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i <3 u

<3 means heart. So it is read as I love you.

This relation can be brought down by simple algebra like,

say, i+5 < 3u+5 => i<3u

The famous style is this,

Solve for i,

9x- 7i < 3 (3x -7u)

= 9x - 7i < 9x - 21u

= -7i < -21u (cancel out the 9x)

simplified: i <3 u !

therefore: I love you

elekinetically. Jokes aside, I know a people who are excellent at Math and some can definitely be classified as genius. What they do is understand the concept rather than learn how to simply answer the question.

This way you’d be surprised that they will figure out extensions of that math question without even properly studying it. The reason is that their concept is so strong and when they link that with their already capable logic the result is a quick and thorough understanding of the topic.

The function f(x)=x3+ln(x+1) is defined over (−1,∞) . The limit at −1 is −∞ , the limit at ∞ is ∞ . The derivative is

f′(x)=3x2+1x+1>0 

so you know that the function is strictly increasing. Therefore the given equation has a single solution. Since f(2)>8 and f(1)<8 , the solution is inside the interval (1,2) .

You can determine an approximation with the desired accuracy with numerical methods.

I assume that, by ‘yn+1’, you mean the (n+1)th derivative of y with respect to x - this is often written (with the parentheses) as a superscript, e.g.

y(n+1) 

In other words, you want to prove that:

ddxn+1(xnln(x))=n!x 

I suggest that you edit the question to make your meaning clearer as, from the two answers submitted before mine, they didn’t understand you.

This doesn’t really count as a proof, but I think its a way to demonstrate why this is true.

From the product rule for differentiation,

dydx=xnddxln(x)+ln(x)ddxxn 

=xnx+nxn−1ln(x)=xn−1+nxn−1ln(x) 

Having differentiated once, we still have to differentiate a further n times.

dn+1ydxn+1=dndxn(xn−1+nxn−1ln(x)) 

From the sum rule for differentiation, we can split this into two parts:

dndxnxn−1+ndndxnxn−1ln(x) 

Let’s look at the first part. When we differentiate an expression than contains a term that is a power of x, we reduce the power by 1. So, if we differentiate xb b times, we end up with a constant [ xb−b=x0 ], and if we differentiate again, we get a zero. In this case, we’re wanting to differentiate xn−1 n times, this means that the term eventually becomes zero. So, our problem simplifies to:

ndndxnxn−1ln(x)=ndn−1dxn−1(ddxxn−1ln(x)) 

Applying the product rule again:

=ndn−1dxn−1(xn−2+(n−1)xn−2ln(x)) 

Applying the sum rule again, the first term again reduces to zero with continued differentiation, so we are left with:

ndn−1dxn−1(n−1)xn−2ln(x) 

As (n−1) is a constant, we can move it outside the differentiation; I’ll also introduce the notation n[2]=n!(n−2)! 

So, we have:

n[2]dn−1dxn−1xn−2ln(x) 

Applying the product rule again:

n[2]dn−2dxn−2(xn−3+(n−2)xn−3ln(x)) 

Applying the sum rule again, the first term again reduces to zero with continued differentiation, so we are left with:

n[3]dn−2dxn−2xn−3ln(x) 

Continuing the pattern we get:

n[4]dn−3dxn−3xn−4ln(x) 

n[5]dn−4dxn−4xn−5ln(x) 

…

n[n−1]d2dx2x1ln(x) 

n[n]ddxln(x)=n[n]x 

As the coefficient is just n! , our answer is:

n!x