{"id":3817,"date":"2022-09-15T12:01:02","date_gmt":"2022-09-15T10:01:02","guid":{"rendered":"https:\/\/erlebnisland-mathematik.de\/?page_id=3817"},"modified":"2023-01-26T14:07:32","modified_gmt":"2023-01-26T13:07:32","slug":"advanced-text-proof-without-words-sum-of-the-square-numbers","status":"publish","type":"page","link":"https:\/\/erlebnisland-mathematik.de\/en\/advanced-text-proof-without-words-sum-of-the-square-numbers\/","title":{"rendered":"Advanced text Proof without words: Sum of the square numbers"},"content":{"rendered":"

[vc_row drowwidth=”sidebar-biest-default sidebar-biest”][vc_column][vc_column_text]<\/p>\n

Proof without words: Sum of the square numbers<\/h1>\n

A “proof without words<\/em>” is so convincing that it is actually taken for a proof, although from the mathematician’s point of view it is not one. However, such a proof is often very suitable for quickly finding a mathematical proof. Why? The presented assumption is not proven universally, but only for some cases, e.g. up to \"n=6\". A mathematically complete proof would have to cover all possible natural numbers<\/em> \"n\", i.e. infinitely many cases. This is what one would do for these examples here with the proof principle of complete induction<\/em>. However, the exhibits are very well suited for making a suitable assumption for a general formula. This is very important, because without a conjecture you also have nothing to prove. Moreover, after careful observation, one can imagine what the next step would look like, then the one after that, and on and on, which is sufficient for a good justification of the facts. Moreover, after careful observation, one can imagine what the next step would look like, then the one after that, and on and on, which is sufficient for a good justification of the facts. Such a proof consists of two parts: the initial step<\/em> and the inductive step<\/em>. First, one shows with a simple case (usually \"n=1\") that the established assumption is correct. This first step is called the initial step. Then, starting from the assumption that the conjecture is correct for an (arbitrary) \"n\", it is shown that the conjecture is then also valid for the successor \"n+1\". This is often illustrated by the image of a ladder on which one has already climbed up to the \"n\"-th step, and from there always gets to the next, \"(n+1)\"-th step. Like the set of natural numbers, this ladder is also thought to be infinite. The exhibit on the sum of odd numbers consists of \"L\"-shaped plastic pieces of different sizes. The first piece consists of one box, the second of 3, the third of 5, etc., the \"n\"th piece of \"2n-1\" boxes. So if you add up the number of all these boxes, you get the sum of the first \"n\" odd numbers. The exhibit even makes it possible for the viewer not to have to do the math himself, but only to add them up. This is because the \"L\"-shaped parts fit together and, in the correct order (one must not leave any gaps), each results in a square. For each added part (corresponding to another summand when adding), the square grows by one box length in each direction. To be more precise: If you add the part of size \"2n-1\", you get a collapsed square with edge length \"n\", i.e. \"n^2\" boxes big. This leads to the formula <\/p>\n

  <\/span>   <\/span>\"\[1+3+5+\cdots+2n-1=\sum_{i=1}^n{(2n-1)}=n^2.\]\"<\/p>\n

[\/vc_column_text][vc_column_text]<\/p>\n

The proof by complete induction<\/h3>\n

[\/vc_column_text][vc_column_text]Now, as an example, we carry out a proof by complete induction of the statement just mentioned about the sum of the first \"n\" odd numbers:<\/p>\n

Claim<\/em>: <\/p>\n

  <\/span>   <\/span>\"\[1+3+5+\cdots+(2n-1)=\sum_{i=1}^n{(2i-1)}=n^2.\]\"<\/p>\n

Initial step<\/em>: For \"n=1\" obviously <\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^n{(2i-1)}=1\]\"<\/p>\n

and also \"n^2=1\". So the assumption for \"n=1\" is correct.<\/p>\n

Now assume that <\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^n{(2i-1)}=n^2\]\"<\/p>\n

is true. We show that <\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^{n+1}{(2i-1)}=(n+1)^2\]\"<\/p>\n

is then true.<\/p>\n

Inductive step<\/em>: Here, the limits of the summation are first shifted so that the assumption can then be used. Now we transform and apply the binomial formula<\/em> so that we get the desired result:
\n<\/em><\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^{n+1}{(2i-1)}=\sum_{i=1}^n{(2i-1)}+(2(n+1)-1)=n^2+2(n+1)-1=n^2+2n+1=(n+1)^2.\]\"<\/p>\n

[\/vc_column_text][vc_column_text]<\/p>\n

A related problem<\/h3>\n

[\/vc_column_text][vc_column_text]We now turn to a similar statement, but this time not about the sum of the first \"n\" odd numbers, but of the first \"n\" numbers:<\/p>\n

There is an anecdote about this task from the school days of Carl Friedrich Gauss<\/em> (1777–1855). In “Spektrum der Wissenschaft Spezial<\/em>“, 2\/2009, it is reported: “It is said about the nine-year-old Gauss that he finished an arithmetic problem in the shortest possible time, which his teacher B\u00fcttner had given the class: \"1+2+3+\cdots + 100 = 5050\"<\/em>. His reasoning is: proceeding from the “outside” to the “inside”, he combines the smallest and the largest number into a sum:<\/em> \"1+100\", \"2+99\", \"3+98\", …, \"50+51\". That gives 50 times the sum 101…. Teacher B\u00fcttner feels that he can’t really teach this boy anything and gives him a textbook on arithmetic, which he works out on his own.” <\/em>This is not the same approach as shown in the corresponding exhibit in the ADVENTURELAND MATHEMATICS, but this story also makes it clear that you can get faster and further with systematic thinking than with simple “calculating on the fly”.<\/em>. But the result is the same: Gauss also obtains (for the case \"n= 100\"):<\/p>\n

  <\/span>   <\/span>\"\[50\cdot 101=\frac{100\cdot 101}{2}=\frac{n(n+1)}{2}.\]\"<\/p>\n

Again we want to demonstrate the induction proof:<\/p>\n

Claim:<\/em> <\/p>\n

  <\/span>   <\/span>\"\[1+2+3+\cdots+n=\sum_{i=1}^n{i}=\frac{n(n+1)}{2}.\]\"<\/p>\n

Initial step<\/em>: For \"n=1\" <\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^n {i} =1\]\"<\/p>\n

and also \"\frac{n(n+1)} {2} =2/2=1\". So the assumption for \"n=1\" is correct.<\/p>\n

Now assume that <\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^n {i} =\frac{n(n+1)} {2} \]\"<\/p>\n

holds.<\/p>\n

This is now used to show that <\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^{n+1} {i} =\frac{(n+1)(n+2)} {2} \]\"<\/p>\n

also applies.<\/p>\n

Inductive step<\/em>: Here, the upper limit of the sum sign is first shifted so that the assumption can then be inserted. Then transform and factor out so that you get the desired result:
\n<\/em><\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^{n+1}{i}=\sum_{i=1}^n{i}+(n+1)=\frac{n(n+1)}{2}+n+1=\frac{n(n+1)+2(n+1)}{2}=\frac{(n+2)(n+1)}{2}.\]\"<\/p>\n

[\/vc_column_text][vc_column_text]<\/p>\n

One final proof<\/h3>\n

[\/vc_column_text][vc_single_image image=”1723″ img_size=”large” alignment=”center”]Figure 1: Exhibit on the sum of the first \"n\" square numbers[\/vc_single_image][vc_column_text]We conclude with a final induction proof on the sum of the first \"n\" square numbers:<\/p>\n

Claim:<\/em> <\/p>\n

  <\/span>   <\/span>\"\[1^2+2^2+3^2+\cdots+n^2=\sum_{i=1}^n{i^2}=\frac{n(n+1)(2n+1)}{6}.\]\"<\/p>\n

Initial step:<\/em> for \"n = 1\", obviously <\/p>\n

  <\/span>   <\/span>\"\[\sum_{i = 1}^{n}{i^2} = 1\]\"<\/p>\n

and also \"\frac{n(n + 1)(2n + 1)}{2}=\frac{1\cdot 2\cdot 3}{6}=1.\" So our assumption is correct for \"n=1\".<\/p>\n

Now assume that <\/p>\n

  <\/span>   <\/span>\"\[\sum_{i=1}^n{i^2}=\frac{n(n+1)(2n+1)} {6} \]\"<\/p>\n

applies.<\/p>\n

Inductive step<\/em>: Here, the upper limits of the summation are first shifted so that the equation of the assumption can then be used. Then it is transformed, factored out and combined again so that the desired result is obtained:<\/p>\n

  <\/span>   <\/span>\"\begin{align*}\sum_{i=1}^{n+1}{i^2}=\sum_{i=1}^n{i^2}+(n+1)^2&=\frac{n(n+1)(2n+1)}{6}+(n+1)^2=\frac{(n+1)(n(2n+1)+6(n+1))}{6}\&=\frac{(n+1)(n+2)(2n+3)}{6}.\end{align*}\"<\/p>\n

[\/vc_column_text][vc_column_text]<\/p>\n

Literature<\/h3>\n

[\/vc_column_text][vc_column_text][1] https:\/\/de.wikipedia.org\/wiki\/Vollst\u00e4ndige_Induktion<\/a>.<\/p>\n

[2] https:\/\/de.wikipedia.org\/wiki\/Gau\u00dfsche_Summenformel<\/a>.[\/vc_column_text][\/vc_column][\/vc_row]<\/p>\n<\/div>","protected":false},"excerpt":{"rendered":"

[vc_row drowwidth=”sidebar-biest-default sidebar-biest”][vc_column][vc_column_text] Proof without words: Sum of the square numbers A “proof without words” is so convincing that it is actually taken for a proof, although from the mathematician’s point of view it is not one. However, such a proof is often very suitable for quickly finding a mathematical proof. Why? The presented assumption … Advanced text Proof without words: Sum of the square numbers<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_acf_changed":false,"footnotes":""},"folder":[],"class_list":["post-3817","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/pages\/3817","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/comments?post=3817"}],"version-history":[{"count":16,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/pages\/3817\/revisions"}],"predecessor-version":[{"id":4667,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/pages\/3817\/revisions\/4667"}],"wp:attachment":[{"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/media?parent=3817"}],"wp:term":[{"taxonomy":"folder","embeddable":true,"href":"https:\/\/erlebnisland-mathematik.de\/en\/wp-json\/wp\/v2\/folder?post=3817"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}