# Use $$and$$ tags NOT $and$ tags for latex.

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• Jun 16th 2013, 09:38 PM
mpx86
Re: Use $$and$$ tags NOT $and$ tags for latex.
$\displaystyle $\sqrt {{b^2} - 4ac}$$
• Aug 8th 2015, 08:43 PM
Igbafe
Re: Use $$and$$ tags NOT $and$ tags for latex.
am quite new to this forum and i have little idea on how to use the editor but please this is my question
Please am stuck with the integration I kindly need assistance in solving this equation, I will really appreciate if anyone could help with the solution or a guide on what i should do

$$C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz[\tex] • Aug 9th 2015, 01:14 AM GLaw The closing tag should be [/TEX], not [\TEX]. \displaystyle C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz Also it is better to write \log rather than log to indicate that it is an operator name rather than the product of three variables l,o,g. \displaystyle C(N)=\int_{0}^{\infty}\log(1+\frac{\rho}{a}z)\frac { N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz • Jul 27th 2017, 07:07 AM Loser66 Re: Use [tex] and$$ tags NOT $and$ tags for latex.
[/Tex]
\int_1^e \dfrac{1}{x}dx=lne-ln1=1
[/Tex]
• Jul 27th 2017, 07:19 AM
Mr.MathType
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
Quote:

Originally Posted by Loser66
[/Tex]
\int_1^e \dfrac{1}{x}dx=lne-ln1=1
[/Tex]

Don't use the / in the opening delimiter. Should be this to begin: [TEX]. So your equation is this:

$\displaystyle \int_1^e \dfrac{1}{x}dx=lne-ln1=1$
• Jul 27th 2017, 07:22 AM
skeeter
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
• Jul 27th 2017, 07:25 AM
Plato
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
Quote:

Originally Posted by Mr.MathType
Don't use the / in the opening delimiter. Should be this to begin: [TEX]. So your equation is this:

$\displaystyle \int_1^e \dfrac{1}{x}dx=lne-ln1=1$

\$\int_1^e \dfrac{1}{x}dx=lne-ln1=1\$ using the dollar sign gives a better rendering.

$\int_1^e \dfrac{1}{x}dx=lne-ln1=1$
• Jul 27th 2017, 08:57 AM
Mr.MathType
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
Actually I just noticed something else that would make it look better: Code the log as \ln, not simply ln. So now we have either $\int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$ or

$\displaystyle \int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$,

depending on whether it's inline or not.
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