# Tag Info

## New answers tagged wolfram-alpha

1

It's interpreting your question as "$x=\frac{{d}x}{dx}$", and $\frac{{d}x}{dx}$ is 1.

1

Wolfram Alpha interprets it as $x = x'(x)$, i.e. $x = \dfrac{dx}{dx} = 1$ It's nothing about slope.

1

You could combine the integrand into one fraction and do $$\int\frac{-e^{-x}}{1-e^{-x}}dx=-\ln|1-e^{-x}|+c$$ since this is of the form $$\int\frac{f'(x)}{f(x)}dx$$

1

However Wolfram|Alpha gives the result x−ln(1−ex), which is imaginary for x≥0 and doesn't the integrand when differentiated. What is going on? Your answer equals $-\log(e^x - 1) + \log(e^x)$ which is Wolfram's form up to some possible weirdness about adding $\log(-1)$ to the constant of integration. The weirdness disappears when differentiating, and ...

7

$$\displaystyle I = \int \left(1-\frac{1}{1-e^{-x}}\right)dx = -\int\frac{e^{-x}}{1-e^{-x}}dx\;,$$ Now let $$(1-e^{-x}) = t\;,$$ Then $\displaystyle (+e^{-x})dx = dt$ So $$\displaystyle I = -\int\frac{1}{t}dt = \ln\left|1-e^{-x}\right| = -\left[\ln\left(\frac{e^{x}-1}{e^x}\right)\right] = -\ln\left|e^x-1\right|+\ln(e^x)$$ So we get \displaystyle I = ...

0

Try asking it to do it numerically more explicitly (putting in a decimal point makes even mathematica assume you're using approximate numeric values, and so it will use approximate numerical methods) and you'll see that the value you get is well within the default tolerance value of the WA engine. Curiously, you'll note the output it gives for "Result" and ...

1

Try something more like this: q1 /. {ToExpression["q" <> "1"] -> 2} The thing your bit of code is doing is trying to find the string q1 instead of the variable with the name q1. Clarification: When you just type the letters q1 in Mathematica, it treats the thing like an expression or a single variable. That is, you can assign it a value such ...

1

Is there a sure way to get values of roots of quintic or higher degree polynomials without brute forcing $($especially non integer roots$)$ ? Some quintics $($both reducible and irreducible$)$ are solvable.

3

There exists no general formula for quintic polynomials from using addition, subtraction, multiplication, division, and $n$th roots. But numerically, we are extremely good at finding roots of polynomials very quickly. If you have a root within a known interval (like you might get by simply evaluating it at a million points and graphing it), then naive ...

0

The reason for Wolfram's astonishingly low term count for the OP example, is that the example given is a special case. Because the initial sin/cos arguments are small integers, the set of unique argument sums as computed in the product-to-sum conversions is a very small set--- just 59 unique arguments, to be exact. So for the OP example, the answer is ...

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