# Limit of $\frac{1}{x^4}\int_{\sin x}^{x} \arctan t dt$

I am trying to find this limit,

$$\lim_{x \rightarrow 0} \frac{1}{x^4} \int_{\sin{x}}^{x} \arctan{t}dt$$

Using the fundamental theorem of calculus, part 1, $\arctan$ is a continuous function, so $$F(x):=\int_0^x \arctan{t}dt$$ and I can change the limit to $$\lim_{x \rightarrow 0} \frac{F(x)-F(\sin x)}{x^4}$$

I keep getting $+\infty$, but when I actually integrate $\arctan$ (integration by parts) and plot the function inside the limit, the graph tends to $-\infty$ as $x \rightarrow 0+$.

I tried using l'Hospital's rule, but the calculation gets tedious.

Can anyone give me hints?

EDIT

I kept thinking about the problem, and I thought of power series and solved it, returned to the site and found 3 great answers. Thank You!

• Numerical evaluation says the limit should be $1/6$. – Simply Beautiful Art Mar 29 '17 at 23:34
• @SimplyBeautifulArt How did you do it? What kind of method did you use for numerical evaluation? – zxcvber Mar 29 '17 at 23:36
• Calculators of course. – Simply Beautiful Art Mar 29 '17 at 23:37

Power series can help, if you know them.

We know that $$\arctan t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+1}}{2n+1},\qquad \lvert t\rvert<1.$$ Therefore an antiderivative for $\arctan t$ is $$F(t):=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+2}}{(2n+1)(2n+2)},\qquad \lvert t\rvert<1.$$ So, as $x\to0$, $$F(x)=\frac{x^2}{2}-\frac{x^4}{12}+O(x^6).$$ Now, recalling that $$\sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{6}+O(x^5)=x+O(x^3),$$ we see that as $x\to0$ (and therefore $\sin x\to0$), \begin{align*} F(\sin x)&=\frac{1}{2}\left(x-\frac{x^3}{6}+O(x^5)\right)^2-\frac{1}{12}\left(x+O(x^3)\right)^4+O[(x+O(x^3))^6]\\ &=\frac{1}{2}\left[x^2-\frac{x^4}{3}+O(x^6)\right]-\frac{1}{12}\left[x^4+O(x^6)\right]+O(x^6)\\ &=\frac{x^2}{2}-\frac{x^4}{4}+O(x^6) \end{align*} So, all told, our limit is $$\lim_{x\to0}\frac{F(x)-F(\sin x)}{x^4}=\lim_{x\to0}\frac{\,\frac{x^4}{6}\,}{x^4}=\frac{1}{6}$$

• I really like the way you used $O(x^n)$ notation here. Thanks! – zxcvber Mar 29 '17 at 23:56

Here's one way to do this. The Taylor expansion for $\arctan(x) = x - x^3/3 + x^5/5 + \cdots$. So then $$\int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12 + x^6/30) - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{12} - \frac{(\sin x)^6}{30}.$$

Using another Taylor expansion, $\sin x = x - x^3/3! + x^5/5! + \cdots$, so the plan is to pay attention only to those terms of degree up to $4$ in these expansions. Then (dropping all terms of degree greater than $4$), $$(\sin x)^2/2 = (x - x^3/3!)^2/2 = x^2/2 - x^4/6$$ and $$(\sin x)^4/12 = (x - x^3/3!)^4/4 = x^4/12.$$ Thus $$\int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12) - (x^2/2 - x^4/6) + x^4/12 = x^4/6.$$

So multiplying by $x^{-4}$ and taking the limit gives $1/6$.

As an aside, l'Hopital's rule would also work.

Since $F(0) = 0$ and everything is smooth, you can apply de l'Hopital and get

$$\lim_{x \to 0}\frac{F(x)-F(\sin x)}{x^4} = \lim_{x \to 0}\frac{\arctan x - \cos x\arctan \sin x}{4x^3}.$$

This last limit can be evaluated using Taylor series:

$$\arctan x = x-\frac{x^3}{3}+O(x^5)$$ and

$$\cos x \arctan \sin x = x-x^3+O(x^5)$$

and the limit you are looking for is equal to

$$\lim_{x \to 0}\frac{x-\dfrac{x^3}{3}-x+x^3 + O(x^5)}{4x^3} = \frac{1}{6}.$$

The ''ugly'' Taylor expansion is obtained combining the Taylor expansions of $\sin$, $\cos$ and $\arctan$. Easier done than said.

• Actually, you just take the derivatives directly instead of trying to combine well-known expansions since that's ugly and we only need a few terms. – Simply Beautiful Art Mar 29 '17 at 23:49
• @SimplyBeautifulArt That's true, but I expect the third derivative of $\cos x \arctan \sin x$ to be even uglier.. – Stefano Mar 29 '17 at 23:51
• Well, the key is that its easy to ask your calculator to take the third derivative ;) Combining Taylor expansions? Not so much. – Simply Beautiful Art Mar 29 '17 at 23:53

Note that by the Mean Value Theorem for integrals we have $$\int_{\sin x}^{x}\arctan t\,dt = (x - \sin x)\arctan c$$ for some $c$ between $x$ and $\sin x$. Then we have $$\lim_{x \to 0}\frac{1}{x^{4}}\int_{\sin x}^{x}\arctan t\,dt = \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\cdot\frac{\arctan c}{x}$$ The first factor on the right tends to $1/6$ (via L'Hospital's Rule or Taylor series) and we show that next factor tends to $1$. For this we assume that $x \to 0^{+}$ so that $\sin x < c < x$ and therefore $$\arctan \sin x < \arctan c < \arctan x$$ and dividing by $x$ we get $$\frac{\arctan \sin x}{\sin x}\cdot\frac{\sin x}{x} < \frac{\arctan c}{x} < \frac{\arctan x}{x}$$ By Squeeze theorem we see that $(\arctan c)/x \to 1$ as $x \to 0^{+}$. A similar argument can be given for $x \to 0^{-}$ and we have the desired limit as $1/6$.


Welcome to our community

youth4ever

New member
Hello my Math friends,

I have a complex integration problem:
The integral to calculate is :

Integrate[Cos [mx]/(x^2 + a^2), {x, -Infinity, Infinity}]
or if particularized for a=1, m=1 the integral will be :

Integrate[Cos [x]/(x^2 + 1), {x, -Infinity, Infinity}]

I know the answer for both of them:
(Pi/a)*(e^-ma) for the first
and
Pi/e for the second.

How this integral can be done through Complex integration ? I don't have a clue and I strongly wish to solve it.
Thanks.

PS: BTW, I would have written with math signs because I don't like the text style of writing math but I don't know how to do it here. Is there a trick ?

Opalg

MHB Oldtimer
Staff member
Hello my Math friends,

I have a complex integration problem:
The integral to calculate is :

Integrate[Cos [mx]/(x^2 + a^2), {x, -Infinity, Infinity}]
or if particularized for a=1, m=1 the integral will be :

Integrate[Cos [x]/(x^2 + 1), {x, -Infinity, Infinity}]

I know the answer for both of them:
(Pi/a)*(e^-ma) for the first
and
Pi/e for the second.

How this integral can be done through Complex integration ? I don't have a clue and I strongly wish to solve it.
Thanks.
Hi youth4ever and welcome to MHB!

Your integral is the real part of $$\displaystyle \int_{-\infty}^\infty \frac{e^{imx}}{x^2+a^2}dx.$$ If you know about contour integration then you should be able to evaluate that integral by using a D-shaped contour.

PS: BTW, I would have written with math signs because I don't like the text style of writing math but I don't know how to do it here. Is there a trick ?
See the LaTeX tips and tutorials forum.

youth4ever

New member
Hi Opalg,

Yes I know that it comes from e^imx/... integral. But I wondered if there is a way to simply integrate the cosine part separately by using a different technique and to double check the answer.
Is this possible ?

Thanks.

ThePerfectHacker

Well-known member
But I wondered if there is a way to simply integrate the cosine part separately by using a different technique and to double check the answer.
Is this possible ?
I think your question is. Rather than making things "complicated", why not just find anti-derivative of the function inside the integral and get the answer. But the problem with what you want is that a lot of these integration problems that use complex analysis to compute do not have anti-derivatives in terms of known standard functions. So what you want cannot be done.

Random Variable

Well-known member
MHB Math Helper
If your interested in a different approach that would confirm the answer you get using contour integration, consider the Laplace transform of $$\displaystyle f(t) = \int_{0}^{\infty} \frac{\cos (mtx)}{a^{2}+x^{2}} \ dx$$

Then by definition,

$$\mathcal{L}_{t} [f(t)](s) = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\cos (mtx)}{a^{2}+x^{2}} \ e^{-st} \ dx \ dt = \int_{0}^{\infty} \frac{1}{a^{2}+x^{2}}\int_{0}^{\infty} \ \cos(mxt) e^{-st} dt \ dx$$

$$= \int_{0}^{\infty} \frac{1}{a^{2}+x^{2}} \frac{s}{s^{2}+(mx)^{2}} \ dx = \frac{s}{s^{2}-a^{2}m^{2}} \int_{0}^{\infty} \left( \frac{1}{a^{2}+x^{2}} - \frac{m^{2}}{s^{2}+m^{2}x^{2}} \right)$$

$$= \frac{s}{s^{2}-a^{2}m^{2}} \left( \frac{\pi}{2a} - \frac{\pi m }{2s} \right) = \frac{\pi}{2a(s+am)}$$

Now undo the transform.

EDIT: What I had previously after this point wasn't quite correct.

$$f(t) = \mathcal{L}^{-1}_{t} \Big[ \mathcal{L}_{t} [f(t)](s) \Big] = \mathcal{L}^{-1}_{t} \Big[ \frac{\pi}{2a(s+am)} \Big]$$

$$= \frac{\pi}{2a} \mathcal{L}^{-1}_{t} \Big[ \frac{1}{s+am} \Big] = \frac{\pi}{2a} e^{-amt}$$

And therefore,

$$\int_{-\infty}^{\infty} \frac{\cos (mx)}{a^{2}+x^{2}} \ dx = 2 f(1) = \frac{\pi}{a} e^{-am}$$

Last edited:

youth4ever

New member
Hello "Random Variable"

And many thanks for the solution. Your solution is incredible. I am amazed.
I have a question although.
How you knew at this point
$$= \int_{0}^{\infty} \frac{1}{a^{2}+x^{2}} \frac{s}{s^{2}+(mx)^{2}} \ dx$$
to amplify the product with
$$\frac{s^{2}-a^{2}m^{2}}{s^{2}-a^{2}m^{2}}$$
and then add and substract $$(mx)^{2}$$ to the numerator
to obtain the sum :
$$= \frac{s}{s^{2}-a^{2}m^{2}} \left( \frac{1}{a^{2}+x^{2}} - \frac{\ m^{2} }{s^{2}+(mx)^{2}} \right)$$
which is easily integrable in x variable? I just need a pattern on how to look because it is not obvious and I want to be able to apply this technique in similar situations.

Thanks.

Last edited:

Random Variable

Well-known member
MHB Math Helper
@youth4ever

I used partial fractions.

Assume that

$$\frac{1}{(a^{2}+x^{2})(s^{2}+m^{2}x^{2})} = \frac{Ax+B}{a^{2}+x^{2}} + \frac{Cx+D}{s^{2}+m^{2}x^{2}}$$

$$= \frac{As^{2}x + Am^{2}x^{3} + Bs^{2} + Bm^{2}x^{2} + Ca^{2}x + Cx^{3} + Da^{2} + Dx^{2}}{(a^{2}+x^{2})(s^{2}+m^{2}x^{2})}$$

If those fractions are equal, their numerators must be the same.

That leads to four equations that must be satisfied.

$$Am^{2} + C = 0$$
$$Bm^{2} + D = 0$$
$$As^{2}+Ca= 0$$
$$Bs^{2}+Da^{2} =1$$

The first and third equations imply that $A$ and $C$ are zero.

Solving the second equation for $D$ and plugging it into equation 4, we get

$$Bs^{2} - Bm^{2} a^{2} = 1 \implies B = \frac{1}{s^{2}-m^{2}a^{2}}$$

Then from equation 2,

$$D = -\frac{m^{2}}{s^{2}-m^{2}a^{2}}$$

youth4ever

New member
@Random Variable
Thanks again.

How you would approach a problem like the following :
It come from the Fourier transform of the sinc function without the coefficient.
$$\int_{0}^{\infty} \frac{sin (ak)}{ak} e^{ikx} \ dk$$
Without the exponential term hanged there that would be easy and is convergent too as the sinc function dies at Infinity. But the complex exponential term introduces another cyclicity to the function so it may be regarded as non convergent because of the 0 and Infinity limits.
However I want to believe that this can be done through complex integration.
Or if not, we could take limits to be from Pi/2 to Pi for example.

What do you think ?

Last edited:


# Are interior points ever limit points as well?

From my understanding of limit points and interior points there is somewhat of an overlap and that a lot of the time interior points are also limit points.

For the reals, a neighborhood, $r>0$, around a point must contain only a single point of the set in question to determine if it's a limit point or not. However, an interior must be completely contained in the set in question, meaning it has a neighborhood that contains at least a point in the set, which also makes it a limit point.

For example: $[0,1)$ in the reals.

From what I understand the set $(0,1)$ in the set of interior points, the set $[0,1]$ in the set of all limit points, and the set $(0,1)$ is the set of all point which are both interior and limit points.

Is this correct or are interior points always not limit points for some reason?

You are right that interior points can be limit points.

Your example was a perfect one: The set $[0,1)$ has interior $(0,1)$, and limit points $[0,1]$. So actually all of the interior points here are also limit points.

Here is an example of an interior point that's not a limit point:

Let $X$ be any set and consider the discrete topology $\mathcal{T} = \mathcal{P}(X)$ on $X$. Then let $A$ be any non-empty subset of $X$. For any $a \in A$, $\{a \}$ is an open set containing $a$, and $a \in \{a \} \subseteq A$, so $a$ is an interior point for $A$. However, $a$ is not a limit point because we can find an open neighborhood of $a$ that doesn't contain any points from $A$ other than $a$ -- namely, $\{a \}$ is the open neighborhood that satisfies this. So $a$ is not a limit point of $A$. (By this same argument, we can show that $A$ has no limit points!)

• You are saying {a} is an open set containing a, but it seems this is not an open set since there is no interval around a that is also in A. – jeffery_the_wind Sep 15 at 11:30
• @jeffery_the_wind You are speaking from the perspective of the Euclidean topology that we are all so used to. But in my answer, I mentioned that $\{a\}$ is open in the discrete topology, which is the topology that consists of all possible subsets of $X$. So every subset of $X$ is open, including $\{a\}$. So the question of which sets are open and which are closed really must be asked/framed in the context of a specific topology, since the answer changes if you change the topology. $\{a\}$ is not open in the Euclidean topology, but it is in the discrete topology. – layman Sep 16 at 3:23

A point $x$ of $A$ can be of one of two mutually exclusive types: a limit point of $A$ or an isolated point of $A$.

If the latter, it means that there exists some open $O$ in $X$ such that $\{x\} = O \cap A$. The negation of this is exactly that every open set $O$ that contains $x$ always intersects points of $A$ unequal to $x$ as well, and this means exactly that $x$ is a limit point of $A$.

E.g. $A = (0,1) \cup \{2,3\}$ (usual topology of the reals) has two isolated points $2$ and $3$ (which are not interior points of $A$), and the rest are limit points of $A$ as well as interior points. There are also limit points $0,1$ that are not in $A$ (showing $A$ is not closed).

So if $A$ has no isolated point, all of the points of $A$ (in particular all its interior points) are limit points of $A$. So often there will quite an overlap between interior and limit points.


# Integral evaluation - analytical vs. numerical

1. May 19, 2005

### broegger

Hi,

Does anyone know a reason why $$\int_{-\infty}^{\infty}\cosh(x)^{-n}dx$$ (n>0) can be evaluated analytically when n = 1,2,3,..., but only numerically when n is non-integer. I don't know if there is a "reason", but I'm using this result in a Quantum Mechanics project and it would be cool if I could give some kind of intuitive reason why this is so.

2. May 19, 2005

### arildno

My best guess:
You may use the residue theorem from complex analysis to gain an exact expression in the case of integer values, whereas this won't work if you have a non-integer.
I might be wrong, though..

3. May 19, 2005

### dextercioby

I've attached the antiderivative and here's your integral

$$\int_{-\infty}^{+\infty} \frac{dx}{\cosh^{n} x}=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n+1}{2}\right)}$$

"n" can be complex,even.

Daniel.

EDIT:The correct result is multiplied by 2.See posts #4 & #7.

#### Attached Files:

• ###### Integrate.gif
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1.6 KB
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Last edited: May 19, 2005
4. May 19, 2005

### Zurtex

O.K, in Mathematica I go that if $\Re (n) > 0$ then:

$$\int_{-\infty}^{+\infty} \cosh^{-n} (x) dx = \frac{ \sqrt{\pi} \, \Gamma\left( \frac{n}{2} \right)}{ \Gamma\left( \frac{n + 1}{2} \right)}$$

5. May 19, 2005

### dextercioby

How do you explain the "1/2" factor discrepancy...?

Daniel.

EDIT:See post #7 for details.

Last edited: May 19, 2005
6. May 19, 2005

### Zurtex

Well if I left n=1 and integrate in mathematica I get $\pi$ and:

$$\Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}$$

Are you sure you did not integrate between 0 and Infinity?

Last edited: May 19, 2005
7. May 19, 2005

### dextercioby

I got it,i plugged only for half a domain

Using [1]

$$\int_{0}^{+\infty} \frac{\sinh^{\mu}x}{\cosh^{\nu}x} \ dx =\frac{1}{2}B\left(\frac{\mu +1}{2},\frac{\nu-\mu}{2}\right)$$ (1)

,provided that

$$\mbox{Re} \left(\mu\right) >-1 \ ; \ \mbox{Re} \left(\mu-\nu\right) <0$$ (2)

Making $\mu=0$ (3) in (1) (admissible according to (2) and implying the condition $\mbox{Re} (\nu) >0$ (4) ),one gets

$$\int_{0}^{\infty} \frac{dx}{\cosh^{\nu} x} \ dx =\frac{1}{2} B\left(\frac{1}{2},\frac{\nu}{2}\right)=\frac{1}{2}\frac{\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{\nu}{2}\right)}{\Gamma\left(\frac{\nu+1}{2}\right)}$$ (5)

Using

$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$ (6)

and

$$\int_{\mathbb{R}} \frac{dx}{\cosh^{\nu} x} \ dx =2 \int_{0}^{\infty} \frac{dx}{\cosh^{\nu} x} \ dx$$ (7)

,one gets exactly the formula posted by Zurtex.

Daniel.

--------------------------------------------------
[1]G & R,5-th edition,Academic Press,CD version,1996,page 388,formula 3.512-2.

8. May 19, 2005

### dextercioby

Yes,i used an online integrator and indeed integrated only half of the domain.

I'll edit and make a reference to your post & mine just above.

Daniel.

9. May 22, 2005

### uart

So just in case anyone is at all unclear about how this relates to the original observation that :
It's because $$\Gamma(x)$$ is easily expressable in terms of elementary functions and constants for whole and half integers but generally needs other forms of numerical calulation for arbitrary values of "x".

Last edited: May 22, 2005
10. May 22, 2005

### dextercioby

Incidentally,both beta & gamma-Euler are tabulated for a certain range of values of their arguments.

So the integral is solvable analytically.Expressing its values in terms of analytical special functions means just that.

Daniel.

11. May 22, 2005

### krab

That means that whether an integral has an analytical solution or not is based on convention (how many special functions are named). I don't find such a distinction very useful. I would echo uart's comment.

12. May 25, 2005

### broegger

Thanks y'all. I've included a bit about the gamma function, interesting stuff...


# Clarify why all logarithms differ by a constant

One of the rules of logarithms that I've been taught is:

$\log_b(x)$ is equivalent to $\dfrac{\log_k(x)}{\log_k(b)}$.

Recently I've also seen another rule that says:

$\log_b(x)$ is equivalent to $\log_b(k)\log_k(x)$.

Are these equivalent (perhaps via some refactoring) ?

-
$\frac1{\log_k(b)}=\frac{\log_k(k)}{\log_k(b)}=\log_b(k)$ – J. M. Dec 13 '10 at 3:28

Divide the second equation by $\log_b(k)$ and shuffle the variables to see that these are the same thing. The reason they're true can be expressed in terms of properties of exponents. Start with the observation that $\log_b(x)=y$ means that $b^y=x$. Then use $b=a^{\log_a(b)}$ and properties of exponents to get $b^y=a^{y\log_a(b)}=x$. The last equation says that $\log_a(x)=y\log_a(b)$. Now substitute $\log_b(x)$ back in for $y$ to get the desired identity.

This fits into the general principle that each of the fundamental properties of logarithms can be seen as a translation of a property of their inverses, the exponential functions. For example, logarithms convert products to sums ($\log_b(xy)=\log_b(x)+\log_b(y)$) because exponential functions convert sums to products ($b^{x+y}=b^xb^y$). Here, the fact that $b^x=a^{x\log_a(b)}$ says that you can change bases of your exponential function by multiplying the input variable by the constant $\log_a(b)$. In other words, exponential functions with different bases are related by horizontal stretches or shrinks of the graphs. When expressed in terms of the inverse functions, input and output switch, and the horizontal stretch (or shrink) changes to a vertical stretch (or shrink). That is, instead of multiplying the input by $\log_a(b)$, you divide the output by $\log_a(b)$ to get $\log_b(x)=\frac{\log_a(x)}{\log_a(b)}$. More succinctly, $g(x)=f(cx)$ implies $g^{-1}(x)=\frac{1}{c}f^{-1}(x)$.

-
After seeing Bill Dubuque's nice one line derivation using the power rule, I thought I'd mention that you can also use the second identity (as proved in my or Arturo's answer) to quickly derive the power rule: $$\log_b(a^x)=\log_b(a)\log_a(a^x)=x\log_b(a).$$ – Jonas Meyer Dec 13 '10 at 5:25

First: They are not "equivalent", they are equal.

That said: remember the meaning of "$\log_b(a) = r$". It means that $r$ is the exponent to which you must raise $b$ in order to get $a$; that is, $\log_b(a)=r$ is equivalent to $b^r=a$.

So... why is $\log_b(x)=\frac{\log_k(x)}{\log_k(b)}$? Because if you let $r=\log_b(x)$, $s=\log_k(x)$ and $t=\log_k(b)$, then that means that $b^r = x$, $k^s = x$, and $k^t=b$. Therefore, $$k^s = x = b^r = (k^t)^r = k^{tr}.$$ So $\log_k(x) = s = rt = \log_b(x)\log_k(b)$, from which you get the equality you want by dividing through by $\log_k(b)$.

For the second equality, $\log_b(x) = \log_b(k)\log_k(x)$, let $r=\log_b(x)$, $s=\log_b(k)$, and $t=\log_k(x)$. Then $b^r = x$, $b^s = k$, and $k^t=s$. So we have: $$b^r = x = k^t = (b^s)^t = b^{st},$$ which means we must have $\log_b(x) = r=st=\log_b(k)\log_k(x)$, as desired.

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This is an excellent explanation. – Quixotic Dec 13 '10 at 14:17

Yes. Your second equation is $\log_b k\cdot \log_k x = \log_b x$. Dividing by $\log_b k$ we get $\log_k x=\frac{\log_b x}{\log_b k}$. This is the same as your first equation with different symbols (an interchange of $k$ and $b$).

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+1,Nice one-liner :) – Quixotic Dec 13 '10 at 14:08

HINT $\:$ The first equation is $\rm\ log_K\$ of $\rm\ B^{log_B\ X}\ =\ X$

and the $\:$ second $\:$ equation is $\rm\ log_B\$ of $\rm\ K^{log_K\ X}\ =\ X$

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When combined together, several gates can make a complex logical evaluation system that has. Conic Sections Transformation. Logic Gate Simulator. Natural deduction proof editor and checker. Go to Daemon Proof Checkeror Quick Help Index. Free Python 3. This site based on the Open Logic Project proof checker. The Logic Daemon. The Propositional Logic Calculator finds all the models of a given propositional formula. Know Your Worth is based on millions of real salaries from Glassdoor users. See Credits. Boolean Algebra expression simplifier & solver. Perfect Proof. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. The system was originally written for UMass’s Intro Logic course, based on Gary Hardegree’s online. Disjunctive normal form (DNF), including perfect. In each step the user. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. Some (importable) sample proofs in the "plain" notation are here. (If you don't want to install this file. They will show you how to use each calculator. See Credits. Line Equations Functions Arithmetic & Comp. 3: Proofs in predicate logic. Modifications by students and faculty at Cal. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. Here, instead, we shall think of this as a proof method, traditionally called “conditional derivation”. Mathematical logic step by step. Com stats: 2613 tutors , 730556 problems solved. One of the most basic rules of deduction in predicate logic says that ( ∀ x P ( x)) P ( a) for any entity a in the domain of discourse of. To typeset these proofs you will need Johann Klüwer's fitch. Place brackets in expressions, given the priority of operations. The Coq Proof Assistant. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. So far in this section, we have been working mostly with propositional logic. For an introduction to logic and proof in this style, consult a textbook such as Kaye , Huth and Ryan , or Bornat. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. Online tool. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. First, we'll look at it in the propositional case, then in the first-order case. Natural deduction proof editor and checker. Proof checker. Proof Checker for forall x: Cambridge and Calgary. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. Build a truth table for the formulas entered. Use symbolic logic and logic algebra. Resolution Refutation. Line Equations Functions Arithmetic & Comp. One of the most basic rules of deduction in predicate logic says that ( ∀ x P ( x)) P ( a) for any entity a in the domain of discourse of. The thing solves algebra, and basic symbolic logic uses, well, I don't want to say the same sort of symbol manipulation because the overlap is imperfect, but both proofs and algebra work by manipulating symbols via a set of well-defined rules. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. Truth Tree Solver. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. The values of the Boolean algebra calculator are denoted with logic 0 & 1. Conjunctive normal form (CNF), including perfect. Right click connections to delete them. All the arguments are syllogisms. Binary numbers multiplication is a part of arithmetic operations in digital electronics. A first prototype of a ProB Logic Calculator is now available online. Propositional logic is also amenable to “deduction,” that is, the development of proofs by writing a series of lines, each of which either is given or is justiﬁed by some previous lines (Section 12. You can select and try out several solver algorithms: the "DPLL better" is the best solver amongst the options. Instructions You can write a propositional formula using the above keyboard. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. for details. Know Your Worth is based on millions of real salaries from Glassdoor users. The Corbettmaths Practice Questions on Algebraic Proof. One Flip application is a proof checker for entering and editing proofs in natural deduction style. Application works on the Chrome browser. The logic language used in this theorem prover is one that was proposed in the author's Master's thesis, back in 1985-1987, at which time it contained most of the features shown here, including the hierarchical type scheme. Know Your Worth is based on millions of real salaries from Glassdoor users. Thus, the argument above is valid, because if all humans are mortal, and if. Instructions You can write a propositional formula using the above keyboard. Place brackets in expressions, given the priority of operations. One of the most basic rules of deduction in predicate logic says that ( ∀ x P ( x)) P ( a) for any entity a in the domain of discourse of. The Conformal Smart Logic Equivalence Checker (LEC) is the next-generation equivalency checking solution. A proposition is a declarative statement which is either true or false. A proof is an argument intended to convince the reader that a general principle is true in all situations. Simplify logical expressions. Try the leading salary calculator the next time you negotiate your salary or ask for a raise, and get paid fairly. This calculator solves linear diophantine equations. φ [ν←τ] where τ is free for ν in φ. This study aid includes: Proof generator. To download DC Proof and for a contact link, visit my homepage. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. It formalizes the rules of logic. Look at line 3. 3: Proofs in predicate logic. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. By using this website, you agree to our Cookie Policy. In mathematics, a Diophantine equation is a polynomial equation in two or more unknowns such that only the integer solutions are searched or studied (an integer solution is a solution such that all the unknowns take integer values). Know Your Worth is based on millions of real salaries from Glassdoor users. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). In other words, it’s based on the mistaken assumption that a lack of evidence is evidence. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. The symbol for this is $$ν$$. (There was the untyped logic language Prolog, and the strongly typed — but general programming language. Use symbolic logic and logic algebra. Modifications by students and faculty at Cal. A variable is a symbol used to represent a logical quantity. Instructors. It is a web application that uses Semantic Tableaux to check the validity of a statement, and provides a proof if it finds that the statement is valid. Line Equations Functions Arithmetic & Comp. This is the mode of proof most of us. Free Python 3. Let x 2K so that xjk. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. Here is a standard example: An argument is valid if and only if the conclusion necessarily follows from the premises. Using the derived rule allowed us to shorten the proof considerably. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). In the dropdown menu, click 'UserDoc'. φ [ν←τ] where τ is free for ν in φ. 3: Proofs in predicate logic. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. 0 is based on classical logic, but it is possible to define your axioms in it. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. Logic Calculator. Propositional Logic • Propositional resolution • Propositional theorem proving •Unification Today we're going to talk about resolution, which is a proof strategy. Boolean Algebra expression simplifier & solver. Oct 24 '18 at 20:18. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. Matrices & Vectors. Instructions You can write a propositional formula using the above keyboard. Proof generator and proof checker for propositional logic in "natural deduction" style. Step through the examples. This study aid includes: Proof generator. Instructors. Learning about Coq. So far in this section, we have been working mostly with propositional logic. Basics Whisky 101. Free Logical Sets calculator - calculate boolean algebra, truth tables and set theory step-by-step This website uses cookies to ensure you get the best experience. Truth Tables, Logic, and DeMorgan's Laws. Get Custom Built Calculator For Your Website. See this pdf for an example of how Fitch proofs typeset in LaTeX look. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. Propositional logic is also amenable to “deduction,” that is, the development of proofs by writing a series of lines, each of which either is given or is justiﬁed by some previous lines (Section 12. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. A drill for the truth functional connectives. Amount of Whisky (fl. Typical applications include the certification of properties of. First, we'll look at it in the propositional case, then in the first-order case. Use symbolic logic and logic algebra. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Natural Deduction examples | rules | syntax | info | download | home: Last Modified : 13-Jun-2021. Truth Tree Solver. Matrices & Vectors. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. Get Custom Built Calculator For Your Website. See Credits. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. The Propositional Logic Calculator finds all the models of a given propositional formula. The assumption set that results is the same as the assumption set for line 3. Using the derived rule allowed us to shorten the proof considerably. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Instructors. Conic Sections Transformation. Logic Gate Simulator. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. A model describes how units of computations, memories, and communications are organized. Back then, the idea of logic languages with types was novel. This study aid includes: Proof generator. See Credits. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. This calculator solves linear diophantine equations. Appeal to ignorance is a logical fallacy in which someone argues either for or against something because there is no contradicting evidence. Natural deduction proof editor and checker. Binary numbers multiplication is a part of arithmetic operations in digital electronics. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. It formalizes the rules of logic. Enter the term you would like to use: τ: Existential Introduction: φ (τ) Eν:φ (ν) Enter the term you would like to replace: τ:. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. Clicking the "Tree Proof" button will pass the statement to wo's Tree Proof Generator. This page is a tutorial and user's guide; there is also a complete reference. LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. Logic Calculator is a free, portable truth table generator for logic formulas i. Matrices & Vectors. Yes, Algebraic Proofs isn't particularly exciting. Solving a classical propositional formula means looking for such values of variables that the formula becomes true. Line Equations Functions Arithmetic & Comp. Enter the term you would like to use: τ: Existential Introduction: φ (τ) Eν:φ (ν) Enter the term you would like to replace: τ:. Logic Calculator. Amount of Whisky (fl. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Computer programs are constantly making decisions based on the current "STATE" of the data held by the program. For modal predicate logic, constant domains and rigid terms are assumed. Get help from our free tutors ===>. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. Oct 24 '18 at 20:18. Combining multiple conditions to form one True/False value is the. Proof generator and proof checker for propositional logic in "natural deduction" style. This allows for submission and automated marking of exercises such as symbol-ization, truth tables, and natural deduction proofs. (2) Logical entailment: displays the truth table of each of the premises along with the. Practice your deduction skills with Proof checker and Random Tasks. Follow the 2 steps guide to find the truth table using the boolean calculator. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. The Coq Proof Assistant. This style of proof is called a resolution proof. Yes, Algebraic Proofs isn't particularly exciting. Take help from sample expressions in the input box or have a look at the boolean functions in the content to understand the mathematical operations used in expressions. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses’s free, online logic teaching software application Carnap (carnap. Thus, the argument above is valid, because if all humans are mortal, and if. Each one has a different shape to show its particular function. Back then, the idea of logic languages with types was novel. This site based on the Open Logic Project proof checker. A model describes how units of computations, memories, and communications are organized. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. Write a symbolic sentence in the text field below. For example, (a -> b) & a becomes true if and only if both a and b are assigned true. proof language Isar [20] which looks like a mixture of English, logic and a programming language, and is based on natural deduction. Valid or Invalid? The rules of this test are simple: it's your job to determine whether an argument is valid or not. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. We applied Demorgan's Law, which is abbreviated DM, to line 2. It executes the logical operations like AND, NAND, OR, NOR, NOT & X-OR. Let x 2K so that xjk. for details. A variable is a symbol used to represent a logical quantity. 0 is based on classical logic, but it is possible to define your axioms in it. Mathematical logic step by step. All in one boolean expression calculator. Propositional logic is also amenable to “deduction,” that is, the development of proofs by writing a series of lines, each of which either is given or is justiﬁed by some previous lines (Section 12. See Credits. Logic Gate Simulator. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. A proof is an argument intended to convince the reader that a general principle is true in all situations. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. 2 Proofs One of the principal aims of this course is to teach the student how to read and, to a lesser extent, write proofs. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. The Gateway to Logic is a collection of web-based logic programs offering a number of logical functions (e. Yes, Algebraic Proofs isn't particularly exciting. Logic gates are symbols that can directly replace an expression in Boolean arithmetic. When combined together, several gates can make a complex logical evaluation system that has. Natural Deduction is a free app published for Windows 10 PC and can be downloaded from Windows Store. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. Yes, Algebraic Proofs isn't particularly exciting. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences. Know Your Worth is based on millions of real salaries from Glassdoor users. Suppose k 2Z and let K = fn 2Z : njkgand S = fn 2Z : njk2g. So far in this section, we have been working mostly with propositional logic. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. The values of the Boolean algebra calculator are denoted with logic 0 & 1. Thus, x 2S. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. (whenever you see $$ν$$ read 'or') When two simple sentences, p and q, are joined in a disjunction statement, the disjunction is expressed symbolically as p $$ν$$ q. When loaded, click 'Help' on the menu bar. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. Online tool. Learning about Coq. Here, instead, we shall think of this as a proof method, traditionally called “conditional derivation”. Natural Deduction is a free app published for Windows 10 PC and can be downloaded from Windows Store. Mathematical logic step by step. This is the mode of proof most of us. Go to Daemon Proof Checkeror Quick Help Index. When combined together, several gates can make a complex logical evaluation system that has. Know Your Worth is based on millions of real salaries from Glassdoor users. The symbol P denotes a sum over its argument for each natural. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Boolean Algebra expression simplifier & solver. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. This allows for submission and automated marking of exercises such as symbol-ization, truth tables, and natural deduction proofs. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses’s free, online logic teaching software application Carnap (carnap. Note that proofs can also be exported in "pretty print" notation (with unicode logic symbols) or LaTeX. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. Get Custom Built Calculator For Your Website. Learn boolean algebra. Yes, Algebraic Proofs isn't particularly exciting. Kevin Klement has done up a prototype of his online natural deduction proof builder/checker that works with the natural deduction system of the Cambridge and Calgary versions of forall x. LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. - Dan Christensen. Simplify logical expressions. Free Ubuntu. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. Using the derived rule allowed us to shorten the proof considerably. All the arguments are syllogisms. This style of proof is called a resolution proof. Use symbolic logic and logic algebra. Boolean Expression Calculator. It executes the logical operations like AND, NAND, OR, NOR, NOT & X-OR. Propositional sequent calculus prover. State University, Monterey Bay. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. A free, simple, online logic gate simulator. Place brackets in expressions, given the priority of operations. Then k2 = (ax)2 = x(a2x) so xjk2. Thus, x 2S. Disjunctive normal form (DNF), including perfect. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. It is a technique of knowledge representation in logical and mathematical form. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. Since this is all about math, I copy some content from wikipedia for a start. Because of its simplicity it is particularly well-suited for mechanical theorem provers. Simplify logical expressions. Proof by induction involves statements which depend on the natural numbers, n = 1,2,3, It often uses summation notation which we now brieﬂy review before discussing induction itself. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Amount of Whisky (fl. You can select and try out several solver algorithms: the "DPLL better" is the best solver amongst the options. Complete your profile, and we will calculate how much you could earn in today's job market. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. This site based on the Open Logic Project proof checker. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Free Python 3. Drag from the hollow circles to the solid circles to make connections. All the arguments are syllogisms. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Yes, Algebraic Proofs isn't particularly exciting. The thing solves algebra, and basic symbolic logic uses, well, I don't want to say the same sort of symbol manipulation because the overlap is imperfect, but both proofs and algebra work by manipulating symbols via a set of well-defined rules. Natural deduction proof editor and checker. One of the most basic rules of deduction in predicate logic says that ( ∀ x P ( x)) P ( a) for any entity a in the domain of discourse of. They will show you how to use each calculator. In each step the user. The specific system used here is the one found in forall x: Calgary Remix. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. Learn more. The symbol P denotes a sum over its argument for each natural. You need to “summarize” what was established after making the desired assumption (the contradiction of the conclusion). Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Complete your profile, and we will calculate how much you could earn in today's job market. Matrices & Vectors. A sequent S is true if and only if there exists a tree of sequents rooted at S where each leaf is an axiom and each internal node is derived from its children by an inference rule. Such proofs can also encode traditional proofs based on modus ponens: the inference P∧(P⇒Q) ⊢ Q can be rewritten as resolution by expanding ⇒ to get P∧(¬P∨Q) ⊢ Q. See Credits. For example, a heart monitoring program might sound an alarm if the pulse is too slow or the blood pressure is too weak. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. ) And that’s it! Enjoy drinking your whisky exactly how you like it. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. Free Windows Terminal Preview. Back then, the idea of logic languages with types was novel. You can select and try out several solver algorithms: the "DPLL better" is the best solver amongst the options. Formal Proof Check. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. Com stats: 2613 tutors , 730556 problems solved. ) Bottle Proof. Videos, worksheets, 5-a-day and much more. Take help from sample expressions in the input box or have a look at the boolean functions in the content to understand the mathematical operations used in expressions. State University, Monterey Bay. techniques as “proof by contradiction” or “proof by contrapositive” (Section 12. The values of the Boolean algebra calculator are denoted with logic 0 & 1. See Credits. Propositional logic in Artificial intelligence. Desired Proof. Kevin Klement has done up a prototype of his online natural deduction proof builder/checker that works with the natural deduction system of the Cambridge and Calgary versions of forall x. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Save your work on device and continue later on. We applied Demorgan's Law, which is abbreviated DM, to line 2. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Logic gates are symbols that can directly replace an expression in Boolean arithmetic. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). Logic Calculator. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. The symbol for this is $$ν$$. Back then, the idea of logic languages with types was novel. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. Resolution Refutation. Since this is all about math, I copy some content from wikipedia for a start. The symbol P denotes a sum over its argument for each natural. The Logic Daemon. 3: Proofs in predicate logic. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. This app is a graphical semantic calculator for a specific kind of. Free Logical Sets calculator - calculate boolean algebra, truth tables and set theory step-by-step This website uses cookies to ensure you get the best experience. Any single variable can have a 1 or a 0 value. Proof Checker for forall x: Cambridge and Calgary. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Logic Gate Simulator. Conic Sections Transformation. The Corbettmaths Practice Questions on Algebraic Proof. Practice your deduction skills with Proof checker and Random Tasks. Propositional sequent calculus prover. Binary numbers multiplication is a part of arithmetic operations in digital electronics. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. ) Bottle Proof. If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. (whenever you see $$ν$$ read 'or') When two simple sentences, p and q, are joined in a disjunction statement, the disjunction is expressed symbolically as p $$ν$$ q. Diana Higgins on Conditional-proof-logic-calculator glennnervi. Combining multiple conditions to form one True/False value is the. A variable is a symbol used to represent a logical quantity. Any single variable can have a 1 or a 0 value. See this pdf for an example of how Fitch proofs typeset in LaTeX look. The Corbettmaths Practice Questions on Algebraic Proof. The symbol for this is $$ν$$. In computer science, and more specifically in computability theory and computational complexity theory, a model of computation is a model which describes how an output of a mathematical function is computed given an input. Binary numbers multiplication is a part of arithmetic operations in digital electronics. By using this website, you agree to our Cookie Policy. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. Online tool. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. Propositional logic (PL) is the simplest form of logic where all the statements are made by propositions. Yes, Algebraic Proofs isn't particularly exciting. Conjunctive normal form (CNF), including perfect. Get help from our free tutors ===>. Actually there are mechanical ways of generating Fitch style proofs. Once you know your perfect proof, this calculator will tell you exactly how much water to add to any amount of whisky to reach it. Boolean formulas are written as sequents. To typeset these proofs you will need Johann Klüwer's fitch. It is also known as argumentum ad ignorantiam (Latin for “argument from ignorance”) and is a type of. Formal Proof Check. This study aid includes: Proof generator. This site based on the Open Logic Project proof checker. Place brackets in expressions, given the priority of operations. At any time get assistance and ideas from Proof generator. Yes, Algebraic Proofs isn't particularly exciting. Natural deduction proof editor and checker. They're especially important in logical arguments and proofs, let's find out why! While the word "argument" may mean a disagreement between two or more people, in mathematical logic, an argument is a sequence or list of statements called premises or assumptions and returns a conclusion. φ [ν←τ] where τ is free for ν in φ. Read from here about the differences between algorithms. Like most proofs, logic proofs usually begin with premises--- statements that you're allowed to assume. Basics Whisky 101. You can also use LaTeX commands. Logic Calculator is a free, portable truth table generator for logic formulas i. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. Conic Sections Transformation. Look at line 3. The Corbettmaths Practice Questions on Algebraic Proof. Back then, the idea of logic languages with types was novel. Modifications by students and faculty at Cal. You need to “summarize” what was established after making the desired assumption (the contradiction of the conclusion). The values of the Boolean algebra calculator are denoted with logic 0 & 1. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). Line Equations Functions Arithmetic & Comp. Use symbolic logic and logic algebra. Propositional logic in Artificial intelligence. It provides a formal language to write mathematical definitions, executable algorithms and theorems together with an environment for semi-interactive development of machine-checked proofs. Right click connections to delete them. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. 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Once you know your perfect proof, this calculator will tell you exactly how much water to add to any amount of whisky to reach it. Modifications by students and faculty at Cal. Try the leading salary calculator the next time you negotiate your salary or ask for a raise, and get paid fairly. The specific system used here is the one found in forall x: Calgary Remix. A proposition is a declarative statement which is either true or false. Basics Whisky 101. Such proofs can also encode traditional proofs based on modus ponens: the inference P∧(P⇒Q) ⊢ Q can be rewritten as resolution by expanding ⇒ to get P∧(¬P∨Q) ⊢ Q. Your Input Approximate the integral $$\int\limits_{0}^{1} \sqrt{\sin^{3}{\left(x \right)} + 1}\, dx$$$with $$n = 5$$$ using the trapezoidal rule. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. The system was originally written for UMass’s Intro Logic course, based on Gary Hardegree’s online. In mathematics, a Diophantine equation is a polynomial equation in two or more unknowns such that only the integer solutions are searched or studied (an integer solution is a solution such that all the unknowns take integer values). In computer science, and more specifically in computability theory and computational complexity theory, a model of computation is a model which describes how an output of a mathematical function is computed given an input. It provides a formal language to write mathematical definitions, executable algorithms and theorems together with an environment for semi-interactive development of machine-checked proofs. The symbol P denotes a sum over its argument for each natural. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Logic Calculator: Desktop application useful to perform logical operations in an arithmetic-calculator fashion, with three modes: 1) Evaluation of logic formulas: displays the truth table along with the models of the given formula. All in one boolean expression calculator. Solving a classical propositional formula means looking for such values of variables that the formula becomes true. Coq is a formal proof management system. A model describes how units of computations, memories, and communications are organized. A variable is a symbol used to represent a logical quantity. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. To download DC Proof and for a contact link, visit my homepage. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. It formalizes the rules of logic. Save your work on device and continue later on. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. To enter logic symbols, use the buttons above the text field, or type ~ for ¬, & for ∧, v for ∨, -> for →, <-> for ↔, (Ax) for ∀x, (Ex) for ∃x, [] for , > for. The only limitation for this calculator is that you have only three atomic propositions to choose from: p,q and r. This site based on the Open Logic Project proof checker. Here is a standard example: An argument is valid if and only if the conclusion necessarily follows from the premises. Proof checker. A proof is an argument intended to convince the reader that a general principle is true in all situations. Note that proofs can also be exported in "pretty print" notation (with unicode logic symbols) or LaTeX. Logic Gate Simulator. If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Yes, Algebraic Proofs isn't particularly exciting. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters "W" and "F" denote the constant values truth and falsehood and that the lower-case letter "v" denotes the disjunction. In chapter 17 we will prove the deduction theorem. The assumption set that results is the same as the assumption set for line 3. In the dropdown menu, click 'UserDoc'. This app is a graphical semantic calculator for a specific kind of. You may add additional sentences to your set by repeating this step. This calculator solves linear diophantine equations. Some (importable) sample proofs in the "plain" notation are here. Modal logic is a type of symbolic logic for capturing inferences about necessity and possibility. Free Windows Terminal Preview. Perfect Proof. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences.


what does $|x-2| < 1$ mean?

I am studying some inequality properties of absolute values and I bumped into some expressions like $|x-2| < 1$ that I just can't get the meaning of them.

Lets say I have this expression

$$|x|<1.$$

This means that $x$ must be somewhere less than $1$ or greater than $-1$ which means that

$$-1 < x < 1.$$

So basically $|x|<1$ and $-1 < x < 1$ are the same thing.

$$|x|<1 \iff -1 < x < 1 \iff\text{"Somewhere less that 1 or greater than -1" or between -1 and 1}$$

Now lets say I have

$$|x-2| < 1.$$

This means that the result of the expression $|x-2|$ must be less than $1$ or greater than $-1$? What does that also mean for $x$? Is it that $x$ must be a value that when we subtract $2$ the result has to stay withing the bound of $-1$ or $1$ or less than zero? If $x =5$ the statement fails because $3 <1$ is false. So it has to determine a boundary of $x$'s that satisfy this equation right?

if $|x| = |-x|$

what can this mean for

$|x-2| = |-x-2|$ or $|x+2|$ or $|-x+2|$ ?

Thank you

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I added the tag intuition to the question because it seems like you're looking for an intuitive way to understand the formula. –  Git Gud Jul 12 '13 at 16:48
Correction to: "Somewhere less that 1 or greater than -1" or between -1 and 1. The second part is correct. The part in quotes should have AND as the conjunction. Otherwise, one could use 4 which is "greater than -1" yet isn't less than 1. –  JB King Jul 12 '13 at 16:59
It means that $x$ is less than 1 away from 2 in either direction on the number line, or any direction on the complex plane. –  Kaz Jul 12 '13 at 19:01

The geometric interpretation, in $\Bbb R$, for $|x-a|<b$ is '$x$ is at a distance smaller than $b$ from $a$'.

In your particular example, $|x-2|<1$, it means that $x$ is at a distance of at most $1$ from $2$ and it (the distance) never reaches $1$.

To interpret $|x-2|=|-x-2|$, I find useful to first note that $|-x-2|=|x-(-2)|$ (why?). The equality $|x-2|=|x-(-2)|$ says that $x$ is at equal distance between $2$ and $-2$.

More generally, $|x-a|=|x-b|$ says that $x$ is at the same distance between $a$ and $b$.

To summarize, read $|x-a|$ as the distance between $x$ and $a$.

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thank you, and |x-2| = |-x-2| ? or |x+2| ? or |-x+2| ? –  themhz Jul 12 '13 at 16:51
@themhz Add that to the question. –  Git Gud Jul 12 '13 at 16:51
so |x-2| => say x=3 then 3-2 = 1 or x=-3 then -3-(-2)=-3+2 =-1 so |-1|=|1|=1 so |x-2| = |-x+2| =|2-x| right? –  themhz Jul 12 '13 at 17:01
"it means that x is at a distance of at most 1 from 2 and it never reaches 2." Of course it reaches 2. It won't reach 1 or 3, though. –  celtschk Jul 12 '13 at 17:04
|x-2| is not the same as |-x-2| as the latter could be expressed as |x+2| which is quite different if you consider evaluating this for a few different values of x. –  JB King Jul 12 '13 at 20:08

The way to think of $$|x-y|$$ is as the distance from $x$ to $y$. For example, is $|x-y| = |y-x|$? Yes, it is, because the distance from $x$ to $y$ is the same as the distance from $y$ to $x$.

Is $$|x-y| + |y-z| = |x-z|?$$ This says that the distance from $x$ to $y$, plus the distance from $y$ to $z$, is equal to the distance from $x$ to $z$. That would mean that $y$ was on the direct path from $x$ to $z$. So we would expect it to be false if $y$ was not on this direct path; say if $x = 2, z=4,$ but $y = 17$. And indeed $|2-17| + |17-4| \ne |2-4|$, so the equation above is not always true. But we might guess from this understanding that $$|x-y| + |y-z| \ge |x-z|,$$ with equality occurring just if $y$ is between $x$ and $z$. And in fact this is always true.

With this idea, what does $$|x|$$ mean? It should be the same as $$|x-0|,$$ which is the distance from $x$ to 0. And that is correct.

Now what does $$|x-2| < 1$$ mean? It means that the distance from $x$ to 2 is less than 1. So another way to write this is $$1\lt x\lt 3.$$

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Hint

Denote $x-2$ by $y$ then $$|x-2|<1\iff |y|<1$$ and you find exactly your first inequality. Can you take it from here?

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that's very intuitive thank you –  themhz Jul 12 '13 at 17:02
You're welcome. –  Sami Ben Romdhane Jul 12 '13 at 18:23

We know that: $$|x-2|= \left\{ \begin{array}{ll} -x+2 & \quad x < 2 \\ x-2 & \quad x \ge 2 \end{array} \right.$$ Now if we have to do $|x-2|<1$ so: $$x\ge2\to x-2<1\to x<3\\\ x<2\to 2-x<1\to x>1$$ This means that , overall, we have $1<x<3$.

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I hope you are feeling better, Babak! ;-) –  amWhy Jul 13 '13 at 0:35

Just as you replace $|x|<1$ with $-1<x<1$, you can replace $|x-2|<1$ with $-1<x-2<1$. Adding $2$ on all three parts leaves the order unchanged, so $-1+2<x-2+2<1+2$, i.e. $1<x<3$.

To see what signs are correct in $|x-2|=|\pm x\pm 2|$, note that the absolute value does not change when we replace its argument with its negative. The argument here is $x-2$, the negative thereof is $-(x-2)$ and that can be simplified to $-x+2$. Thus $|x-2|=|-x+2|$. (Of course in rare cases it may also be true that $|x-2|=|x+2|$, namely precisely when $x=0$)

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Draw a graph of $$f(x) = |x|$$ Then ask yourself, how does one obtain $$g(x) = |x-2|$$ from the graph of $f(x)$. Once you figure out how the graph looks, the question that you asked is simply asking to find all the possible $x$ values such that $g(x) < 1$. You can draw a horizontal line $y=1$ on the graph of $g(x)$ and observe the $x$ values that satisfy.
$|x-2| < 1\iff-1<x-2<1\iff2-1<2+x-2<2+1\iff1<x<3$


# A dance class consists of 22 students, 10 women and 12 men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible?

This is a question from Sheldon Ross.

A dance class consists of 22 students, 10 women and 12 men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible?

So my reasoning is this :

1. I choose 5 women from a pool of 10 in 10C2 ways.
2. I choose 5 men from a pool of 12 in 12C2 ways.

So total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.

But I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!

So I calculate the total ways as 10C2 * 12C2 * 5! * 5!. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.

• Please make your titles informative. We get dozens of "permutations combinations" questions everyday. – Em. Jul 10 '16 at 8:28
• Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$ – Ariana Jul 10 '16 at 8:53
• Did you mean $\binom{10}{5}\binom{12}{5}5!$? – N. F. Taussig Jul 10 '16 at 9:08
• @N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!. – amrx Jul 10 '16 at 9:17
• As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $\binom{10}{5}\binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters. – N. F. Taussig Jul 10 '16 at 9:24

I believe that the problem is that you're counting multiple times. If the couples were ordered (first couple, second couple, etc.) then it would be correct. In this case, though, you can simply go down the list of men and ask yourself "How many different women can I pair with this man?"

• As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely. – Mike Jul 10 '16 at 8:35

After selecting 5 from men and 5 from women, we need to pair them.

Lets assume like finding a pair for each man,

for the 1st guy -- can choose 1 from 5 women

for the 2nd guy -- can choose 1 from 4 women

.

.

for the 5th guy -- can choose 1 from 1 woman

so 5!


# Probability question with similar/distinct dice.

The question:

We're throwing $$3$$ normal dice twice. (Suppose that every vector with length of $$6$$ has an equal probability). What is the probability to get two similar results from the two throws of the $$3$$ dice if:
a) The dice are different.
b) The dice are similar.

My Work:
for (a):
$$\Omega = \{(a_1,a_2,...,a_6) \mid a_i \in \{1,...,6\} ,(i=1,2,...,6)\}$$
if all the dice are different then: $$|\Omega| = 6^6$$
Let $$A\subset\Omega$$, ($$A$$- we get similar results from the two throws), then $$|A|=6^3$$ (first dice throw can be whatever, but the second one must be the same).
So the final answer I got is $$P(A) = \frac {6^3}{6^6}$$.

for (b): (and here comes the confusion):
Things I tried to do and didn't get the right answer:

1. I tried to choose $$3$$ of the dice (since they're all similar) throw them, and sort them, so what I got is $$\binom{6}{3} \cdot 6^3 \cdot 3!$$.
2. I tried to choose $$1$$ of the two throws, and find how many options there are and sort them (same as before, each dice has $$6$$ sides so $$6^3$$ for all), and got $$\binom{2}{1} \cdot 6^3 \cdot 3!$$.

Both of the two ways led me to wrong answers, I'm trying where are my mistakes and why my thinking led me to wrong answers, which can lead me to reach the true answer.

The Final answer is $$\frac {996}{6^6}$$

EDIT:
After deeply thinking about the problem I obtained a new approach (I couldn't reach the answer with it, but I'm not sure if it's my weak combinatorics or if it's a wrong approach).
3. By writing down a couple of examples, I noticed that there's something that I can also use to calculate the probability, we can divide this into $$3$$ possible cases:
First: Getting same number on all of the six dice. $$(2,2,2, 2,2,2)$$
Second: Getting $$2$$ different numbers. $$(2,4,2, 2,4,2)$$
Third: Getting $$3$$ different numbers. $$(1,2,3, 1,2,3)$$
So I could calculate the number of options each case has and divide by the total options $$|\Omega|=6^6$$.
I felt like I've gone too far but I can't see why it wouldn't be true. Any Feedback is really appreciated.
• The word dice is plural; the singular is die. The word dices means to cut into small cubes. – N. F. Taussig Mar 21 at 10:19
• Formatting tip: type $\binom{n}{k}$ to obtain $\binom{n}{k}$. – N. F. Taussig Mar 21 at 10:23
• Your new approach is the right one. However, I don't see how the author obtained $996$ favorable cases. – N. F. Taussig Mar 21 at 10:40
• @N.F.Taussig Thanks alot I'm happy to hear that, could you recognize why the first two approaches reached wrong answers? I still can't understand why they led me to wrong answers and trying to see what problems I have there if it's using binomials the wrong way or just my logic wasn't true. Like for example in approach 1, I chose the $3$ dice of the first throw, and they could be whatever, so the cases that I added in my new approach (according to my understanding) are included in the first approach. seems like I have a big misunderstanding somewhere. – Pwaol Mar 21 at 10:48
• Your first two approaches fail since if your original outcome was, say $(5, 5, 4)$, there are only three ways the second roll could match the first, depending on which die matched the $4$, not $3!$. – N. F. Taussig Mar 21 at 10:52

With the help of N. F. Taussig in the comments, I think I have reached the right solution, So I decided to answer my own question, and would love to hear feedback.

Starting with why my first two approaches are wrong, The number of ways that the second roll could match the first is different whenever there's $$3$$ different numbers or $$2$$ different or all of them are the same, but in my first two approaches, I didn't pay attention to that.

Solution: (Approach 3):
The number of ways to get the same number on all dice, is just simply choosing a number which is $$\binom{6}{1}=6$$.
The number of ways to get two different numbers in the first roll is:
$$\binom{6}{2}$$ choosing $$2$$ numbers, $$\binom{2}{1}$$ choosing the number that will appear twice, $$\binom{3}{1}$$ choosing which place is the single number in the first roll, $$\binom{3}{1}$$ choosing in which place is the single number in the second roll. So in total by multiplying all of them we get $$\binom{6}{2}*2*3*3 = 270$$ ways.

The number of ways to get three different numbers in the first roll is:
$$\binom{6}{3}$$ choosing $$3$$ numbers, $$3!$$ ordering the first roll, $$3!$$ ordering the second roll. in total $$720$$.

Hence we reach the answer by adding them up and dividing by $$|\Omega|$$.

• This is correct. The reason I did not initially obtain $996$ was that I forgot that there were only three dice. – N. F. Taussig Mar 21 at 11:14


# substitution integration question

I want to integrate $\int \sqrt{1 - x^2} dx$.

When I substitute $x = \sin θ$ , I get the right answer. ( $\cos^2\theta$ before integration)

But when I substitute $x = \cos θ$ , I don't get the right answer. ( $-\sin^2\theta$ before integration).

What step is wrong here? If I proceed like this, don't I end up with a wrong answer?

• I get the same result with both substitutions. Can you show us your work so we can see where the problem is? – 5xum Apr 15 '15 at 8:41
• @5xum I'm guessing he/she's having trouble with the derivative of $\cos(\theta)$ having a negative term, and the $\sin(\theta)$ doesn't. – nathan.j.mcdougall Apr 15 '15 at 8:47
• @nathan.j.mcdougall Possibly. But unless he shows his work, we cannot really help him. – 5xum Apr 15 '15 at 8:47
• When I substitute x = sinθ, the equivalent integration function becomes cosθ (cos^2 θ)^0.5. The other substitution yields -sinθ (sin^2 θ)^0.5. Am I wrong in this step? – lgj Apr 15 '15 at 8:54
• Hint: $\sqrt{1-\sin^2 \theta}= |\cos \theta| \ne \cos \theta$. and the same for $\sqrt{1-\cos^2 \theta}= |\sin \theta| \ne \sin \theta$, so the integration requaire a bit more attention. – Emilio Novati Apr 15 '15 at 10:37

The short answer is they become the same answer once you back-substitute $\theta$ for $x$.

It makes more sense if you actually do the integrals.

\begin{align} \int \cos^2 \theta \,d\theta &= \frac{1}{2}\int (1 +\cos 2\theta)\,d\theta \\&= \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= \frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta + C \\&= \frac{1}{2}\arcsin x + \frac{1}{2}x\sqrt{1-x^2} + C \end{align}

\begin{align} \int -\sin^2 \theta \,d\theta &= -\frac{1}{2}\int (1 -\cos 2\theta)\,d\theta \\&= -\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= -\frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta + C \\&= -\frac{1}{2}\arccos x + \frac{1}{2}x\sqrt{1-x^2} + C \end{align}

The key here is that $\arcsin x$ and $-\arccos x$ differ by a constant

$$\arcsin x + C_1 = -\arccos x + \frac{\pi}{2} + C_1 = -\arccos x + C_2$$

What this means is when you take the derivative, the constant goes away, leaving the same function, so the 2 answers you get are equivalent.


# Partial fractions - integration

$$\int \frac{4}{(x)(x^2+4)}$$

By comparing coefficients,

$4A = 4$, $A = 1$

$1 + B = 0$, $B= -1$

$xC= 0$, $C= 0$

where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$.

So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$.

And my final answer is

$\ln|x| - x \ln |x^2 + 4| + C$.

However my answer is wrong , the answer is - $\ln|x| - \frac{1}{2} \ln |x^2 + 4| + C$.

Where have I gone wrong?

• (1) What are $A$, $B$, and $C$? I understand that you must've set up the partial fraction decomposition. But if you don't show your work, how do we know which coefficient is which one? And we don't know whether you set it up correctly. So you really need to show all your work here. (2) How did you get $x\ln|x^2+4|$? This can't possibly be correct here. Remember that when integrating with respect to $x$, $x$ is not a constant. So $\color{red}{\int xf(x)dx\neq x\int f(x)dx}$. – zipirovich Aug 17 '17 at 2:33
• And so we see after the edit that the problem lies not with forming the partial fraction, but indeed with integrating $x/(x^2+4)$. – Graham Kemp Aug 17 '17 at 2:58
• Don't forget $dx$ – gen-ℤ ready to perish Aug 17 '17 at 3:03
• @GrahamKemp: I knew it! :-) – zipirovich Aug 17 '17 at 4:20

Yes, $4 = A(x^2+4)~+~(Bx+C)x \implies A=1, B=-1, C=0$

So

$$\require{enclose}\int \dfrac{4}{x(x^2+4)}~\mathrm d x ~{= \int \dfrac{1}{x}+\dfrac{\enclose{circle}{~-x~}}{(x^2+4)}~\mathrm d x \\ = \ln\lvert x\rvert -\int\dfrac{\tfrac 12\mathrm d (x^2+4)}{(x^2+4)} \\ = \ln\lvert x\rvert -\tfrac 12\ln\lvert x^2+4\rvert+D}$$

• +1 for \enclose{circle}{~-x~}. – Ander Biguri Aug 17 '17 at 11:20

Writing $$\frac{4}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4},$$ you saw that $A=1$, $B=-1$, and $C=0$. So you have $$\int \frac{4}{x(x^2+4)}dx = \int \frac{dx}{x} - \int \frac{x}{x^2+4} dx.$$ To integrate the second term on the right, do a $u$-substitution: let $u=x^2+4$. Then $du=2xdx$. So $$\int \frac{dx}{x} - \int \frac{x}{x^2+4} dx = \int \frac{dx}{x} - \frac{1}{2}\int \frac{du}{u}.$$ After integrating term-by-term, change the $u$ back into $x^2+4$.

Your partial fraction expansion was correct.$$\frac 4{x(x^2+4)}=\frac 1x-\frac x{x^2+4}$$However, the error lies in integrating the second term of the right-hand side. To simplify$$\int\frac x{x^2+4}\, dx$$Make a substituting $z=x^2+4$. The derivative is $dz=2x\, dx$ so $x\, dx=dz/2$. Therefore, the integral transforms into\begin{align*}\int\frac {x\, dx}{x^2+4} & =\frac 12\int\frac {dz}{z}\\ & =\tfrac 12\log z+C\\ & =\tfrac 12\log(x^2+4)+C\end{align*}Hence, we have$$\boxed{\int\frac {4\, dx}{x(x^2+4)}=\log x-\tfrac 12\log(x^2+4)+C}$$

I'm not necessarily sure how you got a factor of $x$ in the second term, but here is my go at it:

$$\frac{4}{x(x^2+4)}= \frac{A}{x} + \frac{Bx+C}{x^2+4}=\frac{A(x^2+4)+x(Bx+C)}{x(x^2+4)}=\frac{Ax^2+Bx^2+Cx+4A}{x(x^2+4)}$$

Then, by equating the numerator of the first and last expressions we have:

$$4=x^2(A+B) +Cx +4A$$

Equating coefficients:

$$A+B=0,C=0,4A=4\Rightarrow A=1,B=-1 \text{ and } C=0$$

Then we have:

$$\int{\frac{4}{x(x^2+4)}}\,dx=\int{\frac{1}{x}-\frac{x}{x^2+4}\,dx} = \int\frac{dx}{x}\, - \int{\frac{x}{x^2+4}\,dx}$$

In the second integral, we let $u=x^2+4$ then $du=2x dx\Rightarrow xdx=\frac{1}{2}du$. Hence

$$\int{\frac{dx}{x}} - \frac{1}{2}\int{\frac{du}{u}}=\ln \left|x\right|+\frac{1}{2}\ln\left|u\right|+C=\ln|x|-\frac{1}{2}\ln|x^2+4|+C$$ as required.


# Solving a three-part log equation using the log laws

Solve $$\log_5(x-1) + \log_5(x-2) - \log_5(x+6)= 0$$

I know that according to log laws, addition with the same base is equal to multiplication and subtraction is equal to division (and vice versa)

By doing this I get $$\log_5{(x-1)(x-2)}=0 (x+6)$$

By moving the $x+6$ to the other side it should become zero, ($x+6 \times 0$), however the answer in the textbook moves the $x+6$ to the other side of the equation and then solves. Why am I wrong? And when would you do my method then?

• To answer your original question: the book is adding $\log_5(x+6)$ to both sides. You wouldn't multiply the equation $x - 2 = 0$ by $2$ to solve for $x$, would you? Same thing here. – pjs36 Mar 10 '16 at 19:25

By the same rules you mentioned you should get $$\log_5 \left[(x-1)(x-2)\right]=\log_5[x+6]$$ which after exponentiation with $5$ becomes $$(x-1)(x-2)=x+6$$

If you want to do it by including all three terms in the logarithm, you get the same: $$\log_5 \left[\frac{(x-1)(x-2)}{x+6}\right]=0$$ but now you can't just multiply by $x+6$ on both sides and expect it to disappear in the denominator in the $\log$, since it is inside the $\log$. What you have to do is to exponentiate both sides to get $$\frac{(x-1)(x-2)}{x+6}=5^0=1$$ which can of course be rewritten as $$(x-1)(x-2)=x+6$$

So both methods lead to the same result.

• actually with the rules i mentioned, the x+6 would be being divided. It would be underneath the (x-1)(x-2) as subtraction is equal to division. The textbook moves it to the other side like you did, i dont understand why – Saad Siddiqui Mar 10 '16 at 18:11
• That (dividing under the log) would work as well. However, you can choose to just use the rule for two of the terms and then move the $\log_5(x+6)$ to the right hand side. – Bobson Dugnutt Mar 10 '16 at 18:13
• If I do the division, then when multiplying x+6 to the other side, it should equal to 0 and thus disappear. This would only leave the left side of my equation equal to zero, This is giving me the wrong answer – Saad Siddiqui Mar 10 '16 at 18:16
• @SaadSiddiqui See my edited answer for an explanation on how to solve it using your method. (Note that you exponentiate with $5$ since it is the base the $\log$ is in.) – Bobson Dugnutt Mar 10 '16 at 18:19
• Oh! Wow yeah that explains why, when using my method and the quadratic formula, I got the wrong answer. I've never been asked to solve after setting up the equation like that in school, only to set it up. Thank you, that helps a ton! I have a similar question which i'll ask in a bit – Saad Siddiqui Mar 10 '16 at 18:26

Laws of logarithm needed:

$$\log_a pq = \log_a p + \log_a q$$ $$\log_a \left(\frac{p}{q}\right) = \log_a p - \log_a q$$

$$\log_b a = 0 \ \implies a = b^0$$

Using these three laws, your equation can be immediately reduced to:

$$\log_5 \left(\frac{(x-1)(x-2)}{x+6}\right) = 0$$

Note: this is the step you got it wrong: $$\left(\frac{(x-1)(x-2)}{x+6}\right) = 5^0 = 1$$

$$x^2 -3x +2 =x+6$$

or we have $$\log_5((x-1)(x-2))=\log_5(x+6)$$ thus we get $$(x-1)(x-2)=x+6$$

• would the x+6 not be being divided? why is it moved to the other side? – Saad Siddiqui Mar 10 '16 at 18:12
• @SaadSiddiqui when you remove log from both sides, $0$ on the R.H.S becomes $1$ – Nikunj Mar 10 '16 at 20:39

Well, the mistake in your approach is the use of the $\log$ properties.

Let's rewrite the initial equation as $\log_5\left((x-1)(x-2)\cdot \frac{1}{x+6}\right)=0.$

Then we have (since $\log_5 1=0$ and since $\log$ is $1-1$) that $(x-1)(x-2)\cdot \frac{1}{x+6}=1.$

Then, just do some calculations to arrive at the solution. Do not forget to take the restrictions $x-1>0,$ $x-2>0$ and $x+6>0$ (which are equivalent to $x>2$) in order the $\log_5$ to be well defined.

• on a previous test, i was able to multiply two log functions with the same base, and then divide by another function of the same base. Something like this: (x-1)(x-2)/(x+6). why does that not work here? – Saad Siddiqui Mar 10 '16 at 18:14