13th Harvard MIT Calculus problem 8

Let f(n)=\sum\limits_{k=2}^{\infty }{\frac{1}{{{k}^{n}}k!}}. Calculate \sum\limits_{k=2}^{\infty }{f(n)} 😀


—> \sum\limits_{k=2}^{\infty }{f(n)}=\sum\limits_{k=2}^{\infty }{\sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}k!}}}

=\sum\limits_{k=2}^{\infty }{\frac{1}{k!}\sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}}}}

we know that \sum\limits_{n=0}^{\infty }{\frac{1}{{{k}^{n}}}}=\frac{1}{{{k}^{0}}}+\frac{1}{{{k}^{1}}}+...........=\frac{1}{1-\frac{1}{k}}=\frac{k}{k-1}

So \sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}}}=\frac{k}{k-1}-1-\frac{1}{k}=\frac{1}{k(k-1)}

—> \sum\limits_{n=2}^{\infty }{f(n)}=\sum\limits_{k=2}^{\infty }{\frac{1}{k!}.\frac{1}{k(k-1)}}

=\sum\limits_{k=2}^{\infty }{\frac{1}{(k-1)!}}.\frac{1}{{{k}^{2}}(k-1)}

–>  =\sum\limits_{k=2}^{\infty }{\frac{1}{(k-1)!}\left( \frac{1}{k-1}-\frac{1}{{{k}^{2}}}-\frac{1}{k} \right)}

=\sum\limits_{k=2}^{\infty }{\left( \frac{1}{(k-1)!(k-1)}-\frac{1}{k.k!}-\frac{1}{k!} \right)}

=\sum\limits_{k=2}^{\infty }{\left( \frac{1}{(k-1)!(k-1)}-\frac{1}{k.k!} \right)}-\sum\limits_{k=2}^{\infty }{\frac{1}{k!}}

*** By Taylor series \sum\limits_{k=0}^{\infty }{\frac{1}{k!}}=e

Therefore \sum\limits_{n=2}^{\infty }{f(n)=\frac{1}{1\cdot 1!}-\left( e-\frac{1}{0!}-\frac{1}{1!} \right)=3-e}


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Mix Limit Derivative and Integral

Anyone can do it?? then compare with my answer =-\frac{1}{4} 😀 I don’t want to post it 😀 I’m not sure my solution is right or wrong 😀 thank you 😀 My answer that I got is completely wrong 😀

We given {{f}^{'}}(x)={{e}^{-{{x}^{2}}}} and \underset{x\to +\infty }{\mathop{\lim }}\,f(x)=0

find the solution of \int\limits_{0}^{+\infty }{{{x}^{2}}f(x)dx} 😀
Here is the link to see the answer:
Answer here

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Nice L’Hopital’s Rule Problem

Find the \underset{x\to 1}{\mathop{\lim }}\,{{\left( \frac{x}{\cos (x-1)} \right)}^{\frac{x}{1-{{x}^{3}}}}}? 😀


\underset{x\to 1}{\mathop{\lim }}\,{{\left( \frac{x}{\cos (x-1)} \right)}^{\frac{x}{1-{{x}^{3}}}}}=\underset{x\to 1}{\mathop{\lim }}\,{{e}^{\frac{x}{1-{{x}^{3}}}\ln \frac{x}{\cos (x-1)}}}=\underset{x\to 1}{\mathop{\lim }}\,{{e}^{x\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1-{{x}^{3}}}\ln \frac{x}{\cos (x-1)}}}

Use The L’Hopital’s Rule for \underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1-{{x}^{3}}}\ln \frac{x}{\cos (x-1)}=\underset{x\to 1}{\mathop{\lim }}\,\frac{{{\left( \frac{x}{\cos (x-1)} \right)}^{'}}}{{{(1-{{x}^{3}})}^{'}}}=\underset{x\to 1}{\mathop{\lim }}\,\left( \frac{\cos (x-1)+x\sin (x-1)}{-3{{x}^{3}}\cos (x-1)} \right)

Plugging x=1 then we get \underset{x\to 1}{\mathop{\lim }}\,\left( \frac{\cos (x-1)+x\sin (x-1)}{-3{{x}^{3}}\cos (x-1)} \right)=-\frac{1}{3}

Therefore \underset{x\to 1}{\mathop{\lim }}\,{{\left( \frac{x}{\cos (x-1)} \right)}^{\frac{x}{1-{{x}^{3}}}}}={{e}^{-\frac{1}{3}}}

Thank you 😀

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Limit Problem 1

Compute \underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x} ? 😀


Since {{\sin }^{2}}x=1-{{\cos }^{2}}x

We multiply the numerator and denominator by 1+\cos x and use the fact that \underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x}=1 Therefore:

\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}(1+\cos x)}{(1-\cos x)(1+\cos x)}=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{x}{\sin x} \right)}^{2}}\underset{x\to 0}{\mathop{\lim }}\,(1+\cos x)=2

**Another solution, using L’Hopital’s rule, is possible: \underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x}{\sin x}=2

Thank you 😀

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The Amazing Number 1,089

1. Choose any three-digit number (where the units and hundreds digits are not the same).

We will do it with you here by arbitrarily selecting 825.

2. Reverse the digits of this number you have selected.

We will continue here by reversing the digits of 825 to get 528.

3. Subtract the two numbers (naturally, the larger minus the smaller).

Our calculated difference is 825 − 528 = 297.

4. Once again, reverse the digits of this difference.

Reversing the digits of 297 we get the number 792.

5. Now, add your last two numbers.

We then add the last two numbers to get 297 + 792 = 1089.

Their result should be the same∗ as ours even though their starting numberswere different from ours.

They will probably be astonished that regardless of which numbers theyselected at the beginning, they got the same result as we did, 1,089.How does this happen? Is this a “freak property” of this number? Did wedo something devious in our calculations?

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12th Harvard MIT Calculus Problem 8

Compute 😀 A=\int\limits_{1}^{\sqrt{3}}{{{x}^{2{{x}^{2}}+1}}+\ln ({{x}^{2{{x}^{2{{x}^{2}}+1}}}})dx} 😀


Using the fact that x={{e}^{\ln x}}, we evaluate the integral as follows:

\int{{{x}^{2{{x}^{2}}+1}}+\ln ({{x}^{2{{x}^{2{{x}^{2}}+1}}}})dx=\int{{{x}^{2{{x}^{2}}+1}}+{{x}^{2{{x}^{2}}+1}}(\ln {{x}^{2}})dx}}

A=\int{{{e}^{\ln (x)(2{{x}^{2}}+1)}}(1+\ln {{x}^{2}})dx}

A=\int{x{{e}^{{{x}^{2}}\ln {{x}^{2}}}}(1+\ln {{x}^{2}})dx}

Noticing that the derivative of {{x}^{2}}\ln {{x}^{2}} is 2x\ln (1+{{x}^{2}}), it follows that the integral evaluates to \frac{1}{2}{{e}^{{{x}^{2}}\ln {{x}^{2}}}}=\frac{1}{2}{{x}^{2{{x}^{2}}}}

Evaluate this from 1 to \sqrt{3} we obtain the answer \int\limits_{1}^{\sqrt{3}}{{{x}^{2{{x}^{2}}+1}}+\ln ({{x}^{2{{x}^{2{{x}^{2}}+1}}}})dx}=\int\limits_{0}^{3\ln 3}{{{e}^{u}}du=13} 😀

Remember to change from x to u 😀

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12th Harvard MIT Calculus Problem 2

The differentiable function F:\mathbb{R}\to \mathbb{R} satisfied F(0)=-1 and \frac{d}{dx}F(x)=\sin (\sin (\sin (\sin (x))))\cos (\sin (\sin (x)))\cos (\sin (x))\cos (x)

Find F(x) as a function of x.


Substitution u=\sin (\sin (\sin (x)))

then du={{(\sin (\sin (x)))}^{'}}\cos (\sin (\sin (x)))dx=\cos (\sin (\sin (x)))\cos (\sin x)\cos xdx

So we get F(x)=\int{\sin udu=-\cos u+c}

since we know that F(0)=-1 then we get c=0

Therefore F(x)=\cos (\sin (\sin (\sin (x))))

😀 Thank you Pablo for comment about F(x)=\cos x 😀

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