Why is the original fraction always positive? I'm confused because by definition ##\sqrt{x^2}=|x|##. So ##x## itself could be a negative number.
Hence in ##\sqrt{\dfrac{1}{2x^3y^5}}##, ##x,y## could be negative and since it's to the third and fifth power respectively, stay negative. Thanks.
Nope. But if ##x,y## can be negative, would this be correct: ##\dfrac{1}{|x||y^2|\sqrt{2xy}}\cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{2x|x||y^2|y}##.
Is there a way to simplify ##2x|x||y^2|y##?
I've multiplied everything out on paper and got ##x=2, \dfrac{15}{4}##, which is correct. However multiplying directly is tedious and from observing this problem I suspect there is a simplification or trick that I missed...
As you can see my instinct is to just plough ahead with the given equations instead of doing substitutions. Is there any specific characteristics to equations that give a hint to which substitution is most likely to succeed or is this just a case of doing loads of problems and memorizing what...
##y+3=3\sqrt{y+7}##
Square both sides:
##\Rightarrow y^2+6y+9=9y+63##
##\Rightarrow y^2-3y-54=0##
##\Rightarrow (y-9)(y+6)=0##
##y=9, -6##
But if you plug in ##y=-6## into the original equation, you get ##-3=3## . So it doesn't work. Why?
Using ##\log_e(n+1)-\log_en=2[\dfrac{1}{2n+1}+\dfrac{1}{3(2n+1)^3}+\dfrac{1}{5(2n+1)^5}+\cdots]## and setting ##n=1##, the result is ##\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+ \cdots##. I thought this was different from ##1-\dfrac{1}{2}+\dfrac{1}{3}- \cdots##.
By definition:
##\log_e(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}- \cdots ## ##(1)##
Replacing ##x## by ##−x##, we have:
##\log_e(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}- \cdots##
By subtraction,
##\log_e(\dfrac{1+x}{1-x})=2(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+ \cdots)##
Put ##...
How did you go from ##\dfrac{n!}{(r-1)!(n-r+1)!}## to ##\dfrac{n(n-1)(n-2)\cdots(n-r+2)}{(r-1)!}##?
I am assuming ##n!=n(n-1)(n-2)\cdots (n-r+2)(n-r+1)(n-r)\cdots 1##. Would this be correct? Thanks.
##r^{th}## term from beginning: ##^nC_{r-1}a^{n-r+1}(2x)^{r-1}##
For the ##r^{th}## term from the end, we first know there are a total of ##n+1## terms in this binomial expansion. Subtracting the (##r^{th}## term from the end) from the total number of terms, ##n+1##, results in ##n+1-r## which...
##(1+x)^n=1+C_1x+C_2x^2+C_3x^3...+C_nx^n##
Let ##x=1##, hence ##2^n=1+C_1+C_2+C_3...+C_n## which is equal to the sum of the coefficients.
I originally thought the sum of the coefficients would be ##2^n-1## since the very first term ##1## is just a number and has no variable. But apparently...