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9

Hint: Consider $(a-b)^2+(b-c)^2+(c-a)^2$. Remark: Using the above hint, you can show that the desired inequality holds for all real numbers $a$, $b$, and $c$.


6

To prove a statement $p \rightarrow q$, we can instead prove its contrapositive $\lnot q \rightarrow \lnot p$, because the two forms of implication are equivalent: an implication is true if and only if its contrapositive is true: $$p\rightarrow q \equiv \lnot q \rightarrow \lnot p.$$ Let $p$ denote "$4$ divides $x^2$." $\lnot p: \;$ "$4$ does not divide ...


5

Suppose $\;\lim a_n=L\;$ exists, then $$(-1)^n=a_n-\frac1n\implies \lim_{n\to\infty}(-1)^n\stackrel{\text{Arith. of limits}}=\lim_{n\to\infty}a_n-\lim_{n\to\infty}\frac1n=L-0=L$$ But it is clear that $\;\{-1,1,-1,1,\;\ldots\}\;$ doesn't converge...


4

Hint $$0\leq (a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$$


4

Let $A_{nw}$ be the event that the "north-west" $2\times2$ block is red, and so on. By inclusion/exclusion the required probability is $$\eqalign{P(A_{nw}\cup A_{ne}\cup A_{se}\cup A_{sw}) &=P(A_{nw})+\cdots\cr &\qquad{}-P(A_{nw}\cap A_{ne})-\cdots\cr &\qquad{}+P(A_{nw}\cap A_{ne}\cap A_{se})+\cdots\cr &\qquad{}-P(A_{nw}\cap ...


4

$$\int_{x=a}^{x=b} f'(x) g(x)\, dx=g(x)f(x)|_{a}^{b}-\int_a^bg'(x)f(x)dx$$ (by integration of parts). This gives your answer. It suffices to prove the statement: if $f$ is integrable on $[a,b]$ then so is $f^2$, then $fg=\dfrac{1}{2}((f+g)^2−f^2−g^2)$is integrable. To show that $f^2$ is integrable use the following: $$\begin{align}U(f^2, P) - L(f^2,P) ...


4

You have some off-by-one errors, but you have the right idea. Note that: \begin{align*} p &= (x_1 + x_2 10 + \cdots + x_n 10^{n-1}) + (x_{n} 10^n + x_{n-1} 10^{n+1} + \cdots + x_1 10^{2n - 1}) \\ &= \sum_{k=1}^n x_k 10^{k-1} + \sum_{k=1}^n x_k 10^{2n-k} \\ &= \sum_{k=1}^n x_k10^{k-1}(1 + 10^{2n - 2k - 1}) \\ &\equiv \sum_{k=1}^n x_k10^{k-1}(1 ...


4

Your answer looks correct. In general, if you have something bounded it is not a norm. I'll explain what I mean: Notice that $\frac{d}{1 + d} \le 1$. Pick any non zero $x$ and consider $\alpha x$, $\alpha \in \mathbb{R}^+.$ Assume by contradiction that it is a norm, then $$\alpha \frac{d}{1+d}(x) = \frac{d}{1+d}(\alpha x) \le 1.$$ Send $\alpha \to \infty$ ...


3

Actually your reasoning solves the question! Note that, without loss of generality, you may order the numbers $a, b, c$ so that c is the greatest of them: $c \geq a , c\geq b$ You can do this because the inequality is symmetric in the variables $a, b, c$; i.e, if this wasn't true, have the greatest of the numbers and $c$ change places. Therefore your last ...


3

You know that $2k +3 \leq 2^k$. Add two to both sides: $2k + 3 + 2 \leq 2^k + 2$. Then you have that $2k + 3 + 2 = 2(k+1) + 3 \leq 2^k +2 < 2^{(k+1)}$. (Why is the last step true?)


2

First, I would recommend you to have a look at Velleman's How to Prove it (2006). In my opinion, this book sheds light proof methods and strategies in an instructive and illuminating way. It is easy to cope with and I'm sure it is exactly what you need until get the feeling of how informal proofs work. Your proof strategy is correct. (1) Whenever we want to ...


2

Using $$ \frac{1}{2k}-\frac{1}{2k+1} = \frac{1}{4k^2+2k} $$ write $M=m/2$ to get $$ T(n) = \sum_{m=2^{n+1}}^\infty \frac{(-1)^m}{m} = \sum_{M=2^n}^\infty \frac{1}{4M^2+2M} $$ from which it follows that $$ \int_{t=2^n}^\infty (2t+1)^{-2}~dt < \sum_{M=2^n}^\infty \frac{1}{(2M+1)^2} < T(n) < \sum_{M=2^n}^\infty \frac{1}{4M^2} < \int_{t=2^n-1}^\infty ...


2

Lemma: $gcd(x,y)=d$ iff $d$ can be expressed as a linear combination of $x$ and $y$ (in $\mathbb{Z}$) We have $gcd(a,b)=d$ $d=ap+bq$ for some $p,q\in \mathbb{Z}$ $d=ap-bp+bp+bq=(a-b)p+b(p+q)=(a-b)p+bq'$ where $q'=p+q\in \mathbb{Z}$ Thus, $gcd(a-b,b)=d$


2

HINT: $T=x\sum_{k=0}^{n}kx^{k-1}$ and now you can use this fact: $$\sum_{k=0}^{n}kx^{k-1}=(\sum_{k=0}^{n}x^{k})'=(\frac{1-x^{n+1}}{1-x})'$$


2

Your proof of reflexivity is fine. For symmetry, you have an error. A correct answer uses $(ab^{-1})^{-1}=ba^{-1}$. See if you can complete the proof. You also have an error in transitivity. Use the fact that $(ab^{-1})(bc^{-1})=ac^{-1}$.


2

$$f^{-1}(f(x))=x\Rightarrow \frac{d}{dx}f^{-1}(f(x))=1\Rightarrow (f^{-1})'\cdot f'(x)=1\Rightarrow (f^{-1})'=\frac{1}{f'(x)}$$ Differentiate both sides of the last expression wrt $x$ again to get: $$(f^{-1})''\cdot f'(x)=-\frac{f''(x)}{(f'(x))^2}\Rightarrow (f^{-1})''=-\frac{f''(x)}{(f'(x))^3}$$


2

You can do: $$\begin{align} f^{-1}(f(x)) &= x \\ (f^{-1})'(f(x)) \ f'(x) &= 1 \\ (f^{-1})''(f(x)) \ (f'(x))^2 + (f^{-1})'(f(x)) \ f''(x) &= 0 \\ (f^{-1})''(f(x)) &= -\frac{(f^{-1})'(f(x)) \ f''(x)}{(f'(x))^2} \\ (f^{-1})''(f(x)) &= -\frac{f''(x)}{(f'(x))^3}\end{align}$$ whenever everything makes sense and is well-defined.


2

You are on the right track, but you should take one more step. Once you get $$(f^{-1}(x))''=-\frac{f''(f^{-1}(x))·(f^{-1}(x))'}{(f'(f^{-1}(x))^{2}},$$ use the formula for the derivative of the inverse function once again $$(f^{-1}(x))'= \frac{1}{f'(f^{-1}(x))}$$ and after the substitution, get ...


2

You're proof seems mostly all right, though I find it hard to follow towards the end. If I were to write a proof along the same lines, I would write it as follows: Let $\{ y_{n} \}_{n=1}^{\infty}$ be a sequence in $X$. $X$ is totally bounded, so we can find finitely many $x_{i} \in X$ ($i = 1, \dots, n $) such that $X = B(x_{1}, \frac{1}{2}) \cup \dots ...


2

The subsequences $(a_{2n})$ and $(a_{2n+1})$ have different limits so the sequence $(a_n)$ is divergent.


2

If a sequence $\{ a_{n} \}_{n = 1}^{\infty}$ converges to a limit $L$, then every subsequence must converge (can you prove this part on your own?), and the subsequences must all converge to the same limit (can you prove this part on your own?). But it is easy to prove that the subsequence $\{ a_{2n} \}_{n = 1}^{\infty}$ of even indices converges to $1$ (can ...


2

Yo can prove that $a_n$ isn't Cauchy sequence. According the definition $a_n$ is a Cauchy sequence if: $$\forall \epsilon>0 \exists N \in \mathbb \forall n,m \geq N \; |a_n-a_m|<\epsilon$$ But for $\epsilon=\frac{1}{2}$ for all $N \in \mathbb{N}$ difference between $a_N$ and $a_{N+1}$ is: ...


2

What we need to show is that $$\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$$ exists and is equal to $$\lim_{x \to x_0} f'(x);$$ this would make $f$ differentiable and $f'$ continuous at $x_0$. But as you said, the thing $$\frac{f(x)-f(x_0)}{x-x_0}$$ is equal to $f'(c)$ for some $c$ that approaches $x_0$ as $x$ does.


2

Notice that by definition of the derivative $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ Therefore $$\lim_{x\to x_0}f'(x)=\lim_{x\to x_0}\{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\}$$ By the assumption of your problem the "inside" limit exists for all $x=x_0\pm\epsilon$ and every $\epsilon>0$. Also the problem states that $\lim_{x\to x_0}f'(x)$ exists ...


2

Let’s dispose of the technical point first. By straight algebra, $$k+\binom{k}2=k+\frac{k(k-1)}2=\frac{2k+k^2-k}2=\frac{k(k+1)}2=\binom{k+1}2\;.$$ Alternatively, $$k+\binom{k}2=\binom{k}1+\binom{k}2=\binom{k+1}2$$ by the Pascal’s triangle identity. You’re quite right that it’s not sufficient to consider splitting your $k+1$ pile into piles of size $1$ ...


2

I think your observation helps if you iterate it more. $$ \begin{align} S_m&=2-m\sum_{n=0}^\infty\frac{n}{2^n(2^n+m)}\\ &=2-m\sum_{n=0}^\infty \frac{n}{2^n}\left(\frac{1}{2^n}-\frac{m}{2^n(2^n+m)}\right)\\ &=2-m\sum_{n=0}^\infty \frac{n}{4^n}+m^2\sum_{n=0}^\infty\frac{n}{4^n(2^n+m)}\\ &=2-\frac49m+m^2\sum_{n=0}^\infty\frac{n}{4^n(2^n+m)}\\ ...


2

Using the fact that $$\sum_{k=0}^{n}k^5=\sum_{k=0}^{n}(k+1)^5-(n+1)^5$$ Using Binomial Expansion $$\sum_{k=0}^{n}k^5=\sum_{k=0}^{n}k^5+5\sum_{k=0}^{n}k^4+10\sum_{k=0}^{n}k^3+10\sum_{k=0}^{n}k^2+5\sum_{k=0}^{n}k+\sum_{k=0}^{n}(1)-(n+1)^5$$ After Rearranging $$5\sum_{k=0}^{n}k^4=(n+1)^5 ...


2

If $x$ were odd, then $x=2k+1$ for some integer $k$, in which case $\,x^2=\;\ldots\;\;?$


2

Both $\cos(x)$ and $-\sin(x)$ are convex functions on the interval $(\pi/2,\pi)$. Since in the endpoints of such interval se have $\cos(x)-\sin(x)=-1$, it follows that $\cos(x)-\sin(x)\leq -1$ all over $[\pi/2,\pi]$. The last inequality implies that $\sin(x)+\cos(x)$ is a decreasing function over $[\pi/2,\pi]$, so, for any $x\in(\pi/2,\pi]$, ...


1

In general, for 2 nonnegative integers $\overline{a_1\ldots a_n}$ and $\overline{b_1\ldots b_m}$, we have $$ \overline{a_1\ldots a_n}-\overline{b_1\ldots b_m}=(\sum a_i)-(\sum b_j)+9A $$ for some integer $A$. Thus, your premise gives you $9|4n$. What does this say about $n$?



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