The value of \[\log 3 + \dfrac{{{{\left( {\log 3} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {\log 3} \right)}^5}}}{{5!}} + ..... + \infty \].
Answer
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Hint: In the given questions , we have given with a series of \[\sinh \left( x \right)\] ( i.e. sine hyperbolic ) which is \[\sinh \left( x \right) = \dfrac{{{e^x} - {e^{ - x}}}}{2}\] , we will be using the expansion of the Taylor series of $\sinh x = x + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} + ..... + \infty $ , on equating the expansion with the expression of \[\sinh \left( x \right)\] we will get the required answer .
Complete step by step answer:
The Taylor series of a function is an infinite sum of all the terms that are expressed in terms of the function's derivatives at a single point . For most common functions, the function and the sum of its Taylor series are equal near this point.
Given: \[\log 3 + \dfrac{{{{\left( {\log 3} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {\log 3} \right)}^5}}}{{5!}} + ..... + \infty \].
Now, using the Taylor expansion for \[\sinh \left( x \right)\] we get,
$\sinh x = x + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} + ..... + \infty $
On comparing given expression with the Taylor expansion, we have \[x = \log 3\] ,
Now using the general exponential term for Taylor expansion, we have,
\[\sinh \left( x \right) = \dfrac{{{e^x} - {e^{ - x}}}}{2}\] .
Now, putting \[x = \log 3\] in the above expression, we get
\[ = \dfrac{{{e^{\log 3}} - {e^{ - \log 3}}}}{2}\]
Using the property of exponent i.e. \[{e^{\log x}} = x\] , we get
\[ = \dfrac{{3 - {e^{ - \log 3}}}}{2}\]
Now, using the property of \[ - \log x = \log \dfrac{1}{x}\] , we get
\[ = \dfrac{{3 - \dfrac{1}{3}}}{2}\]
On solving we get ,
\[ = \dfrac{{\dfrac{8}{3}}}{2}\]
On solving further we get ,
\[ = \dfrac{4}{3}\]
Note: The given expression can not be solved directly , as it is a series so you have to find out a common term for the given series . Moreover, some series are predefined as in the question we have series of \[\sinh \left( x \right)\] ( i.e. sine hyperbolic ) . Also , to solve questions related to \[\log \] you must have prior knowledge about the properties of \[\log \] and the same goes with the \[e\] ( exponent ) . Also, check whether the series is up to infinity \[\left( \infty \right)\] or not , as then the question will be related to AP or GP .
Complete step by step answer:
The Taylor series of a function is an infinite sum of all the terms that are expressed in terms of the function's derivatives at a single point . For most common functions, the function and the sum of its Taylor series are equal near this point.
Given: \[\log 3 + \dfrac{{{{\left( {\log 3} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {\log 3} \right)}^5}}}{{5!}} + ..... + \infty \].
Now, using the Taylor expansion for \[\sinh \left( x \right)\] we get,
$\sinh x = x + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} + ..... + \infty $
On comparing given expression with the Taylor expansion, we have \[x = \log 3\] ,
Now using the general exponential term for Taylor expansion, we have,
\[\sinh \left( x \right) = \dfrac{{{e^x} - {e^{ - x}}}}{2}\] .
Now, putting \[x = \log 3\] in the above expression, we get
\[ = \dfrac{{{e^{\log 3}} - {e^{ - \log 3}}}}{2}\]
Using the property of exponent i.e. \[{e^{\log x}} = x\] , we get
\[ = \dfrac{{3 - {e^{ - \log 3}}}}{2}\]
Now, using the property of \[ - \log x = \log \dfrac{1}{x}\] , we get
\[ = \dfrac{{3 - \dfrac{1}{3}}}{2}\]
On solving we get ,
\[ = \dfrac{{\dfrac{8}{3}}}{2}\]
On solving further we get ,
\[ = \dfrac{4}{3}\]
Note: The given expression can not be solved directly , as it is a series so you have to find out a common term for the given series . Moreover, some series are predefined as in the question we have series of \[\sinh \left( x \right)\] ( i.e. sine hyperbolic ) . Also , to solve questions related to \[\log \] you must have prior knowledge about the properties of \[\log \] and the same goes with the \[e\] ( exponent ) . Also, check whether the series is up to infinity \[\left( \infty \right)\] or not , as then the question will be related to AP or GP .
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