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5

I finally figured it out due to a hint by Sameer Kailasa. $$\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}} \ dx$$ $$u=\frac{\pi}{2}-x \implies du=-dx$$ $$= -\int_{\frac{\pi}{2}}^{0}{\frac{1}{1+\cot^\sqrt{2}(x)}} \ dx = \int_{0}^{\frac{\pi}{2}}{\frac{\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1}} \ dx $$ ...


5

You've shown that if $x$ is odd, then $5x-7$ is even, but that is not what you were supposed to show. You were supposed to show that if $5x-7$ is even, then $x$ is odd. To see the problem with your method, let's use your method to prove this very similar statement: Let $x\in \Bbb Z$. If $4x-2$ is even, then $x$ is odd. Your method now goes like ...


5

Base of induction Since $a_1=1$ and $a_2=2/3$, the inequality $a_{n+1}\le a_n$ is true for $n=1$. Inductive step If $a_{n+1}\le a_n$, then $a_{n+1}^2\le a_n^2$, since all $a_n$ are positive by definition. Hence $$a_{n+2} = \frac13\left(a_{n+1}^2+\frac1{n+1}\right)\le \frac13\left(a_n^2+\frac1n\right) = a_{n+1}$$ which establishes the inductive step. ...


4

Hint: $x_{n+1}$ is a subsequence of $x_n$, so they have the same limit as $n \rightarrow \infty$, provided the limit exists. You already know the sequence is monotone decreasing, so now you'll want to show that it is also bounded below. Once you have that, then a limit $L \neq \pm \infty$ exists by the monotone convergence theorem. At this point, you can ...


3

Consider $f(x) = x^2 + 1$. It is continuous and differentiable on $[-1, 1]$, and has no $0$ on this interval. Moreover, we have that $$(f(0) - f(-1))(f(1) - f(0)) = (1-2)(2-1) = -1 < 0$$ so this satisfies all the assumptions of the question with $a = -1, b=1$, and $c=0$. However, $f(x) \ne 0 $ for all $x\in [-1, 1]$, so that's an example of why your ...


3

It’s not the proof itself that will be used throughout the book: it’s the format of the proof, in which the justification for each equality is given in curly braces immediately after the equals sign. For instance, the third step, $$\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha=2\sin\alpha\cos^2\alpha+\cos2\alpha\sin\alpha\;,$$ is justified by the reason ...


3

$ab^{-1}x=b^{-1}ax=b^{-1}xa=(x^{-1}b)^{-1}a=(bx^{-1})^{-1}a=xb^{-1}a=xab^{-1}$


2

After a quick glance of the linked doc: It seems that the author wants to teach dynamic programming. The approach involves a recursive solution which is then likely to be optimised with memoisation techniques (reuse of once calculated results). Induction is a natural choice for proving a recursive method, because of its related structure (inductive step = ...


2

You want a set $B$ that you can define from $A$ that is big enough so that if $\bigcup x=A$, then $x\in B$. Suppose that $\bigcup x\subseteq A$. If $y\in x$, then every element of $y$ is an element of $A$, so $y\subseteq A$, and $y\in\wp(A)$. This means that $x\subseteq\wp(A)$, and therefore $x\in\wp(\wp(A))$. Thus, you can take $B=\wp(\wp(A))$: it contains ...


2

It may be helpful to notice that if function $f$ is lipschitz and $f(0)=0$ then $f\circ u$ is sobolev with $\partial_i (f\circ u)(x)=f'(u)\partial_i u$. Hence, you could use $$f(x):= \begin{cases} x & |x|<k\\ k & x>k\\ -k & x<-k \end{cases} $$ as your truncation function. Of course $f$ satisfies all conditions I wrote above and of ...


2

The $\sqrt{2}$ is a complete red herring. In fact consider $$I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)}\,dx$$ where $\alpha$ is any nonnegative real number. Then we have \begin{align} I&=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)}\,dx \\&=\int_0^{\frac{\pi}{2}}\frac{1}{1+\frac{\sin^{\alpha}(x)}{\cos^\alpha(x)}}\,dx ...


2

The quadratic formula is all you need to prove this. If you have a quadratic polynomial: $$ ax^2+bx+c=0$$ Then, to find the zeroes, you rearrange the above expression into the quadratic formula: $$ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ From here, you can tell the discriminant, $\Delta =b^2-4ac$ and from $\Delta $ you can tell how many zeroes (where the ...


1

The discriminant of the quadratic $\;y=ax^2+2bx+c\;$ is $$\Delta=4b^2-4ac$$ The above parabola is "hoovering"above the $\;x$- axis (including being tangent to it) iff it has at most one real root (why?), and this happens iff is discriminant is non-positive. End the argument


1

Hint: the vertex $x = -\dfrac{b}{a} \Rightarrow f_{\text{min}} = f\left(-\frac{b}{a}\right) = \cdots$


1

I'm going to use Einstein summation notation since it appears that that is what you are doing. Suppose $$b^i\phi(\overline{e}_i)=\phi(b^i\overline{e}_i)=0$$ This means that there exist $a^i$ such that $$b^i\overline{e}_i=a^ie_i$$ based on the fact that the $e_i$ span the kernel of $\phi$, and $b^i\overline{e}_i$ is in that kernel. It follows that ...


1

Let's proceed by induction on $m+n$. If $m+n=0$, then $m=n=0$ and $2^0\cdot2^0=1=2^{0+0}$. If $m+n\ge1$, then WLOG $n\ge1$ and $$2^m\cdot2^n=2^m\cdot2^{n-1}\cdot2\overset{\text{IH}}=2^{m+n-1}\cdot2=2^{m+n}.$$


1

But then you are not proving the original implication: you are proving its converse.


1

You have shown that $5x-7$ is even when $x$ is odd, but this does not show that it is only even when $x$ is odd - it could also be even when $x$ is even (as for example $6x-8$ would be). Your "proof" would go through for $6x-8$ but it would not be true that if $6x-8$ is even then $x$ is odd.


1

They made the substitution $(a_n + c_n )^2 + (b_n + d_n )^2 = -2a_nb_n - 2b_nc_n - 2c_nd_n - 2d_na_n$ in the equality: $$(a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2) \\= 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) - 2a_nb_n - 2b_nc_n - 2c_nd_n - 2d_na_n$$ to write it as: $$a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 = 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) + (a_n ...


1

Pick two distinct points $x,y∈X/A$. If neither point is $A$, then $x,y∈$$X$ $\backslash$ $A$. With the given information in the problem, particularly that $A⊂X$ is closed, we may deduce that $X$ $\backslash$ $A$ is open in $X$. Then there exist open neighborhoods $U_A$ of $x$ and $V_A$ of $y$ such that $U_A$ and $V_A$ are disjoint from the ...


1

Euler's theorem says $x^{\phi(m)} \equiv 1 \pmod m$, which implies $$x^{\phi(m)+1} = x^{\phi(m)}x \equiv 1x \equiv x \pmod m.$$ Now, $p$ divides $m$, and $m$ divides $x^{\phi(m)+1}-x$, so by transitivity of divisibility, $p$ divides $x^{\phi(m)+1} - x$, which is to say $x^{\phi(m)+1}\equiv x \pmod p$. Update: Euler's theorem can only be applied if $x$ and ...


1

I'm going to deal with the first statement, that $CX$ is contractible. Consider the map $H: X \times{[0,1]} \times{[0,1]}$ $→$ $X \times{[0,1]}$ defined by $H((x,t),s) = (x(1-s)t)$. Then $H$ is continuous and $H((x,0,s) = (x,0)$ for all $x∈X$ and $s∈[0,1]$. Here you show that, whenever two points get identified, namely when they are of the form ...



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