Tag Info

New answers tagged

1

You are correct: as long as both series converge (that is, as long as $\lvert x\rvert\leq\min\{R,S\}$, so that you are inside both radii of convergence), you have $$ \sum_{n=n_0}^{\infty}(a_n+b_n)x^n=\sum_{n=n_0}^{\infty}a_nx^n+\sum_{n=n_0}^{\infty}b_nx^n. $$ But, this doesn't mean that $\min\{R,S\}$ is the best that we can do! For an example of that, think ...


0

Denote a primitive of $f$ by $F$ and add ${1\over2}\bigl(F(a)-F(b)\bigr)$ to both series. Then the first series becomes $${1\over2}\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^kF^{(k)}(b)\bigr){(b-a)^k\over k!}={1\over2}\bigl(F(b)-F(a)\bigr)\ ,$$ and similarly the second series becomes $$-{1\over2}\bigl(F(a)-F(b)\bigr)+\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^k ...


2

Alright, here's another explanation. The idea is that when you're dealing with an analytic function, if you know $f(a), f'(a), f''(a), ...$, then you already have all the information about $f$. If you also happen to know $f(b), f'(b), f''(b), ...$, well that information is totally redundant. Since $$f(x) = ...


2

This sum needs to be from $n=2$ to $\infty$, because otherwise the first 2 terms aren't defined. Assuming this: If we use the Ratio Test, we get $$\lim_{n\to\infty} \left|\frac{(2x-3)^{n+1}}{(n+1)\ln(n+1)}\cdot \frac{n\ln(n)}{(2x-3)^n}\right| = |2x-3| \lim_{n\to\infty} \left|\frac{n\ln(n)}{(n+1)\ln(n+1)}\right| = |2x-3|<1$$ which leads us to conclude ...


0

Using ratio test: $\displaystyle R = \lim_{n \to \infty} |\dfrac{a_{n+1}}{a_n}| = \lim_{n \to \infty} \left(|2x - 3|\times \dfrac{nlnn}{(n+1)ln(n+1)}\right) = |2x - 3|$. So the series converges if $R < 1 \iff |2x - 3| < 1 \iff -1 < 2x - 3 < 1 \iff 2 < 2x < 4 \iff 1 < x < 2$. So the interval of convergence is: $(1, 2)$ Note: please ...


1

If $f(z) = g(z)$ for $z$ in some set of complex numbers, you can substitute any expression $w$ for $z$, and the equation $f(w) = g(w)$ will still be true as long as $w$ is in that same set. There's nothing special about series here.


7

The identity you give is quite pretty! Yes they are equal. Let $$\phi(t) = \sum_{n \ge 0}\frac{(b-a)^{n+1}}{(n+1)!} \left( f^{(n)}(a) \cdot t^{n+1} + f^{(n)}(b) \cdot (-1)^n (1-t)^{n+1}\right).$$ The first series in question is $\frac12(\phi(1)+\phi(0))$ and the second is $\phi(\frac12)$. The "why" is that these are equal because, in fact, $\phi$ is ...


2

Since $a_n = a^n + b^n$, then $$\sum_{n=1}^\infty a_n x^n =\sum_{n=1}^\infty (ax)^n + \sum_{n=1}^\infty (bx)^n .$$ For this to be convergent, both series must be convergent, but these are regular geometric progressions, so the conditions for their convergence are that $|ax|<1$, and $|bx|<1$. So we must have simultaneously $$|x|< \frac{1}{|a|}$$ ...


1

Your series converges when $$\lim_{n \to \infty}\left\vert\sqrt[n]{\left \vert a^n+b^n\right \vert} x \right\vert < 1$$ Hence, the radius of convergence is $$R = \dfrac1{\lim_{n \to \infty} \sqrt[n]{\left \vert a^n+b^n\right \vert}} = \dfrac1{\max(\vert a \vert, \vert b \vert)}$$


3

By Cauchy's-Hadamard formula, with $\;R:=$ convergence radius, with the usual conventions when $\;R=0\,,\,\infty\;$ , we get: $$\frac1R=\lim_{n\to\infty}\sup\sqrt[n]{|a^n+b^n|}$$ and assuming $\;|a|\ge|b|\;$ , we get $$\sqrt[n]{|a^n+b^n|}=|a|\sqrt[n]{1+\left(\frac{|b|}{|a|}\right)^n}\xrightarrow[n\to\infty]{}|a|$$


2

Prove the series converges (absolutely) for $|z|<1$, then the Cauchy product formula says $$\left(\sum_{n=0}^\infty\frac1{4^n}\binom{2n}nz^n\right)^2=\sum_{n=0}^\infty\sum_{k=0}^n\frac1{4^n}\binom{2(n-k)}{n-k}\binom{2k}kz^n$$ Now prove $$\sum_{k=0}^n\binom{2(n-k)}{n-k}\binom{2k}k=4^n$$ So the square of the sum is $$\sum_{n=0}^\infty z^n=\frac1{1-z}$$ ...


0

You can also refer to Abel's continuity theorem. The function $S(x)$ is continuous in its convergence set; in particular if the series converges at $R$ then $S(x)$ is continuous at $R-0$.


0

Let $S_n(x)=\sum\limits_{k=0}^n a_kx^k$ and $L=\sum\limits_{k=0}^\infty a_kR^k$. From $a_k\ge0$, for every $n$ and $0<x<R$ we have $$ S_n(x) \le S(x) \le L. $$ From fixed $n$ and $x\to R-0$ you get $$ S_n(R) =\lim_{x\to R-0} S_n(x) \le \liminf_{x\to R-0} S(x) \le \limsup_{x\to R-0} S(x) \le L. $$ Then, from $n\to\infty$, $$ L=\lim_{N\to\infty} S_n(R) ...


4

Note that $xy$ with $x+y$ fixed is maximized when $x=y$. Thus $$n!=(1\cdot n)(2\cdot(n-1))\cdots(\lfloor (n+1)/2\rfloor\cdot\lceil (n+1)/2\rceil)<\left(\frac{n+1}{2}\right)^n=\frac{(n+1)^n}{2^n}$$ and since $$\left(\frac{n+1}{n}\right)^n=\left(1+\frac{1}{n}\right)^n<e$$ we have ...


3

Consider the series $$f(x) = \sum_{n=0}^{\infty} (-1)^n x^{4n} = \dfrac1{1+x^4}$$ then $$\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{4n+1}}{4n+1} = \int_0^x f(t)dt = \int_0^x \dfrac{dt}{1+t^4} = \int_0^x \dfrac{dt}{(t^2+1)^2-(\sqrt2 t)^2}$$ and some brute force partial fraction to your rescue. Proceed along similar lines for the second problem.


6

Well, to attack the first sum, if we call it $f(x)$, then $$f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{4 n} = \frac1{1+x^4}$$ so that $$f(x) = \int_0^x \frac{dt}{1+t^4}$$ You can use partial fractions to deduce that $$\begin{align}f(x) &= \frac1{2 \sqrt{2}}\int_0^x dt \, \left [\frac{t+\sqrt{2}}{t^2+\sqrt{2} t+1}-\frac{t-\sqrt{2}}{t^2-\sqrt{2} t+1} ...


1

If $|z^2-n^2\pi^2| \ge n^2\pi^2$, then $$\left|\frac{1}{z^2-n^2\pi^2}\right| \le \left|\frac{1}{n^2\pi^2}\right|$$ Therefore, if $z$ is such that $|z^2-n_i^2\pi^2|<n_i^2\pi^2$ for some $\{n_i\} \subset \mathbb{N}$, then $$\sum_{n \not\in \{n_i\}} \left|\frac{1}{z^2-n^2\pi^2}\right| \le \sum_{n \not\in \{n_i\}} ...


0

In general, if $A(z) = \sum_{n \ge 0} a_n z^n$, you get: $$ z \frac{\mathrm{d}}{\mathrm{d} z} A(z) = \sum_{n \ge 0} n a_n z^n $$ So, using $\mathrm{D}$ for "derivative," and $p(n)$ a polynomial: $$ \sum_{n \ge 0} p(n) a_n z^n = p(z \mathrm{D}) A(z) $$ This gives a general method to get sums as yours. Just take care, e.g. $(z \mathrm{D})^2 \ne z^2 ...


0

You just need a sequence $M_{n}$ such that $$|\frac{1}{z^{2} - n^{2}\pi^{2}}| \le M_{n}$$ and se you can take $M_{n} = \frac{1}{n^{2}\pi^{2}}.$ EDIT: Wait, what I wrote is just nonsense. The bound goes the other way in my example. Please ignore it.


0

There are calculus books that say that 0^0 is undefined. The reason for this is tradition; long ago, before continuous functions were well understood, Gauss placed 0^0 in a table of "indeterminate forms", a concept that becomes obsolete once you know the relation between limits and continuous functions. There are numerous places in mathematics where 0^0 is ...


0

When I was young (that is to say a loooong time ago), one of the series the professor asked us to always remember (because of its extreme simplicity) is precisely $$\ln\left(\frac{1+x}{1-x} \right)=2\sum_{n=0}^{\infty}\frac {x^{2n+1}}{2n+1}$$ Thank you for remembering me my youth


1

It is very helpful to keep the following series in mind with problems like this: $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n$$ Integrating, we find $$\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}$$ You can now readily use the fact that $$\ln\frac{7+x}{7-x}=\ln\left(1+\frac{x}{7}\right)-\ln\left(1-\frac{x}{7}\right)$$ The resulting series will ...


2

We will assume that $7-x$ and $7+x$ are positive. Note that $$\ln\left(\frac{7+x}{7-x} \right)=\ln\left(\frac{1+x/7}{1-x/7} \right)=\ln(1+x/7)-\ln(1-x/7).$$ You know the power series expansion of $\ln(1+t)$. Substitute $t=x/7$, $t=-x/7$ and subtract. Remark: The trick goes back at least to Euler. Instead of using $7$, use $3/2$, and let $x=1/2$. Note ...


0

From the binomial theorem, $$(1+x)^{-1}=\sum_{n=0}^{\infty}(-1)^nx^n$$ Letting $x=z^2$, $$(1+z^2)^{-1}=\sum_{n=0}^{\infty}(-1)^nz^{2n}$$ This gives the series expansion around $z=0$. To expand around $z=a$, you can try this: As Mayank Pandey pointed out, $$\frac{1}{z^2 + 1}=\frac{1}{2i}\left(\frac{1}{z+i}-\frac{1}{z-i}\right)$$ For the first term, ...


0

$$\frac{1}{z^2 + 1} = \frac{1/2i}{z - i} - \frac{1/2i}{z + i}$$ In general, you can do this to any poynomial.


0

Hint: $\cos x = O(1)$ and no decreasing bound like $O(1/x)$ can be found for it because $\cos x = 1$ infinitely often as $x$ increases. Actually the question should be asking for $\Theta(f(x))$ for your function $f$, rather than big-O. In terms of big-O, both A and B are correct answers but there's only one correct answer in terms of $\Theta$.


0

Hint: $$0\le\left|\frac{(x+2)\cos^2x}{x^2}\right|\le\left|\frac{x+2}{x^2}\right|\xrightarrow[x\to\pm\infty]{}\;\ldots$$


2

$1$ is an eigenvalue of multiplicity $1$ of a stochastic matrix as long as your graph edges give a path between every pair of vertices $(v,w)$,and the matrix summation will not converge if $1$ is an eigenvalue. Actually, the matrix summation wouldn't converge for any stochastic matrix. You can see this by noting that every power of a stochastic matrix is ...


0

We have $$ x^2\sin(x^4)=\sum_{k=0}^\infty \frac{(-1)^kx^{8k+6}}{(2k+1)!}, $$ and hence $$ \int_0^1 x^2\sin(x^4)\,dx=\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!(8k+7)}. $$ Also, $$ \sin x=x-\frac{x^3}{3!}+\cdots+\frac{(-1)^nx^{2n+1}}{(2n+1)!}\sin^{(2n+1)}(\xi), $$ for some $\xi\in (0,x)$. Hence $$ \left|\,\sin ...


2

Power series with a positive radius of convergence are equal at points other than the center only if all their coefficients match. So $a_n = c a_n b_n$ for all $n$. That means for every nonzero value of $a_n$, you have $1 = cb_n$, so $b_n=1/c$. For values of $n$ for which $a_n=0$, the coefficient $b_n$ could be anything. For example: \begin{align} a ...


3

Recall that the power series for $e^x$ is $$ \sum_{n=0} ^\infty { \frac{x^n}{n!} } . $$ Thus, the power series for $e^{-x^2}$ is $$ \sum_{n=0} ^\infty { \frac{(-x^2)^n}{n!} } = \sum_{n=0} ^\infty { \frac{(-1)^n x^{2n}}{n!} } . $$ Integrating term by term yields $$ F(x) = \int F'(x) = \int \sum_{n=0} ^\infty { \frac{(-1)^n x^{2n}}{n!} } = \sum_{n=0} ...


1

The power series for $F'(X)$ is $$F'(x) = \sum_0^\infty \frac{(-x)^{2n}}{n!} = \sum_0^\infty \frac{(-1)^nx^{2n}}{n!}$$ We now integrate each side and get: $$ \begin{align} F(X) &= \sum_0^\infty \frac{(-1)^n (x)^{2n+1}}{(2n+1)(n!)} \end{align}$$


3

We have $$F'(x)=e^{-x^2}=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}$$ with radius of convergence $+\infty$ hence by integrating term by term we find: $$\int_0^xF'(t)dt=F(x)-F(0)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)n!}$$


0

It looks like this is a notation defined by $$(a_0,a_1,a_2,\ldots) \leftrightarrow a_0 +a_1x+a_2x^2+\cdots$$ or, more briefly, $$(a_n)_{n\in I} \leftrightarrow \sum_{n\in I}a_nx^n$$ where $I$ is the set of nonnegative integers. So your first example is a single term, the second example is a geometric series, and the third term is the derivative of a ...


0

Cauchy's product formula is easy to prove, if you look closely at what formulas with sums mean. Take $A(z) = \sum_{r \ge 0} a_r z^r$ and $B(z) = \sum_{s \ge 0} b_s z^s$. Multiply: \begin{align} A(z) \cdot B(z) &= \left( \sum_{r \ge 0} a_r z^r \right) \cdot \left( \sum_{s \ge 0} b_s z^s \right) \\ &= \sum_{r \ge 0} \sum_{s \ge 0} a_r b_s z^{r ...


5

Probably you know that if $|S_n(R)|<M$ and $a_n>0$, the series $\sum_{n=0}^\infty a_nR^n$ is convergent. If $R=1$, This is Theorem 8.2 of Baby Rudin and for arbitrary $R$ you can see following Theorem that you only need to define $g(x)=S(Rx)-S(R)$. Complex Version of Abel's Limit Theorem. Let $f(z)=\sum_{n=0}^\infty a_nz^n$ have radius of convergence ...


1

$$ \ln(x) = \int\frac{1}{x}dx\\ \frac{1}{x} = \frac{1}{6 - (6 - x)} = \frac{1}{6}\frac{1}{1 - \frac{-(x - 6)}{6}} = \frac{1}{6}\sum_0^\infty (-1)^n\frac{(x - 6)^n}{6^n} $$ This is a geometric series and thus converges when the common ratio: $\frac{x - 6}{6}$ is smaller than one: $$ \left|\frac{x - 6}{6}\right| < 1 \\ |x - 6| < 6 \\ 0 < x < 12 ...


1

If you have two power series $A(x)=\sum_{n\ge 0}A_nx^n$ and $B(x)=\sum_{n\ge 0}B_nx^n$, then their (formal) product is given by the power series $C(x)=A(x)B(x)=\sum_{n\ge 0}C_nx^n$, whereby $C_n=\sum_{k=0}^n A_kB_{n-k}$ This is sometimes called Cauchy's product formula but you may want to verify this yourself. Hence, for the product of $\exp(ax)$ and ...


0

I think the explanation this question is asking for is $$\sum_{n=1}^\infty n c_n x^n = \big(\sum_{n=1}^\infty n c_n x^n\big) + 0 c_0 x^0 = \sum_{n = 0}^\infty n c_n x^n.$$ In other words, simply add the zero term, it is the constant term which when you took the derivative "vanished" and is the reason for the shift in index in the first place. Think about ...


0

I have two simple notes The summation sign starts at $n=1$ i.e. it doesn't contain constant term(I mean of the form $cx^0$) $2a_2\cdot x^0=0\cdot x^0$ Now if we have the following equal polynomials $$a_0+a_1x+a_2x^2+\cdots=b_0+b_1x+b_2x^2+\cdots$$ then by equating coefficients of $x^n$ we get $$x^0:\ \ \ \ \ \ a_0=b_0 \\ x^1:\ \ \ \ \ \ a_1=b_1 \\ ...


0

Sometimes, for example, in physics, we only care about behaviour around a certain point. A not-so-good but common example might be $t<0$ in a plot versus time (which is always neglected in elementary physics). We might want to examine asymptotic behaviour (i.e. getting very far away, very far back in time, very dense, etc.). Other times we want to ...


1

If $(u_N)_N$, $(v_N)_N$ are both convergent sequences, then also the sequence $(u_N- v_N)_N$ converges and the limit is $$\lim_{N\rightarrow \infty} u_N - \lim_{N\rightarrow \infty} v_N = \lim_{N\rightarrow \infty} (u_N - v_N)$$ Now set $u_N=\sum_{n=1}^N a_n x^n$ and $v_N = \sum_{n=1}^N a_n y^n$. This justifies $(1)$. Also $(2)$ is true. If $f(x)$ and ...


0

You can compute $c_n$ directly: $c_n=\frac{f^{(n)}(0)}{n!}$ but it could be quite long and boring. Otherwise you can apply the shortcut suggested by NasuSama: $$ f(x)=\frac{9}{1+100x^2}= \sum_{n=0}^{+\infty}9(-100x^2)^n= \sum_{n=0}^{+\infty}9(-100)^nx^{2n} $$ from which you desume that $c_{2n+1}=0\;\;\forall n\in\mathbb N$ hence $c_1=c_3=c_5=0$. Then ...


1

HINT: Rewrite $f(x)$ as $$\dfrac{9}{1 + 100x^2} = 9 \cdot \dfrac{1}{1 - (-100x^2)}$$ Use the following identity to write $f(x)$ as the power series: $$\dfrac{1}{1 - g(x)} = 1 + g(x) + (g(x))^2 + (g(x))^3 + \cdots = \sum\limits_{n = 0}^{\infty} (g(x))^n$$ Now answer the problem, using the info above. It is easier to write out partial sum of the series ...


2

WLOG: $x<y$. As you probably know, for a continuously differentiable function, you have $$f(x) - f(y) = f'(\zeta)(x-y)$$ for some value $\zeta\in[x,y]$. This means, of course, that $|e^x - e^y| = |e^\zeta||x-y|$ for some $\zeta\in[x,y]$. Since this means that $e^\zeta \geq e^x$, this also means that $$|e^x - e^y| \geq e^x |x-y|,$$ and this means that ...


5

No, there is no such $C$. Assume that there is and take $y = 0$. Then $|e^x - 1| \le C|x|$ for all $x$ which would for example imply that $e^x/x^2 \to 0$ as $x \to \infty$ and that is false. On the other hand, if $x$ and $y$ are restricted to a compact set, then the mean value theorem implies the existence of such a $C$. In other words, $f$ is locally ...


2

Consider the summation $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \text{.}$$ You demonstrated that you already know that you're allowed to replace $x$ with $-x$ to obtain the summation $$e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} \text{.}$$ Intuitively you should be able to throw a k in there as well: $$e^{-kx} = \sum_{n=0}^{\infty} \frac{(-kx)^n}{n!} ...


3

You should use a different index of summation than variables used outside the scope of the summation. $$ e^{-kx}=\sum_{j=0}^\infty\frac{(-kx)^j}{j!} $$


0

For $|x|<1$ $p(x)$ is analytic. But also $p(x)=0$ for $|x|<\delta $. But then $p(x)$ must vanish for all $|x|<1$. Hence $b_{n}=0$.


1

Use the chain rule: $$\ln(1-3x)=\int\dfrac{dx}{1-3x}$$ Let $u=1-3x,du=-3dx$, thus giving $$\ln(1-3x)=\dfrac{-1}{3}\int\dfrac{du}{u}=\dfrac{-1}{3}\ln|1-3x|+C$$ Thus, $$\ln(1-3x)=\int\dfrac{dx}{1-3x}=\sum\dfrac{3^kx^k}{k}$$



Top 50 recent answers are included