latex, napisanie długiego wyrazenia

[Anka20]

Kod: Zaznacz cały

sum_{k=0}^{n} {n choose k} left( k-n frac{x-a}{b-a} 
ight)^{2}left(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} =

= sum_{k=0}^{n} {n choose k} k^{2}left(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} - 2n codt frac{x-a}{b-a} sum_{k=0}^{n} {n choose k} kleft(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} + n^{2}  left(frac{x-a}{b-a}
ight)^{2}sum_{k=0}^{n} {n choose k} k^{2}left(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} stackrel{eqref{5},eqref{6},eqref{1}}{=}

stackrel{eqref{5},eqref{6},eqref{1}}{=} n frac{x-a}{b-a} + n(n-1)left(frac{x-a}{b-a}
ight)^{2} -2n^{2} left(frac{x-a}{b-a}
ight)^{2} + n^{2} left(frac{x-a}{b-a}
ight)^{2} = 

= n frac{x-a}{b-a} + left( n(n-1)-2n^{2}+n^{2}
ight) left(frac{x-a}{b-a}
ight)^{2} = n frac{x-a}{b-a} - n left(frac{x-a}{b-a}
ight)^{2}=

= n frac{x-a}{b-a} left(1- frac{x-a}{b-a}
ight)leq frac{1}{4}n 

Mam takie wyrażenie napisać w Texu ale nie wiem jakiego otoczenia użyć żeby to wyglądało tak ładnie jak w książkach, prosze o pomoc.-- 25 kwi 2012, o 15:23 --zrobilam to tak

Kod: Zaznacz cały

egin{align*}
f(x)sum_{k=0}^{n} {n choose k} left( k-n frac{x-a}{b-a} 
ight)^{2}left(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} = \
&= sum_{k=0}^{n} {n choose k} k^{2}left(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} - \
&-2n codt frac{x-a}{b-a} sum_{k=0}^{n} {n choose k} kleft(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} + n^{2}  left(frac{x-a}{b-a}
ight)^{2}sum_{k=0}^{n} {n choose k} k^{2}left(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} stackrel{eqref{5},eqref{6},eqref{1}}{=} \
&stackrel{eqref{5},eqref{6},eqref{1}}{=} n frac{x-a}{b-a} + n(n-1)left(frac{x-a}{b-a}
ight)^{2} -2n^{2} left(frac{x-a}{b-a}
ight)^{2} + n^{2} left(frac{x-a}{b-a}
ight)^{2} = \
&= n frac{x-a}{b-a} + left( n(n-1)-2n^{2}+n^{2}
ight) left(frac{x-a}{b-a}
ight)^{2} = n frac{x-a}{b-a} - n left(frac{x-a}{b-a}
ight)^{2}= \
&= n frac{x-a}{b-a} left(1- frac{x-a}{b-a}
ight)leq frac{1}{4}n 
end{align*}

ale i tak nie dziala, jak to poprawic?

[szw1710]

Ja widzę to w ten sposób:

Kod: Zaznacz cały

egin{align*}
f(x)
 &=sum_{k=0}^{n}
    inom{n}{k}left(k-nfrac{x-a}{b-a}
ight)^{2}
    left(frac{x-a}{b-a}
ight)^{k}
    left(frac{b-x}{b-a}
ight)^{n-k}=\[1ex]
 &=sum_{k=0}^{n}
    inom{n}{k} k^{2}left(frac{x-a}{b-a}
ight)^{k}
    left(frac{b-x}{b-a}
ight)^{n-k}-\
    &phantom{=;}-2ncdotfrac{x-a}{b-a}sum_{k=0}^{n}inom{n}{k}kleft(frac{x-a}{b-a}
ight)^{k}
             left(frac{b-x}{b-a}
ight)^{n-k}+\[1ex]
 &phantom{=;}+n^{2}left(frac{x-a}{b-a}
ight)^{2}sum_{k=0}^{n}inom{n}{k} k^{2}left(frac{x-a}{b-a}
ight)^{k} left(frac{b-x}{b-a}
ight)^{n-k} stackrel{eqref{5},eqref{6},eqref{1}}{=}\[1ex]
&=n frac{x-a}{b-a} + n(n-1)left(frac{x-a}{b-a}
ight)^{2} -2n^{2} left(frac{x-a}{b-a}
ight)^{2} + n^{2} left(frac{x-a}{b-a}
ight)^{2} = \[1ex]
&= n frac{x-a}{b-a} + left( n(n-1)-2n^{2}+n^{2}
ight) left(frac{x-a}{b-a}
ight)^{2} = n frac{x-a}{b-a} - n left(frac{x-a}{b-a}
ight)^{2}= \[1ex]
&= n frac{x-a}{b-a} left(1- frac{x-a}{b-a}
ight)leqslant frac{1}{4}n
end{align*}