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The equation calculator allows you to take a simple or complex equation and solve by best method possible. Step 2: Click the blue arrow to submit and see the result! The equation solver allows you to enter your problem and solve the equation to see the result. Solve in one variable or many.
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-...
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...
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Solution. Verified by Toppr. We have , yx =ey−x. Taking log both side. ⇒ logyx =logey−x. ⇒ xlogy =y−x.......(1) Differentiating w.r.t x, we get. 1.logy+x 1 ydy dx = dy dx−1. ⇒ dy dx = logy+1 1− x y. = y(logy+1) y− y 1+logy [Using (1)] = (1+logy)2 logy. Was this answer helpful? 86. Similar Questions. Q 1.
In this definition \(e^x\) is the usual exponential function for a real variable \(x\). It is easy to see that all the usual rules of exponents hold: \(e^0 = 1\)
Recall: conditional probability distributions. It all starts with the de nition of conditional probability: P(AjB) = P(AB)=P(B). If X and Y are jointly discrete random variables, we can use this to de ne a probability mass function for X given Y = y.
ディズニーの名曲をたっぷり楽しめる、初級者向けやさしいピアノ曲集が初登場! 「アナと雪の女王」「アラジン」「リトル・マーメイド」などを筆頭とした数々の名作を彩る主題歌・劇中歌をメインに、全36曲の掲載です。. すべての音符に音名カナつきな ...
