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- Thread starter PeggleFrank
- Start date
Also, I have just discovered that
`the set of functions F(G) where G is a field is a Vector Space over G`
You can prove thisfairly easily using `addition` and `multiplication` notation for the field. Firstly define function addition and scalar multiplication as such where `f` and `g` are functions and `a` is a member of `G`
```
[f + g](x) = f(x) + g(x)
[af](x) = af(x)
```
Now we can prove some things about this to satisfy the requirements of vector spaces. Firstly it is easily shown that `F(G)` is an abelian group with vector addition and the zero function (`f(x) = 0`) and thus I will not go into that proof. For the vector space requirements it is easy to see that `af` is an element of `F(G)` or more precisely a function from `G` to `G`. You can also easily prove that `a(f + g) = af + ag` fairly easily
`[a(f + g)](x) = a[f+g](x) = a[f(x) + g(x)] = af(x) + ag(x) =[ af + ag](x)`
`(a + b)f = af + bf` follows a similar process
`[(a + b)f](x) = (a+b)f(x) = af(x) + bf(x) = [af + bf](x)`
associativity as well `a(bf) = (ab)f`
`[a(bf)](x) = a[bf](x) = a[bf(x)] = abf(x) = (ab)f(x) = [(ab)f](x)`
Multiplicative identity is also simple `1f = f`
`[1f](x) = 1 * f(x) = f(x) = f`
Was on my clipboard
`the set of functions F(G) where G is a field is a Vector Space over G`
You can prove thisfairly easily using `addition` and `multiplication` notation for the field. Firstly define function addition and scalar multiplication as such where `f` and `g` are functions and `a` is a member of `G`
```
[f + g](x) = f(x) + g(x)
[af](x) = af(x)
```
Now we can prove some things about this to satisfy the requirements of vector spaces. Firstly it is easily shown that `F(G)` is an abelian group with vector addition and the zero function (`f(x) = 0`) and thus I will not go into that proof. For the vector space requirements it is easy to see that `af` is an element of `F(G)` or more precisely a function from `G` to `G`. You can also easily prove that `a(f + g) = af + ag` fairly easily
`[a(f + g)](x) = a[f+g](x) = a[f(x) + g(x)] = af(x) + ag(x) =[ af + ag](x)`
`(a + b)f = af + bf` follows a similar process
`[(a + b)f](x) = (a+b)f(x) = af(x) + bf(x) = [af + bf](x)`
associativity as well `a(bf) = (ab)f`
`[a(bf)](x) = a[bf](x) = a[bf(x)] = abf(x) = (ab)f(x) = [(ab)f](x)`
Multiplicative identity is also simple `1f = f`
`[1f](x) = 1 * f(x) = f(x) = f`
Was on my clipboard
<insert an ungodly amount of text here>
Is that... the entire text of a chapter or two of the Lord of the Rings?
Why you do dis, computer science professor guy? Why?
Is that... the entire text of a chapter or two of the Lord of the Rings?
Why you do dis, computer science professor guy? Why?
fac' f 0 = 1
fac' f n = n * f (n-1)
fix f = f (fix f)
fac = fix fac'
explicit fixes instead of tail recursion courtesy of https://news.ycombinator.com/item?id=11060649
fac' f n = n * f (n-1)
fix f = f (fix f)
fac = fix fac'
explicit fixes instead of tail recursion courtesy of https://news.ycombinator.com/item?id=11060649
derp.c:5:21: error: expression is not assignable
printf("%d", 85 ++);
~~ ^
1 error generated.
Somebody posted 85 ++ on the GvE thread, so I tried to see if it would compile. It did not.
printf("%d", 85 ++);
~~ ^
1 error generated.
Somebody posted 85 ++ on the GvE thread, so I tried to see if it would compile. It did not.
Input file Size Compressed size Ratio ls -l size .txt.gz Ratio
1 60212 50423 0.837424434 60212 25072 0.416395403
2 69942 58468 0.835949787 69942 28771 0.411355123
3 99862 82373 0.824868318 99949 40503 0.405236671
4 127151 106145 0.83479485 127361 52714 0.413894363
5 122585 100845 0.822653669 122776 49889 0.40634163
6 89207 74184 0.831593933 89250 37033 0.414935574
7 156664 130261 0.831467344 156817 64630 0.412136439
8 127704 107358 0.840678444 127831 53280 0.416800307
9 184639 150881 0.817167554 185225 76138 0.411056823
10 233091 191816 0.822923236 233272 93087 0.399049179
Average 127105.7 105275.4 0.829952157 127263.5 52111.7 0.410720151
Copying a bunch of numbers between Excel and Word for a project that I finished not nearly long enough before the deadline.
1 60212 50423 0.837424434 60212 25072 0.416395403
2 69942 58468 0.835949787 69942 28771 0.411355123
3 99862 82373 0.824868318 99949 40503 0.405236671
4 127151 106145 0.83479485 127361 52714 0.413894363
5 122585 100845 0.822653669 122776 49889 0.40634163
6 89207 74184 0.831593933 89250 37033 0.414935574
7 156664 130261 0.831467344 156817 64630 0.412136439
8 127704 107358 0.840678444 127831 53280 0.416800307
9 184639 150881 0.817167554 185225 76138 0.411056823
10 233091 191816 0.822923236 233272 93087 0.399049179
Average 127105.7 105275.4 0.829952157 127263.5 52111.7 0.410720151
Copying a bunch of numbers between Excel and Word for a project that I finished not nearly long enough before the deadline.
Yes. Although in FORTRAN, it might work. FORTRAN it weird.
<evilcraft:werewolfFlesh:1>.withTag({Id: "84ae15d2-1935-4b8f-80e2-c8bd0fe86f58", Name: "goreae"}) * 1