TeX expression not right

mart_r · 2021-12-18, 16:51

I hope this roughly qualifies as homework help. I've been spending an hour to get this one line into the TeXworks program:

Code:

\Bbb P(\xi_n=1)={1\over\log n} \mbox{ and } \Bbb P(\xi_n=0)=1-{1\over\log n}

What I get is shown in the attachment. I was happy to have the left half of the expression finally work out correctly, but the right half is slowly driving me berserk. What am I doing wrong?

Thanks in advance!

kruoli · 2021-12-18, 17:04

\(\Bbb P(\xi_n=1)={1\over\log n} \mbox{ and } \Bbb P(\xi_n=0)=1-{1\over\log n}\)

Hm... It is interpreted the way you want here. Have you tried:
{\Bbb P}(\xi_n=1)={1\over\log n} \mbox{ and } {\Bbb P}(\xi_n=0)=1-{1\over\log n}
\({\Bbb P}(\xi_n=1)={1\over\log n} \mbox{ and } {\Bbb P}(\xi_n=0)=1-{1\over\log n}\)

mart_r · 2021-12-18, 18:08

I know, it works outside of the program.
But the trick with bracketing the "\Bbb P" part had the desired effect, now the outcome looks like it should. Thanks a lot!

kruoli · 2021-12-18, 21:17

Some clarification for this:
There are some TeX-commands that work like a function with one parameter (which will be the same if written as \x{y} or \x y), multiple parameters (as in \x{y}{z}...) and there are functions that work like a switch (meaning that using them will be applied to the whole current block). I'm leaving out optional parameters here and I might be simplifying things a lot.

2021-12-18, 16:51	#1
mart_r Dec 2008 you know...around... 1100001111₂ Posts	TeX expression not right I hope this roughly qualifies as homework help. I've been spending an hour to get this one line into the TeXworks program: Code: \Bbb P(\xi_n=1)={1\over\log n} \mbox{ and } \Bbb P(\xi_n=0)=1-{1\over\log n} What I get is shown in the attachment. I was happy to have the left half of the expression finally work out correctly, but the right half is slowly driving me berserk. What am I doing wrong? Thanks in advance! Attached Thumbnails

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2021-12-18, 17:04	#2
kruoli "Oliver" Sep 2017 Porta Westfalica, DE 2⁶·17 Posts	\(\Bbb P(\xi_n=1)={1\over\log n} \mbox{ and } \Bbb P(\xi_n=0)=1-{1\over\log n}\) Hm... It is interpreted the way you want here. Have you tried: {\Bbb P}(\xi_n=1)={1\over\log n} \mbox{ and } {\Bbb P}(\xi_n=0)=1-{1\over\log n} \({\Bbb P}(\xi_n=1)={1\over\log n} \mbox{ and } {\Bbb P}(\xi_n=0)=1-{1\over\log n}\)

2021-12-18, 18:08	#3
mart_r Dec 2008 you know...around... 1100001111₂ Posts	I know, it works outside of the program. But the trick with bracketing the "\Bbb P" part had the desired effect, now the outcome looks like it should. Thanks a lot!

2021-12-18, 21:17	#4
kruoli "Oliver" Sep 2017 Porta Westfalica, DE 2⁶·17 Posts	Some clarification for this: There are some TeX-commands that work like a function with one parameter (which will be the same if written as \x{y} or \x y), multiple parameters (as in \x{y}{z}...) and there are functions that work like a switch (meaning that using them will be applied to the whole current block). I'm leaving out optional parameters here and I might be simplifying things a lot.