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Articles by Hasan Saad
Limit of an Integral, Then taking Sum
Yes, you are right about the first step. First of all, let us find the recurrent relation for the $I_n$. Integrating by parts we get: $$I_n=\int^1_0x^ne^x\,dx=e-\int^1_0nx^{n-1}e^x\,dx=e-nI_{n-1}$$ This is equivalent to $I_n+nI_{n-1}=e$. The last equality implies $I_{n+1}+(n+1)I_n=I_n+nI_{n-1}$ or $I_{n+1}=n(I_{n-1}-I_n)$. Now lets analyze our sum. We obtain: $$\sum_{k=1}^n\frac{I_{k+1}}{k}=\sum_{k=2}^n(I_{k-1}-I_k)+I_2=I_1-I_n+I_2$$ Now we must find to which number does $I_n$ tend.
Why L'Hospital's rule doesn't work in this case?
I have a very simple (and maybe stupid) question. Suppose I want to evaluate this limit: It is easy to evaluate this limit using the Squeeze theorem (the answer is $1$). But here both the numerator and the denominator are going to infinity as $x\to \infty$ so I tried using L'Hospital's rule: $$\lim_{x\to \infty} \frac{x}{x-\sin x}=\lim_{x\to \infty} \frac{1}{1-\cos x}$$ However there's no finite $L$ such that $$\lim_{x\to \infty} \frac{1}{1-\cos x}=L$$ which is a contradiction.
Why doesn't L'Hospital's rule work in this case?
I have a very simple (and maybe stupid) question. Suppose I want to evaluate this limit: It is easy to evaluate this limit using the Squeeze theorem (the answer is $1$). But here both the numerator and the denominator are going to infinity as $x\to \infty$ so I tried using L'Hospital's rule: $$\lim_{x\to \infty} \frac{x}{x-\sin x}=\lim_{x\to \infty} \frac{1}{1-\cos x}$$ However there's no finite $L$ such that $$\lim_{x\to \infty} \frac{1}{1-\cos x}=L$$ which is a contradiction.
Is there any way to simplify the following expression into a non-radical form?
The function $$f(x_1,x_2,y_1,y_2,z_1,z_2)=\sqrt{x_{1}^2-2x_{1}x_{2}+x_{2}^2+y_{1}^2-2y_{1}y_{2}+y_{2}^2+z_{1}^2-2z_{1}z_{2}+z_{2}^2}$$ is not a polynomial or even a rational function (over the real numbers). Suppose it was a rational function. Then $$g(x)=f(x,0,0,0,0,0)=\sqrt{x^2}=|x|$$ would also have to be a rational function - of a single variable. So let's write it as $g(x)=\frac{P(x)}{Q(x)}$. Observe that $Q$ has no (real) zeros, so it is either always positive or always negative.
Equivalence Relation between Derivative Being Odd and Function Being Even
In the exercise, I am required to prove that $f'$ is odd $\iff$ $f$ is even Moving from right to left was pretty trivial, however, I couldn't move from left to right. Note that we can only use very basic things like the definition of the derivative and rules of differentiation (No mean value theorem or any other theorems.) I cannot even see the left implying the right, but I hope I will see it after you guys explain how. Thanks in advance.
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