If we want to evaluate $$f(x)=\frac{e^x-1-x}{x^2}$$ then we have to observe its large relative error as $x\to 0$.  My question is that how can we find a method so that we can compute $f(x)$ to full machine precision for all $|x|&lt;1$?  So far, I have been able to use a Taylor series to show that $$f(x)=\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}+\ldots$$ And now I suspect that we have to find the remainder, right? How would I do this?
124228
When evaluating $f(x)=\left(\dfrac{e^x-1-x}{x^2}\right)$, we observe large relative error as $x \rightarrow 0$.   Find a method to evaluate $f(x)$ to full machine precision for all $|x|&lt;1$.
96735
At small values close to $x=1$, you can use taylor expansion for $\ln x$:  $$ \ln x = (x-1) - \frac{1}{2}(x-1)^2 + ....$$  Is there any valid expansion or approximation for large values (or at infinity)?
77397
Given that $$f(1)=2005$$ and $$f(1)+f(2)+...f(n) = n^{2}f(n)$$ for all $n&gt;1$. Determine the value of $f(2004)$.  My progress:  I first substituted $n-1$ into the equation to get $f(1)+f(2)+...+f(n-1) = (n-1)^{2}f(n-1)$ and then I subtracted it from the first equation to get $$f(n) = n^{2}f(n)-(n-1)^{2}f(n-1)$$. I got a little stuck so I checked the answers for a hint and it said that: $$f(n) = \frac{(n-1)^{2}}{n^{2}-1}f(n-1)$$. How is this so? I don't understand the last part.
35132
I just wanted to directly calculate the value of the number $2^{3.1}$ as I was wondering how a computer would do it. I've done some higher mathematics, but I'm very unsure of what I would do to solve this algorithmically, without a simple trial and error.  I noted that   $$ 2^{3.1} = 2^{3} \times 2^{0.1} $$  So I've simplified the problem to an "integer part" (which is easy enough) :  $2^3 = 2\times 2\times 2$, but I'm still very confused about the "decimal part". I also know that :  $$ 2^{0.1} = e^{0.1\log{2}} $$  But that still presents a similar problem, because you'd need to calculate another non-integer exponent for the natural exponential. As far as I can see, the only way to do this is to let:  $$2^{0.1}=a $$   And then trial and error with some brute force approach (adjusting my guess for a as I go). Even Newton's method didn't seem to give me anything meaningful. Does anybody have any idea how we could calculate this with some working algorithm?
103505
I was wondering if there are functions for which $$f'(x) &gt; f(x)$$ for all $x$. Only examples I could think of were $e^x - c$ and simply $- c$ in which $c &gt; 0$. Also, is there any significance in a function that is always less than its derivative?    Edit: Thank you very much for all the replies. It seems almost all functions that apply are exponential by nature...  Are there more examples like - 1/x?  Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration  is always greater than its velocity]
25130
Of course most languages have library functions for this, but suppose I want to do it myself.  Suppose that the float is given like in a C or Java program (except for the 'f' or 'd' suffix), for example "4.2e1", ".42e2" or simply "42". In general, we have the "integer part" before the decimal point, the "fractional part" after the decimal point, and the "exponent". All three are integers.  It is easy to find and process the individual digits, but how do you compose them into a value of type float or double without losing precision?  I'm thinking of multiplying the integer part with 10^n, where n is the number of digits in the fractional part, and then adding the fractional part to the integer part and subtracting n from the exponent. This effectively turns 4.2e1 into 42e0, for example. Then I could use the pow function to compute 10^exponent and multiply the result with the new integer part. The question is, does this method guarantee maximum precision throughout?  Any thoughts on this?
29195
I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is $\displaystyle\frac{x}{\frac{1}{2}}$ is $\displaystyle\frac{2x}{1}$?
112543
I am looking for the numerical approximation of error function, which must be efficient and accurate. Thanks in advance $$\mathrm{erf}(z)=\frac2{\sqrt\pi}\int_0^z  e^{-t^2} \,\mathrm dt$$
54682
If the function $f: \mathbb R \to \mathbb R$ is $n+1$ times differentiable at $x_0$ and $f^{(n+1)}(x_0) \neq 0$, then a form of the remainder in Taylor's Formula is supposedly   $$r_n(x_0;x) = \frac{f^{(n)}\big(x_0 +\theta(x-x_0)\big)}{n!}(x-x_0)^n,$$ where $ 0&lt; \theta &lt; 1$ and $\theta = \theta(x)$ approaches $\frac{1}{n+1}$ as $x$ approaches $x_0.$  How would one derive this?  This seems quite strange to me.   For example, taking $f(x) =e^x$, the normal Taylor expansion for $n=2$ at $0$ would have the form  (Lagrange Remainder): $$e^x = 1 + x + \frac12x^2 +\frac{1}{3!}e^{\xi}x^3,$$ but in this case it is   $$e^x = 1 + x + \frac12x^2 +\frac{1}{2!}e^{\theta(x)}x^2$$.   The properties of $\theta$ do not make much sense. Am I missing something?
33255
Let's say I want to use Maclaurin series to get the series expansion $S(x)$ for $f(x) = e^x$ where $S(x) = c_0x^0 + c_1x^1 + c_2x^2 + c_3x^3 + ...$  $f(0) = S(0) = c_0 = e^0 = 1$ so that's fine.  $f'(0) = S'(0) = c_1 = e^0 = 1$ again, fine.  $f''(0) = S''(0) = 2c_2 = e^0 = 1$ so $c_2 = 1/2$, fine.  $f'''(0) = S'''(0) = 6c_3 = e^0 = 1$ so $c_3 = 1/6$, fine.  $f''''(0) = S''''(0) = 24c_4 = e^0 = 1$ so $c_4 = 1/24$, fine.  And so on, so we conclude that $c_k = \frac{1}{k!}$ so then:  $$S(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$  But now what I don't understand is what allows us to go "Furthermore, $S(x) = f(x)$ for all $x$"! In fact what makes it valid for us to plug any other number other than $x=0$ into this? I see this equation used as a straight-up equivalent to $e^x$ even though we used $x=0$ and nothing else.  So I thought I would try it with a neighborhood of $1$ instead to see what happens:  $S(x) = c_0(x-1)^0 + c_1(x-1)^1 + c_2(x-1)^2 + c_3(x-1)^3 + ...$  $f(1) = S(1) = c_0 = e^1 = e$ so that's fine.  $f'(1) = S'(1) = c_1 = e^1 = e$ again, fine.  $f''(1) = S''(1) = 2c_2 = e^1 = e$ so $c_2 = e/2$, fine.  $f'''(1) = S'''(1) = 6c_3 = e^1 = e$ so $c_3 = e/6$, fine.  $f''''(1) = S''''(1) = 24c_4 = e^1 = e$ so $c_4 = e/24$, fine.  Looks pretty similar:  $$S(x) = \sum_{k=0}^{\infty} \frac{e(x-1)^k}{k!}$$  and, in fact, for a neighborhood of $x=a$:  $$S(x) = \sum_{k=0}^{\infty} \frac{e^a(x-a)^k}{k!}$$  Now we have $e$ inside the function expansion for $e^x$ which seems a little circular?  I guess my question is this: We created the function $S$ to have the same-valued $n$th derivatives as $f(x)$ but only at $x=0$ or $x=1$, or $x=a$, etc, but the series representations look different depending on which neighborhood I pick, and in some cases the expansions seem to include the very number we're trying to describe.  How do we know which is "right" or that it is even "right" to use for all $x$? I know the usual response to this is that it's equivalent when $f(x)$ is analytic but that doesn't help me at all because it says the function is analytic when the Taylor series around $x_0$ converges to the function in a neighborhood around $x_0$ for all $x_0$ in the function's domain. But how do I know this?  Yet again feels circular... just working through these two examples makes me wonder, how do I know these series converge to the function itself? How do I know they're equivalent representations? Am I supposed to avoid the self-referencing?  I'm looking for some context behind how to make sense of these two Taylor series and how I am supposed to know that $e$ is analytic or that I can use $S(x)$ for any $x$ I want even if I only computed it for the neighborhood around $x=0$.
33862
Consider the following integral: $$\mathcal{I}=\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx.$$  I tried to evaluate $\mathcal{I}$ in a closed form (both manually and using Mathematica), but without success.   However, if  is provided with a numerical approximation $\,\mathcal{I}\approx 3.2694067500684...$, it returns a possible closed form:  $$\mathcal{I}\stackrel?=\frac\pi{2\,\ln^2 2}.$$ Further numeric caclulations show that this value is correct up to at least $10^3$ decimal digits. So, I conjecture that this is the exact value of $\mathcal{I}$.     Question: Is this conjecture correct?
94621
Find all functions $f:(0,\infty)\to (0,\infty)$  such that :   $$\frac{(f(w))^2 + (f(x))^2}{f(y^2) + f(z^2)} =\frac{w^2+x^2}{y^2+z^2} $$ for all positive real numbers $x,y,w,z$ satisfying $wx = yz$.   When I saw this question I was just blown away. Can anybody please explain the answer in detail?
87534
I need to calculate $\cos(\pi/20)$ with an error less than $10^{−5}$ of Taylor/Maclaurin series, What I did is like that:  $$|R_n(X)|=\left|\frac{(-1)^nC}{(n+1)!}\right|=\left|\frac{C}{(n+1)!}\right|_{0&lt;C&lt;\frac{\pi}{20}}&lt;\frac{\frac{\pi}{20}}{(n+1)!}=\frac{\pi}{20(n+1)!}&lt;10^{-5}$$  The statement is true when n=7, is my answer correct?
86009
Why is this function continuous over $x$? $$f^{n}(\xi(x))=\frac{-(n!\left(-f(x)+f(x_{0})+f(x_0)(x-x_0)+\ldots+\frac{f^{n-1}(x_0)*(x-x0)^{n-1}}{(n-1)!}\right)}{(x-x_0)^n}$$  I tried solving it via the Heine sequence criteria, but to no avail. I need someone help me with the explanation for this, please! Could the answer be, that since the left side of the equation is the final remainder of a Taylor series and a Taylor series converges to a certain function, then that remainder would ideally converge to zero?
90234
Can someone explain to me this error of the Desmos graphic calculator? Why ? For other values of $x$, the exact value of $f(x)$ is obtained.  Or am I wrong when I assure that $f(\sqrt2)=0$?
100760
So I've spent a few hours trying to find a counter example to this statement, and then gave up and spent additional few hours trying to prove it and Ive finally became frustrated with this problem. :) I seem to get stuck in the scenario where $f\left(\frac{1}{x}\right)$ &lt; 0, I'm just unable to find an epsilon that will satisfy all the conditions.. plus I'm not even sure if the statement is true or false anymore..
56293
Let f be the real-valued function defined on the interval $(-1,1)$ such that $e^{-x}f(x)=2+ \int_0^x\sqrt{t^4+1}dt$ for all $x\in(-1,1)$ and let $f^{-1}$ be the inverse function of $f$. Then find the value of $(f^{-1})'(2)$.  This is my step  1-$f(x)=y$  2-$f^{-1}(y)=x$   3-$f^{-1}(x)=y$  4-$f^{-1}(x)'=\frac{dy}{dx}$  5-  $y=f(x)=2e^{x}+e^{x} \int_0^x\sqrt{t^4+1}dt$  6-  $y'=2e^{x}+e^{x} \int_0^x\sqrt(t^4+1)dt+e^x\sqrt{x^4+1}$  I am lost after this step, please help me as I have jolted down the step.