 0<k<y$<\/p>\n

Rudin defines\u00a0k<\/em>:<\/p>\n

$latex k=\\dfrac{y^n-x}{ny^{n-1}},$<\/p>\n

But never explains why\u00a0$latex 0<k<y$. Here is the reason.<\/p>\n

Proof:<\/strong><\/p>\n

$latex \\text{Since } 1\\le n,\\\\\\\\<\/p>\n

1-n\\le 0\\text{. Also, }x>0\\text{ and }y>0\\therefore y^n>0\\\\\\\\<\/p>\n

\\implies \\dfrac{x}{y^n}>0\\ge 1-n\\\\\\\\<\/p>\n

\\implies \\dfrac{x}{y^n}>1-n\\\\\\\\<\/p>\n

\\implies x>y^n(1-n)<\/p>\n

\\implies -x<y^n(n-1)\\text{ By proposition 1.18(c) with x=-1}\\\\\\\\<\/p>\n

\\implies -x<ny^n-y^n\\\\\\\\<\/p>\n

\\implies y^n-x<ny^n\\\\\\\\<\/p>\n

\\implies \\dfrac{y^n-x}{ny^n}<1\\\\\\\\<\/p>\n

\\implies \\dfrac{y^n-x}{ny^{n-1}.y}<1\\\\\\\\<\/p>\n

\\implies \\dfrac{y^n-x}{ny^{n-1}}<y\\text{ (}y>1\\implies \\dfrac{1}{y}>1\\text{ by proposition 1.18(d)}\\text{)}\\\\\\\\<\/p>\n

\\implies k<y\\\\\\\\<\/p>\n

\\text{Also, } 0<x<y^n\\text { (by assumption) and so }y^n-x>0\\text{ and }ny^{n-1}>0\\\\\\\\<\/p>\n

\\implies \\dfrac{y^n-x}{ny^{n-1}}>0\\implies k>0\\\\\\\\<\/p>\n

\\text{And so }0<k<y$<\/p>\n

Statement:<\/strong><\/p>\n

$latex y^n-t^n\\le y^n-(y-k)^n<kny^{n-1}=y^n-x\\text{ if }t\\ge y-k$<\/p>\n

Proof:<\/strong><\/p>\n

First we’ll show that<\/p>\n

$latex \\text{If }a>b,\\text{ then }a^n>b^n\\text{ when } 0<b<a$<\/p>\n

$latex a.a>b.a\\text{ from proposition 1.18(b)}\\\\\\\\<\/p>\n

\\text{and }b.a>b.b\\\\\\\\<\/p>\n

\\implies a^2>ba>b^2\\text{. Similarly, }a^n>b^n$<\/p>\n

$latex \\text{So if }t\\ge y-k,\\\\\\\\<\/p>\n

t^n\\ge (y-k)^n\\\\\\\\<\/p>\n

\\implies -t^n\\le -(y-k)^n\\text{ Again by proposition 1.18(c) with x=-1}\\\\\\\\<\/p>\n

\\implies y^n-t^n\\le y^n-(y-k)^n\\text{ And so the first part of the inequality is proved}\\\\\\\\<\/p>\n

\\text{ Next, we’ve already seen that }b^n-a^n<(b-a)nb^{n-1}\\\\\\\\<\/p>\n

\\text{Putting }b=y\\text{ and }a=y-k\\\\\\\\<\/p>\n

y^n-(y-k)^n<kny^{n-1}=y^n-x\\text{ from the definition of k. And so}\\\\\\\\<\/p>\n

y^n-t^n\\le y^n-(y-k)^n<kny^{n-1}=y^n-x$<\/p>\n

To me, these are the main missing steps in Theorem 1.21. If I’ve left anything out, let me know!<\/p>\n","protected":false},"excerpt":{"rendered":"

Rudin skips over many essential steps when presenting his proofs. Here are the missing steps for Theorem 1.21.<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21],"tags":[],"class_list":["post-3713","post","type-post","status-publish","format-standard","hentry","category-personal"],"yoast_head":"\nMissing Steps in Theorem 1.21 of Rudin's "Principles of Mathematical Analysis"<\/title>\n<meta name=\"description\" content=\"Rudin skips over many essential steps when presenting his proofs. 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