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In Spivak's Calculus 3rd Edition, there is an exercise to prove the following: $$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$ I can't seem to get the answer. Either I've g...
- You have everything right except the last line. Maybe it is easier to do in this order: $$(x−y)\left(x^{n−1}+x^{n−2}y+\cdots+xy^{n−2}+y^{n−1}\right...
- I think it would be easier for you to recall $$\left(1+x+x^2+\cdots+x^{n-1}\right)(x-1) = x^n-1$$ and put $x=\dfrac{b}{a}$ $$\eqalign{ & \left( {...
- Here is the inductive step, presented more conceptually $$\rm\frac{x^{n+1}-y^{n+1}}{x-y}\: =\ x^n\: +\ y\ \frac{x^n-y^n}{x-y}$$ So, intuitively, pr...
- Your method is sound, you just made a sort of arithmetic mistake. When cancelling or otherwise combining two sequences, try explicitly lining thing...
- The $x^2 y^{n-2}$ term from $x \cdot x y^{n-2}$ is cancelled by the term from $(-y) \cdot x^2 y^{n-3}$. Similarly, the $(-y) \cdot x^{n-2} y$ is...
- Since powers of x and y is always greater than or equal to zero, You can prove it by mathematical induction.
- Sorry but I don't speak English well but I will try to write as correctly as possible. Your demonstration is correct, but your idea of cancellation...
- Since $x^{n-1} + x^{n-2} y + \dots + x y^{n-2} + y^{n-1}$ is a geometric series with $n$ terms and a common factor of $y/x$ , it equals $$ \frac{x...
- Using Induction Hypothesis Let n =k Then , x^k -y^k = (x-y)(x^(k-1) +x^(k-2) y ....+y^k-1) Let n= (k-1) Then by Induction Hypothesis, (x-y){x(x^(k-...
A quick look at the /x/ sound. This is to compliment the Jolly Phonics programme used by many schools around the world. Check out the full list of phonics in the Jolly Phonics order! •...
- 2 min
- 61.4K
- Mr. Andrew's Easy ESL
Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits. About Khan Academy: Khan Academy offers practice exercises, instructional videos...
- 7 min
- 455.4K
- Khan Academy
Jan 7, 2011 · Integral of x^n e^(-x)Instructor: Christine BreinerView the complete course: http://ocw.mit.edu/18-01SCF10License: Creative Commons BY-NC-SAMore information ...
- 11 min
- 86.6K
- MIT OpenCourseWare
Oct 27, 2018 · x^n/e^xの極限. より一般に、任意の正の整数 $n$ に対して 公式2： $\displaystyle\lim_{x\to\infty}\dfrac{x^n}{e^x}=0$ が成立します。 公式1と同様に証明することができます。
We dive into proving the formula for the derivative of x^n by skillfully applying the binomial theorem. Together, we expand (x + Δx)^n, simplify the expression, and take the limit as Δx approaches zero to reveal the power rule for derivatives. Created by Sal Khan.
Find a value of n such that y=x^ne^x satisfies the equation xy'=(x-10)· (Use symbolic notation and fractions where needed.)
